THE CLASSICAL MOMENT PROBLEM AS A

SELF-ADJOINT FINITE DIFFERENCE OPERATOR

Barry Simon^{∗}

Division of Physics, Mathematics, and Astronomy California Institute of Technology Pasadena, CA 91125

November 14, 1997

Abstract. This is a comprehensive exposition of the classical moment problem using methods from the theory of finite difference operators. Among the advantages of this approach is that the Nevanlinna functions appear as elements of a transfer matrix and convergence of Padé approximants appears as the strong resolvent convergence of finite matrix approximations to a Jacobi matrix. As a bonus of this, we obtain new results on the convergence of certain Padé approximants for series of Hamburger.

§1. Introduction

The classical moment problem was central to the development of analysis in the period from 1894 (when Stieltjes wrote his famous memoir [38]) until the 1950’s (when Krein completed his series on the subject [15, 16, 17]). The notion of measure (Stieltjes integrals), Padé approximants, orthogonal polynomials, extensions of positive linear functionals (Riesz-Markov theorem), boundary values of analytic functions, and the Herglotz-Nevanlinna-Riesz representation theorem all have their roots firmly in the study of the moment problem.

This expository note attempts to present the main results of the theory with two main ideas in mind. It is known from early on (see below) that a moment problem is associated to a certain semi-infinite Jacobi matrix, A. The first idea is that the basic theorems can be viewed as results on the self-adjoint extensions of A. The second idea is that techniques from the theory of second-order difference and differential equations should be useful in the theory.

Typeset by AMS-TEX

∗ This material is based upon work supported by the National Science Foundation under Grant No. DMS-9401491. The Government has certain rights in this material.

To appear in Advances in Mathematics

Of course, neither of these ideas is new. Both appear, for example, in Stone’s treatment [39], in Dunford-Schwartz [5], and in Akhiezer’s brilliant book on the subject [1]. But, normally, (and in [1], in particular), these ideas are presented as applications of the basic theory rather than as the central tenets. By placing them front and center, many parts of the theory become more transparent — in particular, our realization of the Nevanlinna matrix as a transfer matrix makes its properties (as an entire function of order 1 and minimal type) evident (see Theorem 4.8 below).

Two basic moment problems will concern us here:

Hamburger Moment Problem. Given a sequence γ_{0}, γ_{1}, γ_{2}, … of reals, when is there a measure, dρ, on (−∞, ∞) so that

γ_{n} = ∫_{−∞}^{∞} x^{n} dρ(x)  (1.1)

and if such a ρ exists, is it unique? We let M_{H}(γ) denote the set of solutions of (1.1).

Stieltjes Moment Problem. Given a sequence γ_{0}, γ_{1}, γ_{2}, … of reals, when is there a measure, dρ on [0, ∞) so that

γ_{n} = ∫_{0}^{∞} x^{n} dρ(x)  (1.2)

and if such a ρ exists, is it unique? We let M_{S}(γ) denote the set of solutions of (1.2).

We will not attempt a comprehensive historical attribution of classical results; see Akhiezer [1] and Shohat-Tamarkin [35] for that.

We will always normalize the γ’s so that γ_{0} = 1. By replacing γ_{n} by γ_{n}/γ_{0}, we can reduce general γ’s to this normalized case. We will also demand that dρ have infinite support, that is, that ρ not be a pure point measure supported on a finite set. This eliminates certain degenerate cases.

One immediately defines two sesquilinear forms, H_{N} and S_{N}, on ℂ^{N} for each N by

H_{N}(α, β) = ∑_{\begin{matrix}n=0,1,...,N−1 \\ m=0,1,...,N−1 \end{matrix}} ᾱ_{n}β_{m}γ_{n+m}  (1.3)

S_{N}(α, β) = ∑_{\begin{matrix}n=0,1,...,N−1 \\ m=0,1,...,N−1 \end{matrix}} ᾱ_{n}β_{m}γ_{n+m+1}  (1.4)

and corresponding matrices H_{N} and S_{N} so that H_{N}(α, β) = ⟨α, H_{N}β⟩ and S_{N}(α, β) = ⟨α, S_{N}β⟩ in the usual Euclidean inner product. Our inner products are linear in the second factor and anti-linear in the first.

A standard piece of linear algebra says:

Lemma 1.1. An N × N Hermitean matrix A is strictly positive definite if and only if each submatrix A^{[1,J]} = (a_{ij})_{1≤i,j≤J} has det(A^{[1,J]}) > 0 for J = 1, 2, …, N.

Proof. If A is strictly positive definite, so is each A^{[1,J]}, so their eigenvalues are all strictly positive and so their determinants are all strictly positive.

For the converse, suppose that A^{[1,N−1]} is positive definite. By the min-max principle, the (N − 1) eigenvalues of A^{[1,N−1]} interlace the N eigenvalues of A^{[1,N]} ≡ A. If all the eigenvalues of A^{[1,N−1]} are positive, so are the N − 1 largest eigenvalues of A. If the N^{th} eigenvalue were non-positive, det(A) ≤ 0. Thus, A^{[1,N−1]} strictly positive definite and det(A^{[N,N]}) > 0 imply A^{[1,N]} is strictly positive definite. An obvious induction completes the proof. □

This immediately implies

Proposition 1.2. {H_{N}}_{N=1}^{∞} are strictly positive definite forms if and only if det(H_{N}) > 0 for N = 1, 2, . . .. Similarly, {S_{N}}_{N=1}^{∞} are strictly positive definite forms if and only if det(S_{N}) > 0 for N = 1, 2, . . ..

Suppose that the γ_{n} obey (1.1). Then by an elementary calculation,

∫ | ∑_{n=0}^{N−1} α_{n}x^{n}|^{2} dρ(x) = H_{N}(α, α)  (1.5)

∫ x| ∑_{n=0}^{N−1} α_{n}x^{n}|^{2} dρ(x) = S_{N}(α, α).  (1.6)

Taking into account that if ∫ |P (x)|^{2} dρ(x) = 0, ρ must be supported on the zeros of P, we have:

Proposition 1.3. A necessary condition that (1.1) holds for some measure dρ on (−∞, ∞) with infinite support is that each sesquilinear form H_{N} is strictly positive definite. A necessary condition that there be a dρ supported on [0, ∞) is that each H_{N} and each S_{N} be strictly positive definite.

Suppose now that each H_{N} is strictly positive definite. Let ℂ[X] be the family of complex polynomials. Given P (X) = ∑_{n=0}^{N−1} α_{n}X^{n}, Q(X) = ∑_{n=0}^{N−1} β_{n}X^{n} (we suppose the upper limits in the sums are equal by using some zero α’s or β’s if need be), define

⟨P, Q⟩ = H_{N}(α, β).  (1.7)

This defines a positive definite inner product on ℂ[X], and, in the usual way, we can complete ℂ[X] to a Hilbert space H^{(γ)} in which ℂ[X] is dense. ℂ[X] can be thought of as abstract polynomials or as infinite sequences (α_{0}, α_{1}, …, α_{N−1}, 0, …) which are eventually 0 via α ∼ ∑_{j=0}^{N−1} α_{j}X^{j}.

We will start using some basic facts about symmetric and self-adjoint operators on a Hilbert space — specifically, the spectral theorem and the von Neumann theory of self-adjoint extensions. For an exposition of these ideas, see Chapters VII, VIII, and X of Reed-Simon, volumes I and II [33, 34] and see our brief sketch at the start of Section 2.

Define a densely defined operator A on H^{(γ)} with domain D(A) = ℂ[X] by

A[P (X)] = [XP (X)].  (1.8)

In the sequence way of looking at things, A is the right shift, that is, A(α_{0}, α_{1}, …, α_{N}, 0, …) = (0, α_{0}, α_{1}, …, α_{N}, 0, …). This means that

⟨P, A[Q]⟩ = S_{N}(α, β)  (1.9)

and, in particular,

⟨1, A^{n}1⟩ = γ_{n}.  (1.10)

Since H_{N} and S_{N} are real-symmetric matrices, A is a symmetric operator, that is, ⟨A[P], Q⟩ = ⟨P, A[Q]⟩. Moreover, if we define a complex conjugation C on ℂ[X] by C(∑_{n=0}^{N−1} α_{n}X^{n}) = ∑_{n=0}^{N−1} ᾱ_{n}X^{n}, then CA = AC. It follows by a theorem of von Neumann (see Corollary 2.4 in Section 2) that A has self-adjoint extensions.

If each S_{N} is positive definite, then ⟨P, A[P]⟩ ≥ 0 for all P, and it follows that A has a non-negative self-adjoint extension A_{F}, the Friedrichs extension (discussed further in Section 3). We thus see:

Proposition 1.4. If all H_{N} are positive definite, then A has self-adjoint extensions. If all S_{N} are positive definite, then A has non-negative self-adjoint extensions.

Let Ã be a self-adjoint extension of A. By the spectral theorem, there is a spectral measure dμ̃ for Ã with vector [1] ∈ H^{(γ)}, that is, so that for any bounded function of Ã,

⟨1, f(Ã)1⟩ = ∫ f(x) dμ̃(x).  (1.11)

Since 1 ∈ D(A^{N}) ⊂ D(Ã^{N}), (1.11) extends to polynomially bounded functions and we have by (1.10) that,

γ_{N} = ∫ x^{N} dμ̃(x).

We see therefore that a self-adjoint extension of A yields a solution of the Hamburger moment problem. Moreover, a non-negative self-adjoint extension has supp(dμ̃) ⊂ [0, ∞) and so yields a solution of the Stieltjes moment problem. Combining this with Propositions 1.2, 1.3, and 1.4, we have the first major result in the theory of moments.

Theorem 1 (Existence). A necessary and sufficient condition for there to exist a measure dρ with infinite support obeying (1.1) is that det(H_{N}) > 0 for N = 1, 2, . . .. A necessary and sufficient condition that also dρ be supported on [0, ∞) is that both det(H_{N}) > 0 and det(S_{N}) > 0 for N = 1, 2, . . ..

Historically, existence was a major theme because the now standard tools on existence of measures were invented in the context of moment problems. We have settled it quickly, and the bulk of this paper is devoted to uniqueness, especially the study of cases of nonuniqueness. Non-uniqueness only occurs in somewhat pathological situations, but the theory is so elegant and beautiful that it has captivated analysts for a century.

Henceforth, we will call a set of moments {γ_{n}}_{n=0}^{∞} with det(H_{N}) > 0 for all N a set of Hamburger moments. If both det(H_{N}) > 0 and det(S_{N}) > 0 for all N, we call them a set of Stieltjes moments.

We will call a solution of the moment problem which comes from a self-adjoint extension of A a von Neumann solution. The name is in honor of the use below of the von Neumann theory of self-adjoint extensions of densely-defined symmetric operators. This name, like our use of Friedrichs solution and Krein solution later, is not standard, but it is natural from the point of view of self-adjoint operators. As far as I know, neither von Neumann nor Friedrichs worked on the moment problem per se. While Krein did, his work on the moment problem was not in the context of the Krein extension we use to construct what we will call the Krein solution. What we call von Neumann solutions, Akhiezer calls N-extremal and Shohat-Tamarkin call extremal. This last name is unfortunate since we will see there exist many solutions which are extreme points in the sense of convex set theory, but which are not von Neumann solutions (and so, not extremal in the Shohat-Tamarkin sense).

Given the connection with self-adjoint extensions, the following result is reasonable (and true!):

Theorem 2 (Uniqueness). A necessary and sufficient condition that the measure dρ in (1.1) be unique is that the operator A of (1.8) is essentially self-adjoint (i.e., has a unique self-adjoint extension). A necessary and sufficient condition that there be a unique measure dρ in (1.1) supported in [0, ∞) is that A have a unique non-negative self-adjoint extension.

This result is surprisingly subtle. First of all, it is not obvious (but true, as we will see) that distinct self-adjoint extensions have distinct spectral measures dμ̃, so there is something to be proven before multiple self-adjoint extensions imply multiple solutions of the moment problem. The other direction is even less clear cut, for not only is it not obvious, it is false that every solution of the moment problem is a von Neumann solution (Reed-Simon [34] has an incorrect proof of uniqueness that implicitly assumes every solution comes from a self-adjoint extension). As we will see, once there are multiple solutions, there are many, many more solutions than those that come from self-adjoint extensions in the von Neumann sense of looking for extensions in \overline{D(A)}. But, as we will see in Section 6, there is a sense in which solutions are associated to self-adjoint operators in a larger space.

We also note we will see cases where the Stieltjes problem has a unique solution but the associated Hamburger problem does not.

The Hamburger part of Theorem 2 will be proven in Section 2 (Theorems 2.10 and 2.12); the Stieltjes part will be proven in Sections 2 and 3 (Theorems 2.12 and 3.2). If there is a unique solution to (1.1), the moment problem is called determinate; if there are multiple solutions, it is called indeterminate. It is ironic that the English language literature uses these awkward terms, rather than determined and undetermined. Stieltjes was Dutch, but his fundamental paper was in French, and the names have stuck. Much of the interesting theory involves analyzing the indeterminate case, so we may as well give some examples that illuminate non-uniqueness.

Example 1.1. Let f be a non-zero C^{∞} function on ℝ supported on [0, 1]. Let g(x) = f̂(x), with f̂ the Fourier transform of f. Then

∫_{−∞}^{∞} x^{n}g(x) dx = \sqrt{2π} (−i)^{n} \frac{d^{n}f}{dx^{n}} (0) = 0.

Let dρ_{1}(x) = (Re g)_{+}(x) dx, the positive part of Re g, and let dρ_{2} = (Re g)_{−}(x) dx. By the above,

∫_{−∞}^{∞} x^{n}dρ_{1}(x) = ∫_{−∞}^{∞} x^{n} dρ_{2}(x)

for all n. Since dρ_{1} and dρ_{2} have disjoint essential supports, they are unequal and we have non-unique solutions of the moment problem. (We will see eventually that neither is a von Neumann solution.) The moments from ρ_{1}, ρ_{2} may not be normalized. But we clearly have non-uniqueness after normalization.

This non-uniqueness is associated to non-analyticity in a Fourier transform and suggests that if one can guarantee analyticity, one has uniqueness. Indeed,

Proposition 1.5. Suppose that {γ_{n}}_{n=0}^{∞} is a set of Hamburger moments and that for some C, R > 0,

|γ_{n}| ≤ CR^{n}n!  (1.12a)

Then the Hamburger moment problem is determinate.

If {γ_{n}}_{n=0}^{∞} is a set of Stieltjes moments and

|γ_{n}| ≤ CR^{n}(2n)!  (1.12b)

then the Stieltjes moment problem is determinate.

Proof. Let dρ obey (1.1). Then x^{2n} ∈ L^{1}(ℝ, dρ), and by the monotone convergence theorem,

∫_{−∞}^{∞} cosh(\frac{x}{2R}) dρ(x) = lim_{N→∞} ∫_{−∞}^{∞} ∑_{n=0}^{N}(\frac{x}{2R})^{2n} \frac{1}{(2n)!} dρ(x)

≤ C lim_{N→∞} ∑_{n=0}^{N}(\frac{1}{2})^{2n} = \frac{4}{3}C < ∞.

Thus, e^{αx} ∈ L^{1}(ℝ, dρ(x)) for |α| < \frac{1}{2R}. It follows that F_{ρ}(z) ≡ ∫ e^{izx}dρ(x) has an analytic continuation to \{z | |Im z| < \frac{1}{2R}\}. If μ is a second solution of (1.1), F_{μ}(z) is also analytic there. But −i^{n} \frac{d^{n}F_{ρ}}{dz} (0) = γ_{n}, so the Taylor series of F_{ρ} and F_{μ} at z = 0 agree, so F_{ρ} = F_{μ} in the entire strip by analyticity.

This implies that μ = ρ by a variety of means. For example, in the topology of bounded uniformly local convergence, (i.e., f_{n} → f means sup ∥f_{n}∥_{∞} < 0 and f_{n}(x) → f(x) uniformly for x in any [−κ, κ]), linear combinations of the {e^{iyx} | y ∈ ℝ} are dense in all bounded continuous functions. Or alternatively, G_{μ}(z) = ∫ \frac{dμ(x)}{x−z} = i ∫_{−∞}^{0} e^{−iyz}F_{μ}(y) dy for Im z > 0, and μ can be recovered as a boundary value of G_{μ}.

The (2n)! Stieltjes result follows from the n! result and Proposition 1.6 below. □

We will generalize Proposition 1.5 later (see Corollary 4.5).

Proposition 1.6. Let {γ_{n}}_{n=0}^{∞} be a set of Stieltjes moments. Let

Γ_{2m} = γ_{m},  m = 0, 1, …
Γ_{2m+1} = 0,  m = 0, 1, ….

If {Γ_{j}}_{j=0}^{∞} is a determinate Hamburger problem, then {γ_{n}}_{n=0}^{∞} is a determinate Stieljes problem.

Proof. Let dρ solve the Stieltjes problem. Let

dμ(x) = \frac{1}{2}[χ_{[0,∞)}(x) dρ(x^{2}) + χ_{(−∞,0]}(x) dρ(x^{2})].

Then the moments of μ are Γ. Thus uniqueness for the Γ problem on (−∞, ∞) implies uniqueness for the γ problem on [0, ∞). □

Remark. We will see later (Theorem 2.13) that the converse of the last assertion in Proposition 1.6 is true. This is a more subtle result.

Example 1.2. This example is due to Stieltjes [38]. It is interesting for us here because it is totally explicit and because it provides an example of non-uniqueness for the Stieltjes moment problem. Note first that

∫_{0}^{∞} u^{k}u^{− ln u} sin(2π ln u) du = 0.

This follows by the change of variables v = −\frac{k+1}{2} + ln u, the periodicity of sin( ·), and the fact that sin is an odd function. Thus for any θ ∈ [−1, 1],

\frac{1}{\sqrt{π}} ∫_{0}^{∞} u^{k}u^{− ln u}[1 + θ sin(2π ln u)] du = e^{\frac{1}{4} (k+1)^{2}}

(by the same change of variables) so γ_{k} = e^{\frac{1}{4} (k+1)^{2}} is an indeterminate set of Stieltjes moments. Notice for θ ∈ (−1, 1), if dρ_{θ}(u) is the measure with these moments, then sin(2π ln(u))/1+θ sin(2π ln(u)) is in L^{2}(dρ_{θ}) and orthogonal to all polynomials, so dρ_{θ}(u) is a measure with all moments finite but where the polynomials are not L^{2} dense. As we shall see, this is typical of solutions of indeterminate moment problems which are not von Neumann solutions.

Looking at the rapid growth of the moments in Example 1.2, you might hope that just as a condition of not too great growth (like Proposition 1.5) implies determinacy, there might be a condition of rapid growth that implies indeterminacy. But that is false! There are moments of essentially arbitrary rates of growth which lead to determinate problems (see the remark after Corollary 4.21 and Theorem 6.2).

Example 1.3. There is a criterion of Krein [18] for indeterminacy that again shows the indeterminacy of Example 1.2, also of the example of the moments of exp(−|x|^{α}) dx (on (−∞, ∞)) with α < 1 and also an example of Hamburger [8], the moments of

χ_{[0,∞)}(x) exp(−\frac{π\sqrt{x}}{ln^{2} x + π^{2}} ) dx.

Proposition 1.7 (Krein [18]). Suppose that dρ(x) = F (x) dx where 0 ≤ F (x) ≤ 1 and either

(i) supp(F) = (−∞, ∞) and

∫_{−∞}^{∞} −\frac{ln(F (x))}{1 + x^{2}} dx < ∞  (1.13a)

or (ii) supp(F) = [0, ∞) and

∫_{0}^{∞} −\frac{ln(F (x))}{(1 + x)} \frac{dx}{\sqrt{x}} < ∞.  (1.13b)

Suppose also that for all n:

∫_{−∞}^{∞} |x|^{n}F (x) dx < ∞.  (1.13c)

Then the moment problem (Hamburger in case (i), Stieltjes in case (ii)) with moments

γ_{n} = \frac{∫ x^{n}F (x) dx}{∫ F (x) dx}

is indeterminate.

Remarks. 1. Hamburger’s example is close to borderline for (1.13b) to hold.

2. Since ∫_{−∞}^{∞} x^{2n} exp(−|x|^{α}) dx = 2α^{−1}Γ(\frac{2n+1}{α}) ∼ (\frac{2n}{α})! and (1.13a) holds for F (x) = exp(−|x|^{α}) if α < 1, we see that there are examples of Hamburger indeterminate moment problems with growth just slightly faster than the n! growth, which would imply by Proposition 1.5 that the problem is determinate. Similarly, since ∫_{0}^{∞} x^{n} exp(−|x|^{α}) dx = α^{−1}Γ(\frac{n+1}{α}) ∼ (\frac{n}{α})! and (1.13b) holds for F (x) = exp(−|x|^{α}) if α < \frac{1}{2}, we see there are examples of Stieltjes indeterminate problems with growth just slightly faster than the (2n)! of Proposition 1.5.

3. Since F (x) = \frac{1}{2}e^{−|x|} has moments γ_{2n} = (2n)! covered by Proposition 1.5, it is a determinate problem. The integral in (1.13a) is only barely divergent in this case. Similarly, F (x) = χ_{[0,∞)}(x)e^{−\sqrt{x}} is a Stieltjes determinate moment problem by Proposition 1.5 and the integral in (1.13b) is barely divergent. This says that Krein’s conditions are close to optimal.

4. Krein actually proved a stronger result by essentially the same method. F need not be bounded (by a limiting argument from the bounded case) and the measure defining γ_{n} can have an arbitrary singular part. Moreover, Krein proves (1.13a) is necessary and sufficient for {e^{iαx}}_{0≤α<∞} to not be dense in L^{2}(ℝ, F (x) dx).

5. Krein’s construction, as we shall see, involves finding a bounded analytic function G in ℂ_{+} ≡ {z ∈ ℂ | Im z > 0} so that |G(x + i0)| ≤ F (x). From this point of view, the fact that F (x) cannot decay faster than e^{−|x|} is connected to the Phragmén-Lindelöf principle.

6. The analog of Krein’s result for the circle instead of the half plane is due to Szego.

Proof. Suppose we can find G with Re G ≢ 0 with |G(x)| ≤ F (x) so that

∫ x^{n}G(x) dx = 0.

Then both \frac{F (x) dx}{∫ F (x) dx} and \frac{[F (x)+Re G(x)] dx}{∫ F (x) dx} solve the γ_{n} moment problem, showing indeterminacy.

In case (i), define for z ∈ ℂ_{+} (see (1.19) and the definition of Herglotz function below):

Q(z) = \frac{1}{π} ∫ (\frac{1}{x − z} − \frac{x}{1 + x^{2}} )[− ln(F (x))] dx,

which is a convergent integral by (1.13a). Then, Im Q(z) ≥ 0 and by the theory of boundary values at such analytic functions [13], lim_{ε↓0} Q(x + iε) ≡ Q(x + i0) exists for a.e. x ∈ ℝ and Im Q(x + i0) = − ln F (x).

Let G(z) = exp(iQ(z)) for z ∈ ℂ_{+} and G(x) = exp(iQ(x + i0)). Then the properties of Q imply |G(z)| ≤ 1, lim_{ε↓0} G(x + i0) = G(x) for a.e. x in ℝ, and |G(x)| = F (x). A standard contour integral argument then shows that for any ε > 0 and n > 0,

∫ x^{n}(1 − iεx)^{−n−2}G(x) dx = 0.

Since (1.13c) holds and |G(x)| ≤ F (x), we can take ε ↓ 0 and obtain ∫ x^{n}G(x) dx = 0. In case (ii), define

Q(z) = \frac{2z}{π} ∫_{0}^{∞} −\frac{ln F (x^{2})}{x^{2} − z^{2}} dx.

Again, the integral converges by (1.13b) and Im Q(z) > 0. Moreover, Q(−z̄) = −\overline{Q(z)}. We have for a.e. x ∈ ℝ, Im Q(x + i0) = − ln F (x^{2}). Thus, if H(z) = exp(iQ(z)) for z ∈ ℂ_{+} and H(x) = exp(iQ(x + i0)) for x ∈ ℝ and G(x) = Im H(\sqrt{x}) for x ∈ [0, ∞), then, as above,

∫_{−∞}^{∞} x^{2n+1}H(x) dx = 0,

so since H(−x) = \overline{H(x)}, ∫_{0}^{∞} x^{2n+1}G(x^{2}) dx = 0 or ∫_{0}^{∞} y^{n}G(y) dy = 0. Since |H(x)| = F (x^{2}), we have that |G(x)| ≤ F (x). □

Additional examples of indeterminate moment problems (specified by their orthogonal polynomials) can be found in [9, 10].

One point to note for now is that in all these examples, we have non-uniqueness with measures dρ of the form dρ(x) = G(x) dx (i.e., absolutely continuous). This will have significance after we discuss Theorem 5 below.

To discuss the theory further, we must look more closely at the operator A given by (1.8). Consider the set {1, X, X^{2}, … } in H^{(γ)}. By the strict positivity of H_{N}, these elements in H^{(γ)} are linearly independent and they span H^{(γ)} by construction. Thus by a Gram-Schmidt procedure, we can obtain an orthogonal basis, {P_{n}(X)}_{n=0}^{∞}, for H^{(γ)}. By construction,

P_{n}(X) = c_{nn}X^{n} + lower order,  with c_{nn} > 0  (1.14a)
⟨P_{n}, P_{m}⟩ = 0,  m = 0, 1, 2, …, n − 1  (1.14b)
⟨P_{n}, P_{n}⟩ = 1.  (1.14c)

These are, of course, the well-known orthogonal polynomials for dρ determined by the moments {γ_{n}}_{n=0}^{∞}. Note that often the normalization condition (1.14c) is replaced by c_{nn} ≡ 1, yielding a distinct set of “orthogonal” polynomials. There are explicit formulae for the P_{n}(X) in terms of determinants and the γ’s. We discuss them in Appendix A.

By construction, {P_{j}}_{j=0}^{n} is an orthonormal basis for the polynomials of degree n. The realization of elements of H^{(γ)} as ∑_{n=0}^{∞} λ_{n}P_{n}(X) with ∑_{n=0}^{∞} |λ_{n}|^{2} < ∞ gives a different realization of H^{(γ)} as a set of sequences λ = (λ_{0}, …) with the usual ℓ^{2}({0, 1, 2, … }) inner product. ℂ[X] corresponds to these λ’s with λ_{n} = 0 for n sufficiently large. But we need to bear in mind this change of realization as sequences.

Note that span[1, …, X^{n}] = span[P_{0}, …, P_{n}(X)]. In particular, XP_{n}(X) has an expansion in P_{0}, P_{1}, …, P_{n+1}. But ⟨XP_{n}, P_{j}⟩ = ⟨P_{n}, XP_{j}⟩ = 0 if j < n − 1 since then XP_{j} is of degree at most n − 1, and (1.14b) holds. Thus for suitable sequences, {a_{n}}_{n=0}^{∞}, {b_{n}}_{n=0}^{∞}, and {c_{n}}_{n=0}^{∞} (with P_{−1}(X) ≡ 0),

XP_{n}(X) = c_{n}P_{n+1}(X) + b_{n}P_{n}(X) + a_{n−1}P_{n−1}(X)  (1.15)

for n = 0, 1, 2, . . .. Notice that by (1.14a), c_{n} > 0 and

c_{n} = ⟨P_{n+1}, XP_{n}⟩ = ⟨P_{n}, XP_{n+1}⟩ = a_{n}.

(1.15) thus becomes

XP_{n}(X) = a_{n}P_{n+1}(X) + b_{n}P_{n}(X) + a_{n−1}P_{n−1}(X).  (1.15′)

Since {P_{n}(X)} are an orthonormal basis for H^{(γ)}, this says that in this basis, A is given by a tridiagonal matrix, and D(A) is the set of sequences of finite support.

Thus, given a set {γ_{n}}_{n=0}^{∞} of Hamburger moments, we can find b_{0}, b_{1}, … real and a_{0}, a_{1}, … positive so that the moment problem is associated to self-adjoint extensions of the Jacobi matrix,

A = ( \begin{matrix}b_{0} & a_{0} & 0 & 0 & … \\ a_{0} & b_{1} & a_{1} & 0 & … \\ 0 & a_{1} & b_{2} & a_{2} & … \\ 0 & 0 & a_{2} & b_{3} & … \\ … & … & … & … & … \end{matrix} ) .  (1.16)

There are explicit formulae for the b_{n}’s and a_{n}’s in terms of the determinants of the γ_{n}’s. We will discuss them in Appendix A.

Conversely, given a matrix A of the form (1.16), we can find a set of moments for which it is the associated Jacobi matrix. Indeed, if δ_{0} is the vector (1, 0, …, 0, …) in ℓ_{2}, then

γ_{n} = (δ_{0}, A^{n}δ_{0}).

Thus the theory of Hamburger moments is the theory of semi-infinite Jacobi matrices.

So far we have considered the P_{n}(X) as abstract polynomials, one for each n. It is useful to turn this around. First, replace the abstract X by an explicit complex number. For each such z, consider the semi-infinite sequence π(z) = (P_{0}(z), P_{1}(z), P_{2}(z), …). (1.15′) now becomes (with P_{−1}(z) interpreted as 0):

a_{n}P_{n+1}(z) + (b_{n} − z)P_{n}(z) + a_{n−1}P_{n−1}(z) = 0,  n ≥ 0  (1.17)

so P_{n}(z) obeys a second-order difference equation. If, for example, π(x_{0}) ∈ ℓ^{2}, then x_{0} is formally an eigenvalue of the Jacobi matrix A. (Of course, π is never a finite sequence because it obeys a second-order difference equation and π_{0}(z) = 1. Thus, π ∉ D(A). It may or may not happen that π ∈ D(Ā), so the formal relation may or may not correspond to an actual eigenvalue. It will always be true, though, as we shall see, that if π(x_{0}) ∈ ℓ^{2}, then (A^{∗} − x_{0})π(x_{0}) = 0.)

For convenience, set a_{−1} = 1. For any z ∈ ℂ, the solutions of the equation

a_{n}u_{n+1} + (b_{n} − z)u_{n} + a_{n−1}u_{n−1} = 0,  n ≥ 0  (1.18)

are two-dimensional, determined by the initial data (u_{−1}, u_{0}).

P_{n}(z) corresponds to taking

u_{−1} = 0, u_{0} = 1.

There is a second solution, Q_{n}(z), taking initial conditions

u_{−1} = −1, u_{0} = 0.

We will also define ξ(z) = (Q_{0}(z), Q_{1}(z), …). It can be seen by induction that for n ≥ 1, Q_{n}(X) is a polynomial of degree n − 1. We will see later (Proposition 5.16) that the Q’s are orthogonal polynomials for another moment problem.

As we will see, this normalization is such that if some combination η(z) ≡ tπ(z)+ξ(z) ∈ ℓ^{2}, then (A^{∗} −z)η = δ_{0} and ⟨δ_{0}, η⟩ = t. Here δ_{n} is the Kronecker vector with 1 in position n and zeros elsewhere. We will have a lot more to say about Q_{n}(z) in Section 4.

There is a fundamental result relating the solution vectors π, ξ to determinacy of the moment problem. In Section 4 (see Proposition 4.4 and Theorem 4.7), we will prove

Theorem 3. Fix a set of moments and associated Jacobi matrix. Then the following are equivalent:

(i) The Hamburger moment problem is indeterminate.

(ii) For some z_{0} ∈ ℂ with Im z_{0} ≠ 0, π(z_{0}) ∈ ℓ_{2}.

(iii) For some z_{0} ∈ ℂ with Im z_{0} ≠ 0, ξ(z_{0}) ∈ ℓ_{2}.

(iv) For some x_{0} ∈ ℝ, both π(x_{0}) and ξ(x_{0}) lie in ℓ_{2}.

(v) For some x_{0} ∈ ℝ, both π(x_{0}) and \frac{∂π}{∂x} (x_{0}) lie in ℓ_{2}.

(vi) For some x_{0} ∈ ℝ, both ξ(x_{0}) and \frac{∂ξ}{∂x}(x_{0}) lie in ℓ_{2}.

(vii) For all z ∈ ℂ, both π(z) and ξ(z) lie in ℓ_{2}.

Remarks. 1. Appendix A has explicit determinantal formulae for ∑_{n=0}^{N} |P_{n}(0)|^{2} and ∑_{n=0}^{N} |Q_{n}(0)|^{2} providing “explicit” criteria for determinacy in terms of limits of determinants.

2. Theorem 3 can be thought of as a discrete analog of Weyl’s limit point/limit circle theory for self-adjoint extensions of differential operators; see [34].

This implies that if all solutions of (1.18) lie in ℓ^{2} for one z ∈ ℂ, then all solutions lie in ℓ^{2} for all z ∈ ℂ.

A high point of the theory, discussed in Section 4, is an explicit description of all solutions of the moment problem in the indeterminate case in terms of a certain 2 × 2 matrix valued entire analytic function. It will suffice to only vaguely describe the full result in this introduction.

Definition. A Herglotz function is a function Φ(z) defined in ℂ_{+} ≡ {z ∈ ℂ | Im z > 0} and analytic there with Im Φ(z) > 0 there.

These are also sometimes called Nevanlinna functions. It is a fundamental result (see, e.g., [2]) that given such a Φ, there exists c ≥ 0, d real, and a measure dμ on ℝ with ∫ \frac{dμ(x)}{1+x^{2}} < ∞ so that either c ≠ 0 or dμ ≠ 0 or both, and

Φ(z) = cz + d + ∫ [\frac{1}{x − z} − \frac{x}{1 + x^{2}} ] dμ(x).  (1.19)

Theorem 4. The solutions of the Hamburger moment problem in the indeterminate case are naturally parametrized by Herglotz functions together with the functions Φ(z) = t ∈ ℝ ∪ {∞}. This later set of solutions are the von Neumann solutions.

If the Stieltjes problem is also indeterminate, there is t_{0} > 0 so that the solutions of the Stieltjes problem are naturally parametrized by Herglotz functions, Φ(z), which obey Φ(x+i0) ∈ (t_{0}, ∞) for x ∈ (−∞, 0) together with [t_{0}, ∞)∪{∞}. The later set of solutions are the von Neumann solutions.

We will prove Theorem 4 in Section 4 (see Theorems 4.14 and 4.18). From the explicit Nevanlinna form of the solutions, one can prove (and we will in Section 4; see Theorems 4.11 and 4.17)

Theorem 5. In the indeterminate case, the von Neumann solutions are all pure point measures. Moreover, for any t ∈ ℝ, there exists exactly one von Neumann solution, μ^{(t)}, with μ^{(t)}({t}) > 0. Moreover, for any other solutions, ρ, of the moment problem, ρ({t}) < μ^{(t)}({t}).

Remark. The parametrization μ^{(t)} of this theorem is inconvenient since t ↦ μ^{(t)} is many to one (in fact, infinity to one since each μ^{(t)} has infinite support). We will instead use a parametrization t ↦ μ_{t} given by

∫ \frac{dμ_{t}(x)}{x} = t,

which we will see is a one-one map of ℝ ∪ {∞} to the von Neumann solutions.

Examples 1.1 and 1.2 revisited. As noted, the explicit non-unique measures we listed were absolutely continuous measures. But there are, of necessity, many pure point measures that lead to the same moments. Indeed, the measures μ_{t} associated to the von Neumann solutions are each pure point. As we will see, there are many other pure point measures with the given moments. If ν is a Cantor measure, then it can be proven that ∫ μ_{t} dν(t) will be a singular continuous measure with the given moments. In the indeterminate case, the class of measures solving (1.1) is always extremely rich.

Given a set of moments {γ_{n}}_{n=0}^{∞} and c ∈ ℝ, one can define a new set of moments

γ_{n}(c) = ∑_{j=0}^{n} \binom{n}{j}c^{j}γ_{n−j}.  (1.20)

For the Hamburger problem, the solutions of the {γ_{n}}_{n=0}^{∞} and each {γ_{n}(c)}_{n=0}^{∞} problem are in one-one correspondence. If μ solves the {γ_{n}}_{n=0}^{∞} problem, then dρ(x − c) = dρ_{c}(x) solves the {γ_{n}(c)}_{n=0}^{∞} problem, and vice-versa. But translation does not preserve the condition supp(ρ) ⊂ [0, ∞), so that as c decreases, the set of solutions of the Stieltjes problem shrinks. We will show that for any indeterminate Stieltjes moment problem, there is always a c_{0}, so that for c > c_{0}, {γ_{n}(c)}_{n=0}^{∞} is an indeterminate Stieltjes problem. For c < c_{0}, {γ_{n}(c)}_{n=0}^{∞} has no solution dρ with supp(dρ) ⊂ [0, ∞), and {γ_{n}(c_{0})} is a determinate Stieltjes problem (but, of course, an indeterminate Hamburger problem). This means there are lots of examples of moments which are determinate Stieltjes but indeterminate Hamburger. The existence of c_{0} is intimately connected to Theorem 5. Among all the von Neumann solutions, there is a distinguished one, dμ_{F} (the Friedrichs solution), with f_{0} = inf(supp(μ_{t})) maximal. One just takes c_{0} = −f_{0}. That f_{0} is a pure point of dμ_{F} is important in the analysis.

Another topic in the moment problem concerns the theory of Padé approximants, or what is esssentially the same thing — continued fractions. Typical is Theorem 6 below, which we will prove in Section 5.

Consider a sequence of numbers {κ_{n}}_{n=0}^{∞},

κ_{n} = (−1)^{n} ∫_{0}^{∞} x^{n} dρ(x)

for some ρ. We are interested in “summing” the formal series (called a series of Stieltjes)

∑_{n=0}^{∞} κ_{n}z^{n},  (1.21)

which is formally

∫_{0}^{∞} \frac{dρ(x)}{1 + xz} .  (1.22)

If the series (1.21) converges, then (1.22) is analytic in a circle, |z| < R, which implies that ρ is supported in [0, \frac{1}{R}]. Thus, if dρ does not have compact support, then the series (1.21) will not converge for any z ≠ 0.

Definition. The [N, M] Padé approximant to the series (1.21) is the unique rational function of the form

f^{[N,M]}(z) ≡ \frac{A^{[N,M]}(z)}{B^{[N,M]}(z)} ,  (1.23)

where A is a polynomial of degree N and B is a polynomial of degree M, and (as z → 0)

f^{[N,M]}(z) − ∑_{n=0}^{N+M} κ_{n}z^{n} = O(z^{N+M+1}).  (1.24)

Note that A/B has (N + 1) + (M + 1) − 1 = N + M + 1 free parameters and {κ_{n}}_{n=0}^{N+M} is N + M + 1 free numbers. There is an explicit solution for A, B in terms of determinants; see Baker and Graves-Morris [4]. We will say more about the definition of Padé approximants in Section 5.

Theorem 6. Let ∑ κ_{n}z^{n} be a series of Stieltjes. Then for any z ∈ ℂ\(−∞, 0), lim_{N→∞} f^{[N−1,N]}(z) ≡ f_{−}(z) exists and lim_{N→∞} f^{[N,N]}(z) = f_{+}(z) exists. The convergence is uniform on compact subsets of ℂ\(−∞, 0]. Moreover, f_{+} = f_{−} if and only if the associated Stieltjes problem is determinate. If the problem is indeterminate and ρ is any solution of the associated Stieltjes problem, then

f_{−}(x) ≤ ∫_{0}^{∞} \frac{dρ(y)}{1 + xy} ≤ f_{+}(x)  (1.25)

for any x ∈ [0, ∞). In any event, for x > 0, f^{[N,N]}(x) is monotone decreasing in N and f^{[N−1,N]}(x) is monotone increasing.

In terms of the language of Section 3, f_{−} is associated to the Friedrichs solution and f_{+} to the Krein solution. For an interesting application of Theorem 6, see [22, 36]. In Section 5, we will also discuss series of Hamburger and prove the new result that the f^{[N,N]}(z) Padé approximants always converge in that case.

As a by-product of the proof of Theorem 6, we will find

Theorem 7. Let {γ_{n}}_{n=0}^{∞} be a set of Stieltjes moments. Then the Stieltjes problem is indeterminate if and only if

∑_{n=0}^{∞} |P_{n}(0)|^{2} < ∞  (1.26)

and

sup_{n}|\frac{Q_{n}(0)}{P_{n}(0)}| < ∞.  (1.27)

Theorem 8. Let {γ_{n}}_{n=0}^{∞} be a set of Stieltjes moments. Then the Stieltjes problem is determinate while the Hamburger problem is indeterminate if and only if

∑_{n=0}^{∞} |Q_{n}(0)|^{2} < ∞  (1.28)

and

lim_{n→∞}|\frac{Q_{n}(0)}{P_{n}(0)}| = ∞.  (1.29)

We will see that (1.26)/(1.27) are equivalent to a criterion of Stieltjes.

Theorem 6 is proven in Section 5 as parts of Theorem 5.2, Proposition 5.6, Proposition 5.8, and Proposition 5.11. Theorem 7 is Theorem 5.21 and Theorem 8 is Proposition 5.22.

In Section 6, which synthesizes and extends results in Akhiezer [1], we will study when the closure of the polynomials has finite codimension, and among our results proven as part of Theorem 6.4 is Theorem 9. Let ρ lie in M^{H}(γ) and let H_{0} be the closure of the set of polynomials in H_{ρ} ≡ L^{2}(ℝ, dρ). Then H_{0}^{⊥} has finite dimension if and only if the Herglotz function, Φ, in the Nevanlinna parametrization of Theorem 4 is a real rational function (i.e., a ratio of real poynomials). Equivalently, if and only if the measure μ of (1.19) has finite support.

In fact, we will see that dim(H_{0}^{⊥}) is the degree of the rational function Φ.

One consequence of this will be the following, proven in Appendix B (see Theorems B.1 and B.4).

Theorem 10. M^{H}(γ) is a compact convex set (in the weak topology as a subset of the dual of C(ℝ ∪ {∞})). Its extreme points are dense in the set.

Remark. Do not confuse this density with the fact that the Krein-Millman theorem says that the convex combinations of the extreme points are dense. Here we have density without convex combinations, to be compared with the fact that in many other cases (e.g., probability measures on a compact Hausdorff space), the extreme points are closed.

Here is a sketch of the contents of the rest of this paper. In Section 2, we review von Neumann’s theory of self-adjoint extensions and use it to prove the uniqueness result (Theorem 2) in the Hamburger case. In Section 3, we review the Birman-Krein-Vishik theory of semi-bounded self-adjoint extensions and use it to prove the uniqueness result (Theorem 3) in the Stieltjes case. In Section 4, we turn to a detailed analysis of the polynomials P and Q in terms of transfer matrices for the difference equation associated to the Jacobi matrix and prove Theorems 3, 4, and 5. In Section 5, we discuss Padé approximations and prove Theorems 6, 7, and 8. In Section 6, we discuss solutions, ρ, where the closure of the polynomials has finite codimension in L^{2}(ℝ, dρ). Appendix A is devoted to a variety of explicit formulae in terms of determinants of moments, and Appendix B to the structure of M^{H}(γ) as a compact convex set. Appendix C summarizes notation and some constructions.

§2. The Hamburger Moment Problem as a Self-Adjointness Problem

Let us begin this section with a brief review of the von Neumann theory of selfadjoint extensions. For further details, see [2, 5, 33, 34]. We start out with a densely defined operator A on a Hilbert space H, that is, D(A) is a dense subset of H and A : D(A) → H a linear map. We will often consider its graph Γ(A) ⊂ H × H given by Γ(A) = {(φ, Aφ) | φ ∈ D(A)}. Given operators A, B, we write A ⊂ B and say B is an extension of A if and only if Γ(A) ⊂ Γ(B).

One defines a new operator A^{∗} as follows: η ∈ D(A^{∗}) if and only if there is a ψ ∈ H so that for all φ ∈ D(A), ⟨ψ, φ⟩ = ⟨η, Aφ⟩. We set A^{∗}η = ψ. In essence, A^{∗}, called the adjoint of A, is the maximal object obeying

⟨A^{∗}η, φ⟩ = ⟨η, Aφ⟩  (2.1)

for all η ∈ D(A^{∗}), φ ∈ D(A).

An operator is called closed if and only if Γ(A) is a closed subset of H × H and it is called closable if and only if \overline{Γ(A)} is the graph of an operator, in which case we define Ā, the closure of A, by Γ(Ā) = \overline{Γ(A)}. Thus, D(Ā) = \{φ ∈ H | ∃φ_{n} ∈ D(A), so that φ_{n} → φ and Aφ_{n} is Cauchy} and one sets Aφ = lim Aφ_{n}.

Adjoints are easily seen to be related to these notions: A^{∗} is always closed; indeed,

Γ(A^{∗}) = {(Aφ, −φ) ∈ H × H | φ ∈ D(A)}^{⊥}.  (2.2)

A^{∗} is densely defined if and only if A is closable, in which case (by (2.2)), Ā = (A^{∗})^{∗}.

An operator is called symmetric if A ⊂ A^{∗} (equivalently, if ⟨φ, Aψ⟩ = ⟨Aφ, ψ⟩ for all φ, ψ ∈ D(A)), self-adjoint if A = A^{∗}, and essentially self-adjoint if Ā is self-adjoint (equivalently, if A^{∗} is self-adjoint). Notice that symmetric operators are always closable since D(A^{∗}) ⊃ D(A) is dense. Note also that if A is symmetric, then A^{∗} is symmetric if and only if A is essentially self-adjoint.

Von Neumann’s theory solves the following fundamental question: Given a symmetric operator A, when does it have self-adjoint extensions, are they unique, and how can they be described? If B is a self-adjoint extension of A, then B is closed, so Ā ⊂ B. Thus, looking at self-adjoint extensions of A is the same as looking at self-adjoint extensions of Ā. So for now, we will suppose A is a closed symmetric operator.

Define K_{±} = ker(A^{∗} ∓ i), that is, K_{+} = {φ ∈ H | A^{∗}φ = iφ}. They are called the deficiency subspaces. Using ∥(A ± i)φ∥^{2} = ∥Aφ∥^{2} + ∥φ∥^{2}, it is easy to see that if A is closed, then Ran(A ± i) are closed, and since ker(A^{∗} ∓ i) = Ran(A ± i)^{⊥}, we have

Ran(A ± i) = K_{±}^{⊥}.  (2.3)

Let d_{±} = dim(K_{±}), the deficiency indices of A.

Place the graph norm on D(A^{∗}), that is, ∥φ∥_{A^{∗}}^{2} = ⟨φ, φ⟩ + ⟨A^{∗}φ, A^{∗}φ⟩. This norm comes from an inner product, ⟨φ, ψ⟩_{A^{∗}} = ⟨φ, ψ⟩ + ⟨A^{∗}φ, A^{∗}ψ⟩.

Proposition 2.1. Let A be a closed symmetric operator. Then

D(A^{∗}) = D(A) ⊕ K_{+} ⊕ K_{−},  (2.4)

where ⊕ means orthogonal direct sum in the ⟨ · , · ⟩_{A^{∗}} inner product.

Proof. If φ ∈ K_{+} and ψ ∈ K_{−}, then ⟨φ, ψ⟩_{A^{∗}} = ⟨φ, ψ⟩ + ⟨iφ, −iψ⟩ = 0 so K_{+}⊥_{A^{∗}}K_{−}. If φ ∈ D(A) and ψ ∈ K_{±}, then ⟨A^{∗}φ, A^{∗}ψ⟩ = ⟨Aφ, ±iψ⟩ = ⟨φ, ±iA^{∗}ψ⟩ = −⟨φ, ψ⟩, so D(A)⊥_{A^{∗}}K_{+} ⊕ K_{−}.

Let η ∈ D(A^{∗}). By (2.3), Ran(A+i)+K_{+} = H so we can find φ ∈ D(A) and ψ ∈ K_{+}, so (A^{∗} + i)η = (A + i)φ + 2iψ. But then (A^{∗} + i)[η − φ − ψ] = 0, so η − φ − ψ ∈ K_{−}, that is, η ∈ D(A) + K_{+} + K_{−}. □

Corollary 2.2. (i) Let A be a closed symmetric operator. Then A is self-adjoint if and only if d_{+} = d_{−} = 0.

(ii) Let A be a symmetric operator. Then A is essentially self-adjoint if and only if d_{+} = d_{−} = 0.

If A ⊂ B, then B^{∗} ⊂ A^{∗}, so if B is symmetric, then A ⊂ B ⊂ B^{∗} ⊂ A^{∗}. Thus, to look for symmetric extensions of A, we need only look for operators B with A ⊂ B ⊂ A^{∗}, that is, for restriction of A^{∗} to D(B)’s with D(A) ⊂ D(B). By Proposition 2.1, every such D(B) has the form D(A) + S with S ⊂ K_{+} + K_{−}. On D(A^{∗}) × D(A^{∗}), define the sesquilinear form (sesquilinear means linear in the second factor and anti-linear in the first)

Q(φ, ψ) = ⟨φ, A^{∗}ψ⟩_{H} − ⟨A^{∗}φ, ψ⟩_{H}.

Proposition 2.3. Let A be a closed symmetric operator. Then

(i) The operators B with A ⊂ B ⊂ A^{∗} are in one-one correspondence with subspaces S of K_{+} + K_{−} under D(B) = D(A) + S.

(ii) B is symmetric if and only if Q ↾ D(B) × D(B) = 0.

(iii) B is symmetric if and only if Q ↾ S × S = 0.

(iv) B is closed if and only if S is closed in K_{+} ⊕ K_{−} in D(A^{∗}) norm.

(v) φ ∈ D(A^{∗}) lies in D(B^{∗}) if and only if Q(φ, ψ) = 0 for all ψ ∈ D(B).

(vi) Let J : K_{+} ⊕ K_{−} → K_{+} ⊕ K_{−} by J(φ, ψ) = (φ, −ψ). If D(B) = D(A) + S, then

D(B^{∗}) = D(A) + J[S]^{⊥}, where ^{⊥} is in K_{+} ⊕ K_{−} in the D(A^{∗}) norm. (vii) K_{+}(B) = K_{+} ∩ S^{⊥}, K_{−}(B) = K_{−} ∩ S^{⊥} (with ^{⊥} in the ⟨ , ⟩_{A^{∗}} inner product).

Proof. We have already seen (i) holds and (ii) is obvious. (iii) holds since if φ ∈ D(A) and ψ ∈ D(A^{∗}), then Q(φ, ψ) = 0 by definition of A^{∗} and symmetry of A. Thus, if φ_{1}, φ_{2} ∈ D(A) and ψ_{1}, ψ_{2} ∈ S, then Q(φ_{1} + ψ_{1}, φ_{2} + ψ_{2}) = Q(ψ_{1}, ψ_{2}). (iv) is immediate if one notes that Γ(B) = Γ(A) ⊕ {(φ, A^{∗}φ) | φ ∈ S} with ⊕ in H × H norm and that the H × H norm on {(φ, A^{∗}φ) | φ ∈ S} is just the D(A^{∗}) norm. (v) follows from the definition of adjoint.

To prove (vi), let η = (η_{1}, η_{2}), φ = (φ_{1}, φ_{2}) ∈ K_{+} ⊕ K_{−}. Then, direct calculations show that

Q(η, φ) = 2i[⟨η_{1}, φ_{1}⟩_{H} − ⟨η_{2}, φ_{2}⟩_{H}]  (2.5)
⟨η, φ⟩_{A^{∗}} = 2[⟨η_{1}, φ_{1}⟩_{H} + ⟨η_{2}, φ_{2}⟩_{H}].  (2.6)

Thus, Q(η, φ) = 0 if and only if η⊥Jφ. (v) thus implies (vi).

To prove (vii), note that by (vi), K_{±}(B) = K_{±} ∩ D(B^{∗}) = K_{±} ∩ J[S]^{⊥} = K_{±} ∩ S^{⊥} since φ ∈ K_{±} lies in J[S]^{⊥} if and only if Jφ lies in S^{⊥} if and only if φ lies in S^{⊥} by Jφ = ±φ if φ ∈ K_{±}. □

With these preliminaries out of the way, we can prove von Neumann’s classification theorem:

Theorem 2.4. Let A be a closed symmetric operator. The closed symmetric extensions B of A are in one-one correspondence with partial isometries of K_{+} into K_{−}, that is, maps U : K_{+} → K_{−} for which there are closed subspaces U_{+} ⊂ K_{+} and U_{−} ⊂ K_{−} so that U is a unitary map of U_{+} → U_{−} and U ≡ 0 on U_{+}^{⊥}. If B corresponds to U, then D(B) = D(A) + {φ + Uφ | φ ∈ U_{+}}. Moreover, K_{±}(B) = K_{±} ∩ U_{±}^{⊥} (H orthogonal complement). In particular, A has self-adjoint extensions if and only if d_{+} = d_{−} and then the self-adjoint extensions are in one-one correspondence with unitary maps U from K_{+} to K_{−}.

Proof. Let B be a symmetric extension of A. Let φ ∈ D(B) ∩ (K_{+} ⊕ K_{−}), say φ = (φ_{1} + φ_{2}). By (2.5), Q(φ, φ) = 2i(∥φ_{1}∥^{2} − ∥φ_{2}∥^{2}), so B symmetric implies that ∥φ_{1}∥ = ∥φ_{2}∥, so D(B) ∩ (K_{+} ⊕ K_{−}) = {some (φ_{1}, φ_{2})}, where ∥φ_{1}∥ = ∥φ_{2}∥. Since D(B) is a subspace, one cannot have (φ_{1}, φ_{2}) and (φ_{1}, φ′_{2}) with φ_{2} ≠ φ′_{2} (for then (0, φ_{2} − φ′_{2}) ∈ D(B) so ∥φ_{2} − φ′_{2}∥ = 0). Thus, D(B) ∩ (K_{+} ⊕ K_{−}) is the graph of a partial isometry. On the other hand, if U is a partial isometry, Q(φ + Uφ, ψ + Uψ) = 2i(⟨φ, ψ⟩ − ⟨Uφ, Uψ⟩) = 0 so each U does yield a symmetric extension.

That K_{±}(B) = K_{±} ∩ U_{±}^{⊥} follows from (vii) of the last proposition if we note that if φ ∈ K_{+}, then φ ⊥_{A^{∗}} {(ψ + Uψ) | ψ ∈ U_{+}} if and only if φ ⊥_{A^{∗}} {ψ | ψ ∈ U_{+}} if and only if φ ⊥_{H} U_{+} (by (2.6)).

Thus B is self-adjoint if and only if U is a unitary from K_{+} to K_{−}, which completes the proof. □

Recall that a map T is anti-linear if T (aφ+bψ) = āT (φ)+b̄T (ψ) for a, b ∈ ℂ, φ, ψ ∈ H; that T is anti-unitary if it is anti-linear, a bijection, and norm-preserving; and that a complex conjugation is an anti-unitary map whose sequence is 1.

Corollary 2.5. Let A be a symmetric operator. Suppose there exists a complex conjugation C : H → H so that C : D(A) → D(A) and CAφ = ACφ for all φ ∈ D(A). Then A has self-adjoint extensions. If d_{+} = 1, every self-adjoint extension B is real, that is, obeys C : D(B) → D(B) and CBφ = BCφ for all φ ∈ D(B). If d_{+} ≥ 2, there are non-real self-adjoint extensions.

Proof. C is an anti-unitary from K_{+} to K_{−} so d_{+} = d_{−}.

We claim that if B is self-adjoint and is associated to U : K_{+} → K_{−}, then B is real if and only if CUCU = 1 for it is easy to see that C : D(A^{∗}) → D(A^{∗}) and CA^{∗}φ = A^{∗}Cφ, so B is real if and only if C maps D(B) to itself. Thus, C(φ + Uφ) must be of the form ψ + Uψ. Since ψ must be CUφ, this happens if and only if Cφ = UCUφ, that is, CUCUφ = φ. This proves our claim.

If d_{+} = 1 and φ ∈ K_{+}, CUφ = e^{iθ}φ for some θ. Then (CUCU)φ = Ce^{iθ}Uφ = e^{−iθ}CUφ = φ showing that every self-adjoint extension is real.

If d_{+} ≥ 2, pick φ, ψ ∈ K_{+} with φ ⊥ ψ and let Uφ = Cψ, Uψ = iCφ. Then CUCUφ = CUCCψ = CUψ = CiCψ = −iφ, so CUCU ≢ 1. Thus, the B associated to this U will not be real. □

Next we analyze a special situation that will always apply to indeterminate Hamburger moment problems.

Theorem 2.6. Suppose that A is a closed symmetric operator so that there exists a complex conjugation under which A is real. Suppose that d_{+} = 1 and that ker(A) = {0}, dim ker(A^{∗}) = 1. Pick φ ∈ ker(A^{∗}), Cφ = φ, and η ∈ D(A^{∗}), not in D(A) + ker(A^{∗}). Then ⟨φ, A^{∗}η⟩ ≠ 0 and ψ = {η − [⟨η, A^{∗}η⟩/⟨φ, A^{∗}η⟩]φ}/⟨φ, A^{∗}η⟩ are such that in φ, ψ basis, ⟨ · , A^{∗} · ⟩ has the form

⟨ · , A^{∗} · ⟩ = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) .  (2.7)

The self-adjoint extensions, B_{t}, can be labelled by a real number or ∞ where

D(B_{t}) = D(A) + {α(tφ + ψ) | α ∈ ℂ}  t ∈ ℝ
= D(A) + {αφ | α ∈ ℂ}  t = ∞.

The operators B_{t} are independent of which real ψ in D(A^{∗})\D(A) is chosen so that (2.7) holds.

Proof. If ⟨φ, A^{∗}η⟩ = 0, then the ⟨ · , A^{∗} · ⟩ matrix with basis φ, η would have the form ( \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} ) with a ∈ ℝ, in which case Q ≡ 0 on the span of φ and η, which is incompatible with the fact that d_{+} = 1. Thus, ⟨φ, A^{∗}η⟩ ≠ 0 and (2.7) holds by an elementary calculation.

Once (2.7) holds, it is easy to see that the subspaces S of D(A^{∗}) with D(A) ⊂ S, dim(S/D(A)) = 1, and Q ↾ S × S = 0 are precisely the D(B_{t}).

If ψ̃ is a second ψ for which (2.7) holds, then Q(ψ − ψ̃, ρ) = 0 for all ρ ∈ D(A^{∗}), so ψ − ψ̃ ∈ D(A) and the D(B_{t})’s are the same for ψ and ψ̃. □

Remarks. 1. It is easy to see that if ker(A) = {0}, then dim ker(A^{∗}) ≤ d_{+}, so if d_{+} = 1 and ker(A) = \{0\}, dim ker(A^{∗}) is either 0 or 1. The example of A = −\frac{d^{2}}{dx^{2}} on L^{2}(0, ∞) with D(A) = C_{0}^{∞}(0, ∞) shows that it can happen that A is real, d_{+} = 1, but dim ker(A^{∗}) = 0.

2. One example where this formalism is natural is if A ≥ α > 0 (for α ∈ ℝ), in which case dim ker(A^{∗}) = d_{+}. One takes η = A_{F}^{−1}φ, where A_{F} is the Friedrichs extension. We will discuss this further in Section 3. A second example, as we will see, is the indeterminate moment problem, in which case one can take φ = π(0), ψ = ξ(0).

We can now turn to the analysis of the symmetric operator A on H^{(γ)} described in Section 1. In terms of the explicit basis P_{n}(X) in H^{(γ)}, we know A has the Jacobi matrix form (1.16). We will first explicitly describe A^{∗} and the form Q. Given any sequence s = (s_{0}, s_{1}, …), we define a new sequence F(s) (think of F for “formal adjoint”) by

F(s)_{n} = \{ \begin{matrix}b_{0}s_{0} + a_{0}s_{1} & if n = 0 \\ a_{n−1}s_{n−1} + b_{n}s_{n} + a_{n}s_{n+1} & if n ≥ 1. \end{matrix}

Given any two sequences, s_{n} and t_{n}, we define their Wronskian,

W (s, t)(n) = a_{n}(s_{n+1}t_{n} − s_{n}t_{n+1}).  (2.8)

The Wronskian has the following standard property if

a_{n}s_{n+1} + c_{n}s_{n} + a_{n−1}s_{n−1} = 0  (2.9)

and

a_{n}t_{n+1} + d_{n}t_{n} + a_{n−1}t_{n−1} = 0.  (2.10)

Then multiplying (2.9) by t_{n} and subtracting (2.10) multiplied by s_{n}, we see that

W (s, t)(n) − W (s, t)(n − 1) = (d_{n} − c_{n})s_{n}t_{n}.  (2.11)

In particular, if d = c, then W is constant.

Theorem 2.7. Let A be the Jacobi matrix (1.16) with D(A) = {s | s_{n} = 0 for n sufficiently large}. Then

D(A^{∗}) = {s ∈ ℓ_{2} | F(s) ∈ ℓ_{2}}  (2.12)

with

A^{∗}s = F(s)  (2.13)

for s ∈ D(A^{∗}). Moreover, if s, t ∈ D(A^{∗}), then

lim_{n→∞} W (s̄, t)(n) = ⟨A^{∗}s, t⟩ − ⟨s, A^{∗}t⟩.  (2.14)

Proof. Since the matrix (1.16) is a real-symmetric matrix, it is easy to see that for any t ∈ D(A), and any sequences

⟨s, At⟩ = ⟨F(s), t⟩.  (2.15)

Since t and At are finite sequences, the sum in ⟨ · , · ⟩ makes sense even if s or F(s) are not in ℓ_{2}. Since D(A) is dense in ℓ_{2}, (2.15) says that s ∈ ℓ_{2} lies in D(A^{∗}) precisely if F(s) ∈ ℓ_{2} and then (2.13) holds. That proves the first part of the theorem.

For the second, we perform a calculation identical to that leading to (2.11):

∑_{n=0}^{N} [ \overline{Fs_{n}} t_{n} − \overline{s_{n}} (Ft_{n})] = W (s̄, t)(0) + ∑_{n=1}^{N} W (s̄, t)(n) − W (s̄, t)(n − 1)
= W (s̄, t)(N).  (2.16)

If s, t ∈ D(A^{∗}), then the left side of (2.16) converges to ⟨A^{∗}s, t⟩ − ⟨s, A^{∗}t⟩ as N → ∞, so (2.14) holds. □

Lemma 2.8. If φ ∈ D(A^{∗}), (A^{∗} − z)φ = 0 for some z and φ_{0} = 0, then φ ≡ 0.

Proof. Essentially, this is so because (A^{∗} − z)s = 0 is a second-order difference equation with s_{−1} ≡ 0, so solutions are determined by s_{0}. Explicitly, suppose A^{∗}s = zs and s_{0} = 0. Then

s_{n+1} = a_{n}^{−1}[(−b_{n} + z)s_{n} − a_{n−1}s_{n−1}],  n ≥ 1
= a_{0}^{−1}[(−b_{0} + z)s_{0}],  n = 0.

By a simple induction, s ≡ 0. □

Corollary 2.9. The operator A associated to a Hamburger moment problem always has either deficiency indices (1, 1) or deficiency indices (0, 0).

Proof. We have already seen that d_{+} = d_{−} so it suffices to show that ker(A^{∗} − i) has dimension at most 1. By Lemma 2.8, the map from ker(A^{∗} − i) to ℂ that takes s ↦ s_{0} is one-one, so ker(A^{∗} − i) is of dimension 0 or 1. □

We are now ready to prove the more subtle half of the Hamburger part of Theorem 2.

Theorem 2.10 (one quarter of Theorem 2). Let A be essentially self-adjoint. Then the Hamburger moment problem has a unique solution.

Proof. Pick z with Im z > 0 and a measure ρ obeying (1.1). Since A is essentially selfadjoint, (A − i)[D(A)] is dense in ℓ^{2}, and by an identical argument, (A − z)[D(A)] is dense in ℓ^{2}. Thus, there exists a sequence R_{n}^{(z)}(X) of polynomials so that

∥(X − z)R_{n}^{(z)}(X) − 1∥ → 0

in H^{(γ)}, and thus

∫ [(x − z)R_{n}^{(z)}(x) − 1]^{2} dρ(x) → 0

since dρ(x) realizes H^{(γ)} on polynomials. Now \frac{1}{x−z} is bounded for x ∈ ℝ since Im z > 0. Thus

∫ |R_{n}^{(z)}(x) − \frac{1}{x − z} |^{2} dρ(x) → 0.

It follows that

G_{ρ}(x) ≡ ∫ \frac{dρ(x)}{x − z} = lim_{n→∞} ∫ R_{n}^{(z)}(x) dρ(x)

is independent of ρ. Since G_{ρ}(x) determines ρ (because ρ({a}) = lim_{ε↓0} ε Im G_{ρ}(a + iε) and ρ((a, b)) + ρ([a, b]) = \frac{2}{π} lim_{ε↓0} ∫_{a}^{b} Im G_{ρ}(y + iε) dy), all ρ’s must be the same. □

Theorem 2.11. Let A be a Jacobi matrix with D(A) the sequences of finite support. Suppose A is not essentially self-adjoint and B, F are distinct self-adjoint extensions of A. Then

(δ_{0}, \frac{1}{B − i} δ_{0}) ≠ (δ_{0}, \frac{1}{F − i} δ_{0}).  (2.17)

Remarks. 1. The proof shows for any z ∈ ℂ\ℝ, (δ_{0}, (B − z)^{−1}δ_{0}) ≠ (δ_{0}(F − z)^{−1}δ_{0}). It also works for z ∈ ℝ as long as z is in the resolvent set for both B and F.

2. In Section 4, we will have a lot more to say about the possible values of (δ_{0}, \frac{1}{B−z} δ_{0}) as B runs through the self-adjoint extensions of A.

Proof. We first claim that δ_{0} ∉ Ran(A−i). For suppose on the contrary that there exists η ∈ D(A) and δ_{0} = (A − i)η and that (A^{∗} + i)φ = 0. Then (δ_{0}, φ) = ((A − i)η, φ) = (η, (A^{∗}+i)φ) = 0. By Lemma 2.8, φ ≡ 0. Thus, if δ_{0} ∈ Ran(A−i), then ker(A^{∗}+i) = {0}, so A is essentially self-adjoint. By hypothesis, this is false, so δ_{0} ∉ Ran(A − i).

Thus, (B − i)^{−1}δ_{0} and (F − i)^{−1}δ_{0} are in D(A^{∗})\D(A). Since dim(D(B)/D(A)) = dim(D(F)/D(A)) = 1, if these vectors were equal, D(B) would equal D(F), so B = A^{∗} ↾ D(B) would equal F = A^{∗} ↾ D(F). Thus, (B − i)^{−1}δ_{0} ≠ (F − i)^{−1}δ_{0}.

Let η = (B − i)^{−1}δ_{0} − (F − i)^{−1}δ_{0}. Then

(A^{∗} − i)η = (A^{∗} − i)(B − i)^{−1}δ_{0} − (A^{∗} − i)(F − i)^{−1}δ_{0} = δ_{0} − δ_{0} = 0,

so η ≠ 0 implies η_{0} ≠ 0 by Lemma 2.8. Thus, (δ_{0}, η) = (δ_{0}, (B − i)^{−1}δ_{0}) − (δ_{0}, (F − i)^{−1}δ_{0}) ≠ 0. □

As a corollary of Theorem 2.11, we have

Theorem 2.12 (two quarters of Theorem 2). A Hamburger moment problem for which A is not essentially self-adjoint is indeterminate. A Stieltjes moment problem for which A has multiple non-negative self-adjoint extensions is indeterminate.

Proof. Theorem 2.11 implies that distinct self-adjoint extensions lead to distinct spectral measures since

(δ_{0}, (B − i)^{−1}δ_{0}) = ∫ \frac{dμ^{B}(x)}{x − i} ,

where μ^{B} is the solution to (1.1) associated to B. Positive self-adjoint extensions yield solutions of the Stieltjes moment problem. □

With this machinery available, we can prove the converse of Proposition 1.6:

Theorem 2.13. Let {γ_{n}}_{n=0}^{∞} be a set of moments for which the Stieltjes problem has solutions. Let

Γ_{2m} = γ_{m},  m = 0, 1, …  (2.18a)
Γ_{2m+1} = 0,  m = 0, 1, ….  (2.18b)

Then {Γ_{j}}_{j=0}^{∞} is a determinate Hamburger problem if and only if {γ_{n}}_{n=0}^{∞} is a determinate Stieltjes problem.

Proof. The proof of Proposition 1.6 shows there is a one-one correspondence between solutions of the Stieltjes problem for γ and those solutions dρ of the Hamburger problem for Γ with dρ(−x) = dρ(x). We will call such solutions symmetric. Thus it suffices to show that if the Hamburger problem for Γ is indeterminate, then there are multiple symmetric solutions.

Let U map ℂ(X) to itself by U[P (X)] = P (−X). By Γ_{2m+1} = 0, U is unitary, and so extends to a unitary map from H^{(Γ)} to itself. Clearly, U maps D(A) = ℂ(X) to itself and UAU^{−1} = −A.

Thus, U maps D(A^{∗}) to itself and UA^{∗}U^{−1} = −A^{∗}. This means that U maps K_{+} to K_{−}.

Let C be the complex conjugation under which A is real. Then UC = CU is an anti-unitary map of K_{+} to itself. So if φ is a unit vector in K_{+}, then UCφ = e^{iθ}φ for some real θ. Thus, UC(e^{iθ/2}φ) = e^{−iθ/2}UCφ = e^{iθ/2}φ so there exists ψ ≠ 0 in K_{+} with UCψ = ψ.

In particular, since UC = CU, U(ψ ± Cψ) = ±(ψ ± Cψ) and therefore, if B = A^{∗} ↾ D(B) and F = A^{∗} ↾ D(F), where D(B) = D(A)+[ψ+Cψ] and D(F) = D(A)+[ψ−Cψ], then U leaves both D(B) and D(F) invariant, and so B and F are distinct extensions with UBU^{−1} = −B and UF U^{−1} = −F. Their spectral measures thus yield distinct symmetric extensions. □

Remark. Since U(ψ +e^{iθ}Cψ) = e^{iθ}(ψ +e^{−iθ}Cψ), we see that if θ ≠ 0, π, then U does not leave D(B_{θ}) invariant. Thus among von Neumann solutions of the Γ problem, exactly two are symmetric.

As a consequence of our identification of A^{∗}, we can prove one small part of Theorem 3. Recall that π(x_{0}) is the sequence {P_{n}(x_{0})}_{n=0}^{∞} and ξ(x_{0}) is the sequence {Q_{n}(x_{0})}_{n=i}^{∞}.

Theorem 2.14. Let {γ_{n}}_{n=0}^{∞} be a set of Hamburger moments. If either of the following holds, then the Hamburger moment problem is indeterminate.

(i) For some x_{0} ∈ ℝ, π(x_{0}) and \frac{∂π}{∂x} (x_{0}) lie in ℓ_{2}.

(ii) For some x_{0} ∈ ℝ, ξ(x_{0}) and \frac{∂ξ}{∂x}(x_{0}) lie in ℓ_{2}.

Remark. Since P_{n}(z) is a polynomial, \frac{∂π}{∂x} makes sense as a sequence.

Proof. In terms of the formal adjoint F, we have

F(π(x_{0})) = x_{0}π(x_{0})  (2.19)
F(\frac{∂π}{∂x} (x_{0})) = π(x_{0}) + x_{0} \frac{∂π}{∂x} (x_{0})  (2.20)
F(ξ(x_{0})) = δ_{0} + x_{0}ξ(x_{0})  (2.21)
F(\frac{∂ξ}{∂x} (x_{0})) = ξ(x_{0}) + x_{0} \frac{∂ξ}{∂x_{0}} .  (2.22)

Thus, if (i) holds, we conclude π, \frac{∂π}{∂x} are in D(A^{∗}) and

(A^{∗} − x_{0})π(x_{0}) = 0, (A^{∗} − x_{0})\frac{∂π}{∂x} |_{x=x_{0}} = π(x_{0}).

If A^{∗} were self-adjoint, then

∥π(x_{0})∥^{2} = ⟨ π, (A^{∗} − x_{0})\frac{∂π}{∂x} |_{x=x_{0}} ⟩ = ⟨(A^{∗} − x_{0})π, \frac{∂π}{∂x} |_{x=x_{0}} ⟩ = 0,

which is impossible since P_{0}(x_{0}) = 1. Thus, A^{∗} is not self-adjoint and the problem is indeterminate.

If (ii) holds, ξ, \frac{∂ξ}{∂x} are in D(A^{∗}) and

(A^{∗} − x_{0})ξ(x_{0}) = δ_{0}, (A^{∗} − x_{0})\frac{∂ξ}{∂x}|_{x=x_{0}} = ξ(x_{0}).

If A^{∗} were self-adjoint, then

∥ξ(x_{0})∥^{2} = ⟨ ξ(x_{0}), (A^{∗} − x_{0})\frac{∂ξ}{∂x}|_{x=x_{0}} ⟩

= ⟨ (A^{∗} − x_{0})ξ(x_{0}), \frac{∂ξ}{∂x}|_{x=x_{0}} ⟩

= \frac{∂Q_{0}(x)}{∂x}|_{x=x_{0}} = 0

since Q_{0}(x_{0}) ≡ 0. This is impossible since Q_{1}(x_{0}) ≠ 0 and so again, the problem is indeterminate. □

While we have used the theory of essential self-adjointness to study the moment problem, following Nussbaum [30], one can turn the analysis around:

Definition. Let B be a symmetric operator on a Hilbert space H and let φ ∈ C^{∞}(B) = ∩_{n}D(B^{n}). φ is called a vector of uniqueness for B if and only if the Hamburger moment problem for

γ_{n} = \frac{(φ, B^{n}φ)}{∥φ∥^{2}}  (2.23)

is determinate.

Theorem 2.15 (Nussbaum [30]). Let B be a (densely defined) symmetric operator on a Hilbert space H. Suppose that D(B) contains a dense subset of vectors of uniqueness. Then B is essentially self-adjoint.

Proof. Let U be the vectors of uniqueness for B. Let φ ∈ U and let A_{φ} be the restriction of B to the closure of the space spanned by {B^{n}φ}_{n=0}^{∞}. A_{φ} is unitarily equivalent to the Jacobi matrix for the moment problem (2.23). Thus, A_{φ} is essentially self-adjoint on H_{φ}, the closure of the span of \{B^{n}φ\}_{n=0}^{∞}. Therefore, \overline{(A_{φ} + i)[D(A_{φ})]} = H_{φ} and, in particular, φ ∈ \overline{(A_{φ} + i)[D(A_{φ})]} ⊂ \overline{(B + i)[D(B)]}. It follows that U ⊂ \overline{(B + i)[D(B)]} and thus, \overline{Ran(B + i)} = H. Similarly, \overline{Ran(B − i)} = H, so B is essentially self-adjoint by Corollary 2.2. □

The motivation for Nussbaum was understanding a result of Nelson [28].

Definition. Let B be a symmetric operator. φ ∈ C^{∞} is called analytic if ∥B^{n}φ∥ ≤ CR^{n}n!

semi-analytic if ∥B^{n}φ∥ ≤ CR^{n}(2n)!
quasi-analytic if ∑_{n=1}^{∞} ∥B^{2n}φ∥^{1/2n} < ∞.
Stieltjes if ∑_{n=1}^{∞} ∥B^{n}φ∥^{1/2n} < ∞.

Corollary 2.16. If B is a symmetric operator, then any analytic or quasi-analytic vector is a vector of uniqueness. If B is also bounded from below, then any semi-analytic or Stieltjes vector is a vector of uniqueness. In particular,

(i) (Nelson [28]) If D(B) contains a dense set of analytic vectors, then B is essentially self-adjoint.

(ii) (Nussbaum [31], Masson-McClary [23]) If D(B) contains a dense set of semi-

analytic vectors and is bounded from below, then B is essentially self-adjoint. (iii) (Nussbaum [30]) If D(B) contains a dense set of quasi-analytic vectors, then B

is essentially self-adjoint. (iv) (Nussbaum [31], Masson-McClary [23]) If D(B) contains a dense set of Stieltjes

vectors and is bounded from below, then B is essentially self-adjoint.

Proof. The first two assertions follow Proposition 1.5, Corollary 3.4, and Corollary 4.5. Theorem 2.14 completes the proof. □

§3. The Stieltjes Moment Problem as a Self-Adjointness Problem

One goal in this section is to prove the remaining part of Theorem 2, namely, that if the operator A has a unique non-negative self-adjoint extension, then the Stieltjes problem is determinate. In the indeterminate case, we will also introduce two distinguished nonnegative von Neumann solutions: the Friedrichs and Krein solutions.

The name Friedrichs and Krein for the associated self-adjoint extensions is standard. We will naturally carry over the names to the measures dμ_{F} and dμ_{K} that solve the Stieltjes problem. Those names are not standard. Krein did important work on the moment problem, but not in the context of what we will call the Krein solution because of his work on self-adjoint extensions.

We begin with a very brief summary of the Birman-Krein-Vishik theory of extensions of strictly positive operators (see [3] for a complete exposition). Suppose A is a closed symmetric operator and for some α > 0, (φ, Aφ) ≥ α∥φ∥^{2} for all φ ∈ D(A). One closes the quadratic form analogously to closing the operator, that is, φ ∈ Q(A_{F}) if and only if there is φ_{n} ∈ D(A) so that φ_{n} → φ and (φ_{n} − φ_{m}, A(φ_{n} − φ_{m})) → 0 as n → ∞. One then sets (φ, A_{F} φ) = lim(φ_{n}, Aφ_{n}). The quadratic form (φ, A_{F} φ) on Q(A_{F}) is closed, that is, Q(A_{F}) is complete in the norm \sqrt{(φ, A_{F} φ)}.

Not every quadratic form abstractly defined is closable (e.g., on L^{2}(ℝ, dx), the form with form domain C_{0}^{∞}(ℝ) given by φ ↦ |φ(0)|^{2} is not closable), but it is not hard to see that if a quadratic form comes from a non-negative symmetric operator, it is closable (see Theorem X.23 in [34] or [12], Chapter VI).

While a closed symmetric operator need not be self-adjoint, it is a deep result (see [12, 34]) that a closed positive quadratic form is always the quadratic form of a selfadjoint operator. It follows that A_{F} is the quadratic form of a self-adjoint operator, the Friedrichs extension of A. It follows that inf spec(A_{F}) = α > 0.

Define N = ker(A^{∗}) which is closed since A^{∗} is a closed operator. Then D(A^{∗}) = D(A) + N + A_{F}^{−1}[N], where + means independent sum. In particular, N ≠ {0} if A is not self-adjoint (which we suppose in the rest of this paragraph). The Krein extension is the one with D(A_{K}) = D(A) + N. Then Q(A_{K}) = Q(A_{F}) + N. Since N ⊂ D(A_{K}), we have that inf spec(A_{K}) = 0 so A_{K} ≠ A_{F}. The set of non-negative self-adjoint extensions of A is precisely the set of self-adjoint operators B with A_{K} ≤ B ≤ A_{F} (where 0 ≤ C ≤ D if and only if Q(D) ⊂ Q(C) and (φ, Cφ) ≤ (φ, Dφ) for all φ ∈ Q(D); equivalently, if and only if (D + 1)^{−1} ≤ (C + 1)^{−1} as bounded operators). These B’s can be completely described in terms of closed but not necessarily densely defined quadratic forms, C, on N. One can think of C = ∞ on Q(C)^{⊥}. A_{F} corresponds to C ≡ ∞ (i.e., Q(C) = {0}) and A_{K} to C ≡ 0. In general, Q(A_{C}) = Q(A_{F}) + Q(C) and if φ ∈ Q(A_{F}) and η ∈ Q(C), then (φ + η, A_{C}(φ + η)) = (φ, A_{F} φ) + (η, Cη).

Two aspects of Jacobi matrices (defined on sequences of finite support) make the theory special. First, dim(N) ≤ 1. Second, by Theorem 5 (which we will prove in the next section without using any results from the present section), if A is not essentially self-adjoint, A_{F} has a discrete spectrum with the property that if t ∈ spec(A_{F}), then μ_{F} ({t}) > μ({t}) for any other solution, μ, of the Hamburger moment problem. Moreover, if t ∈ spec(A_{F}), t ∉ spec(B) for any other self-adjoint extension, B.

Proposition 3.1. Let A be a Jacobi matrix defined on the sequences of finite support, and suppose that A has deficiency indices (1, 1) and that A is non-negative (i.e., that the moment problem associated to A has at least some solutions of the Stieltjes moment problem). Then A has a unique non-negative self-adjoint extension if and only if 0 is an eigenvalue in the spectrum of A_{F}.

Proof. Let α = inf spec(A_{F}). By hypothesis, α ≥ 0, and by Theorem 5, α is a discrete, simple eigenvalue of A_{F}. We want to show that if α > 0, then A has additional nonnegative self-adjoint extensions; and contrary-wise, if α = 0, then it has no additional non-negative self-adjoint extensions.

If α > 0, A ≥ α > 0, then A has a Krein extension distinct from the A_{F}, so there are multiple non-negative self-adjoint extensions.

Suppose α = 0 and that B is some non-negative self-adjoint extension. Since A_{F} is the largest such extension, 0 ≤ B ≤ A_{F}, so

(A_{F} + 1)^{−1} ≤ (B + 1)^{−1} ≤ 1.  (3.1)

Since α = 0, ∥(A_{F} + 1)^{−1}∥ = 1, so (3.1) implies that ∥(B + 1)^{−1}∥ = 1. Since B has discrete spectrum by Theorem 5, this means 0 is an eigenvalue of B. But, by Theorem 5 again, A_{F} is the unique self-adjoint spectrum with a zero eigenvalue. Thus, B = A_{F}, that is, there is a unique non-negative self-adjoint extension. □

Theorem 3.2 (last quarter of Theorem 2). Suppose that A is a Jacobi matrix on the vectors of finite support and that A is non-negative. If A has a unique non-negative self-adjoint extension, then the associated Stieltjes moment problem is determinate.

Proof. Clearly, if A is essentially self-adjoint, then the Hamburger problem is determinate and so, a fortiori, the Stieltjes problem is determinate. Thus, we need only consider the case where A has multiple self-adjoint extensions, but only one that is non-negative.

Then A has deficiency indices (1, 1). By Proposition 3.1, α = inf spec(A_{F}) = 0. Moreover, by Theorem 5, 0 is a pure point of μ_{F}. Let φ = P_{{0}}(A_{F})δ_{0}. By Lemma 2.8, φ ≠ 0. Then there exists φ_{n} ∈ D(A) so φ_{n} → φ with ∥φ_{n}∥ = ∥φ∥ and (φ_{n}, A_{F} φ_{n}) → (A_{F} φ, φ) = 0. If μ_{F} (\{0\}) = τ, then ∥φ∥^{2} = ⟨φ, δ_{0}⟩ = τ so ∥φ_{n}∥ = \sqrt{τ} and ⟨φ_{n}, δ_{0}⟩ → τ.

Suppose ρ is some solution of the Stieltjes moment problem. Since φ_{n} → φ in L^{2}(dμ_{F}), φ_{n} is Cauchy in L^{2}(dρ), and so φ_{n} → f in L^{2}(dρ) and ∥f∥ = \sqrt{τ}. Since ∫ xφ_{n}(x)^{2} dρ → 0, we conclude that ∫ xf^{2} dρ = 0. Since ρ is a measure supported on [0, ∞), we conclude f(x) = 0 for x ≠ 0, and thus ρ has a pure point at zero also. Since (φ_{n}, δ_{0}) → τ, we see that ∫ f dρ = lim ∫ φ_{n}(x) dρ(x) = lim(φ_{n}, δ_{0}) = τ. But ∫ f^{2} dρ = ∥f∥^{2} = τ and ∫ f dρ = τ. Thus f(0) = 1 and so ρ({0}) = τ. But Theorem 5 asserts that any solution of the Hamburger problem distinct from μ_{F} has μ_{F} ({0}) < τ. We conclude that ρ = μ_{F}, that is, that the solution of the Stieltjes problem is unique. □

We will call the solution of the Stieltjes problem associated to the Friedrichs extension of A the Friedrichs solution. If the Stieltjes problem is indeterminate, then A ≥ α > 0 by Proposition 3.1 and Theorem 3.2. Moreover, since the Hamburger problem is indeterminate, N = ker(A) = span[π(0)] has dimension 1, so the Krein extension exists and is distinct from the Friedrichs extension. We will call its spectral measure the Krein solution of the Stieltjes problem.

As discussed in the introduction, every indeterminate Stieltjes problem has associated to it an example of a determinate Stieltjes problem where the associated Hamburger problem is indeterminate. For by Proposition 3.1 and Theorem 3.2, the indeterminate Stieltjes problem is associated to an A where the bottom of the spectrum of A_{F} is some f_{0} > 0. The moment problem \{γ_{m}(−f_{0})\} with γ_{m}(−f_{0}) = ∑_{j=0}^{m} \binom{m}{j}γ_{j}(−f_{0})^{m−j} will be determinate Stieltjes (since the associated A_{F} is the old A_{F} − f_{0}) but indeterminate Hamburger.

We summarize with

Theorem 3.3. Suppose that {γ_{n}}_{n=0}^{∞} is a set of Stieltjes moments. Then, if γ is Hamburger indeterminate, there is a unique c_{0} ≤ 0 so that γ(c_{0}) is Stieltjes determinate. Moreover, γ(c) is Stieltjes indeterminate if c > c_{0} and γ(c) are not Stieltjes moments if c < c_{0}. In particular, if γ(c) is Stieltjes determinate for two values of c, then γ is Hamburger determinate.

Corollary 3.4. If {γ_{n}}_{n=0}^{∞} is a set of Stieltjes moments and obeys

|γ_{n}| ≤ CR^{n}(2n)!  (3.2)

then {γ_{n}}_{n=0}^{∞} is Hamburger determinate.

Proof. By (3.2),

|γ_{n}(c)| ≤ CR^{n}(2n)! ∑_{j=0}^{n} \binom{n}{j}c^{j} = C[R(c + 1)]^{n}(2n)!

so by Proposition 1.5, {γ_{n}}_{n=0}^{∞} is Stieltjes determinate for all c ≥ 0. By Theorem 3.3, it is Hamburger determinate.

§4. Transfer Matrices and the Indeterminate Moment Problem

In this section, we will analyze the indeterminate Hamburger problem in great detail, using the second-order difference equation associated to the Jacobi matrix A. In particular, we will prove Theorems 3, 4, and 5.

Throughout, we will fix a set of Hamburger moments {γ_{m}}_{m=0}^{∞}. It will be convenient to define a linear functional E( ·) on polynomials P (X) by E(X^{m}) ≡ γ_{m}. Where we have polynomials P of several variables X, Y, …, we will indicate by E_{X}( ·) the obvious map to polynomials of Y, … with

E_{X}(X^{m}Y^{n_{1}} …) = γ_{m}Y^{n_{1}} ….

In (1.14), we defined the set of polynomials P_{n}(X) by the three conditions, deg(P_{n}(X)) = n, E(P_{n}(X)P_{m}(X)) = δ_{nm}, and P_{n}(X) = c_{nn}X^{n} + ⋯ with c_{nn} > 0. We noted that with {a_{n}}_{n=0}^{∞}, {b_{n}}_{n=0}^{∞}, and a_{−1} ≡ 1, the elements of the Jacobi matrix associated to γ, the sequence u_{n} = P_{n}(z); n = 0, 1, 2, … obeys

a_{n}u_{n+1} + (b_{n} − z)u_{n} + a_{n−1}u_{n−1} = 0  (4.1)

with

u_{−1} = 0,  u_{0} = 1.  (4.2)

We also introduced the sequence of polynomials Q_{n}(X) of degree n − 1 by requiring that u_{n} = Q_{n}(z) solve (4.1) with the initial conditions

u_{−1} = −1,  u_{0} = 0.  (4.3)

Notice that the Wronskian W (Q_{·}(z), P_{·}(z))(−1) = a_{−1}[Q_{0}(z)P_{−1}(z)−Q_{−1}(z)P_{0}(z)] = 1. Thus, by (2.11) we have:

Proposition 4.1.

a_{k−1}[Q_{k}(z)P_{k−1}(z) − Q_{k−1}(z)P_{k}(z)] ≡ 1

The following is a useful formula for the Q’s in terms of the P ’s. Note that since \frac{X^{n}−Y^{n}}{(X−Y)} = ∑_{j=0}^{n−1} X^{j}Y^{n−1−j} for any polynomial P, \frac{P (X)−P (Y)}{X−Y} is a polynomial in X and Y.

Theorem 4.2. For n ≥ 0,

E_{X}(\frac{P_{n}(X) − P_{n}(Y)}{X − Y}) = Q_{n}(Y).

Proof. \frac{[P_{n}(X)−P_{n}(Y)]}{(X−Y)} is a polynomial in X and Y of degree n − 1, so we can define polynomials

R_{n}(Y) ≡ E_{X}(\frac{P_{n}(X) − P_{n}(Y)}{X − Y}).

Subtract the equations,

a_{n}P_{n+1}(X) + b_{n}P_{n}(X) + a_{n−1}P_{n−1}(X) = XP_{n}(X)
a_{n}P_{n+1}(Y) + b_{n}P_{n}(Y) + a_{n−1}P_{n−1}(Y) = Y P_{n}(Y)

and see that for n = 0, 1, 2, … with R_{−1}(Y) ≡ 0:

a_{n}R_{n+1}(Y) + b_{n}R_{n}(Y) + a_{n−1}R_{n−1}(Y) = E_{X}(\frac{XP_{n}(X) − Y P_{n}(Y)}{X − Y})

= E_{X}(\frac{Y P_{n}(X) − Y P_{n}(Y)}{X − Y} + P_{n}(X))

= Y R_{n}(Y) + δ_{n0}.

Replace the abstract variable Y with a complex variable z and let R̃_{n}(z) = R_{n}(z) if n ≥ 0 but R̃_{−1}(z) = −1. Then

a_{n}R̃_{n+1}(z) + b_{n}R̃_{n}(z) + a_{n−1}R̃_{n−1}(z) = zR̃_{n}(z)

(since the δ_{n0} term becomes −a_{n−1}R̃_{n−1}(z) for n = 0). Moreover,

R̃_{−1}(z) = −1,  R̃_{0}(z) = E_{X}(\frac{1 − 1}{X − Y} ) = 0

so by uniqueness of solutions of (4.1) with prescribed initial conditions, R̃_{n}(z) = Q_{n}(z), that is, for n ≥ 0, R_{n} = Q_{n}, as claimed. □

One consequence of Theorem 4.2 is that P_{n}(X) and Q_{n}(X) have the same leading coefficient, that is, if P_{n}(X) = c_{nn}X^{n} + ⋯, then Q_{n}(X) = c_{nn}X^{n−1} + · · ·.

The following calculation is part of the standard theory [1] and plays a critical role in what follows:

Theorem 4.3. Let ρ solve the moment problem and assume z ∈ ℂ\ℝ. Set ζ = G_{ρ}(z) ≡ ∫ \frac{dρ(x)}{x−z}. Then

⟨(x − z)^{−1}, P_{n}(x)⟩_{L^{2}(dρ)} = Q_{n}(z) + ζP_{n}(z).  (4.4)

In particular,

∑_{n=0}^{∞} |Q_{n}(z) + ζP_{n}(z)|^{2} ≤ \frac{Im ζ}{Im z}  (4.5)

with equality if ρ is a von Neumann solution.

Remark. We will eventually see (Proposition 4.15 and its proof) that the “if” in the last sentence can be replaced by “if and only if.”

Proof.

∫ \frac{P_{n}(x)}{x − z} dρ(x) = ζP_{n}(z) + ∫ \frac{P_{n}(x) − P_{n}(z)}{x − z} dρ(x) = ζP_{n}(z) + Q_{n}(z)

by Theorem 4.2. This is just (4.4). (4.5) is just Parseval’s inequality for the orthonormal set {P_{n}(x)}_{n=0}^{∞} in L^{2}(dρ) if we note that

∫ \frac{dρ(x)}{|x − z|^{2}} = \frac{1}{z − z̄} ∫ dρ(x)[\frac{1}{x − z} − \frac{1}{x̄ − z̄}] = \frac{Im ζ}{Im z} .

If ρ comes from a self-adjoint extension, by construction of H^{(γ)}, {P_{n}(x)}_{n=0}^{∞} is an orthonormal basis for L^{2}(dρ), so equality holds in (4.5). □

Remark. In H^{(γ)}, thought of as limits of polynomials, the vector with components Q_{n}(z) + ζP_{n}(z) is represented by

∑_{n}[Q_{n}(z) + ζP_{n}(z)]P_{n}(X).

This product of P ’s may seem strange until one realizes that ∑_{n=1}^{N} P_{n}(y)P_{n}(x) is a reproducing kernel in L^{2}(dρ) for polynomials of degree N. This links the construction to ideas in Landau [19].

Given Theorem 2, the following proves the equivalence of parts (i), (ii), (iii), and (vii) for z ∈ ℂ\ℝ of Theorem 3.

Proposition 4.4. Suppose that z_{0} ∈ ℂ\ℝ. Then the following are equivalent: (i) The Jacobi matrix A is not essentially self-adjoint. (ii) π(z_{0}) = {P_{n}(z_{0})}_{n=0}^{∞} is in ℓ^{2}. (iii) ξ(z_{0}) = {Q_{n}(z_{0})}_{n=0}^{∞} is in ℓ^{2}. (iv) Both π(z_{0}) and ξ(z_{0}) are in ℓ^{2}. Moreover, when any of these conditions holds, there is a closed disk of positive radius,

D(z_{0}) , in the same half-plane as z_{0} so that for any solution ρ of the moment problem, ζ = G_{ρ}(z_{0}) ∈ D(z_{0}). The values of ζ when ρ is a von Neumann solution lie on ∂D(z_{0}) and fill this entire boundary. Every point in D(z_{0}) is the value of G_{ρ}(z_{0}) for some ρ solving the moment problem.

In addition, if these conditions hold, π(z_{0}) and ξ(z_{0}) lie in D(A^{∗})\D(A).

Remarks. 1. One can show (using (2.11)) that the center of the disk is

lim_{n→∞} −\frac{Q_{n}(z_{0}) \overline{P_{n−1}(z_{0})} − Q_{n−1}(z_{0}) \overline{P_{n}(z_{0})}}{P_{n}(z_{0})\overline{P_{n−1}(z_{0})} − P_{n−1}(z_{0}) \overline{P_{n}(z_{0})}}

and the radius is

\frac{1}{2|Im z_{0}|} \frac{1}{∥π(z_{0})∥^{2}}  (4.6)

but these explicit formulae will play no role in our discussions.

2. We will see later (Theorem 4.14) that if ρ is not a von Neumann solution, then ζ ∈ D(z_{0})^{int}.

Proof. Since A is real, A fails to be essentially self-adjoint if and only if there is a non-zero solution of (A^{∗} − z_{0})u = 0. By Theorem 2.6 and the unique solubility of second-order difference operators given u_{−1} = 0 and u_{0}, every such solution has u_{n} = u_{0}P_{n}(z_{0}) so (i) is equivalent to (ii). Let ζ = G_{ρ}(z_{0}) for some von Neumann solution. Then ξ(z_{0}) + ζπ(z_{0}) ∈ ℓ^{2}, so (ii) is equivalent to (iii) or (iv).

If (i)–(iv) hold, then (4.5) has the form

a|ζ|^{2} + bζ + b̄ζ̄ + c ≤ 0,

where a = ∥π(z_{0})∥^{2}, c = ∥ξ(z_{0})∥^{2}, and b = 2⟨ξ(z_{0}), π(z_{0})⟩ − \frac{i}{2 Im z_{0}} . The set where this inequality holds is always a disk, D(z_{0}), although a priori, it could be empty depending on the values of a, b, c. However, by Theorem 2.11, we know that there are multiple ζ’s obeying (4.5) so the disk must have strictly positive radius. By Theorem 4.3, ζ’s for von Neumann solutions obey equality in (4.5) and so lie in ∂D(z_{0}).

The self-adjoint extensions are parametrized by a circle (unitaries from K_{+} to K_{−}) in such a way that the map from the parametrization to values of (δ_{0}, (B_{t} − z_{0})^{−1}δ_{0}) is continuous. By Theorem 2.11, this map is one-one. Every one-one continuous map of the circle to itself is surjective, so all of ∂D(z_{0}) occurs.

Given a point ζ in D(z_{0})^{int}, find ζ_{0} and ζ_{1} in ∂D(z_{0}) and θ ∈ (0, 1) so ζ = θζ_{0}+(1−θ)ζ_{1}. If μ_{0}, μ_{1} are the von Neumann solutions that have G_{μ_{i}}(z_{0}) = ζ_{i} (i = 0, 1), then ρ = θμ_{0} + (1 − θ)μ_{1} is a measure solving the moment problem with G_{ρ}(z_{0}) = ζ.

If (iv) holds, then π(z̄_{0}), ξ(z_{0}) ∈ D(A^{∗}). By (2.14) and Proposition 1.4, ⟨A^{∗}π(z̄_{0}), ξ(z_{0})⟩ − ⟨π(z̄_{0}), A^{∗}ξ(z_{0})⟩ = −1, so neither π(z̄_{0}) nor ξ(z_{0}) lies in D(A). □

Remark. The Cayley transform way of looking at self-adjoint extensions says that for any z_{0} ∈ ℂ\ℝ,

(B_{t} − z̄_{0})(B_{t} − z_{0})^{−1} = V (z_{0}) + e^{iθ(z_{0},t)}X(z_{0}),

where V (z_{0}) is the t-independent map (Ā−z̄_{0})(Ā−z_{0})^{−1} from \overline{Ran(A − z_{0})} to \overline{Ran(A − z̄_{0})} and X(z_{0}) is any isometry from ker(A^{∗} − z̄_{0}) to ker(A^{∗} − z_{0}) extended to H as a partial isometry. θ is a parametrization depending on z_{0}, t (and the choice of X(z_{0})), but the theory guarantees that as t varies through all possible values, e^{iθ} runs through the full circle in an injective manner. Now (B_{t} − z̄_{0})(B_{t} − z_{0})^{−1} = 1 + 2i(Im z_{0})(B_{t} − z_{0})^{−1}, so

(δ_{0}, (B_{t} − z_{0})^{−1}δ_{0}) = [2i(Im z_{0})]^{−1}[−1 + (δ_{0}, V (z_{0})δ_{0}) + e^{iθ(z_{0},t)}(δ_{0}, X(z_{0})δ_{0})]

is seen directly to be a circle. Since one can take X(z_{0}) to be

(δ_{n}, X(z_{0})δ_{m}) = \frac{P_{m}(z̄_{0})P_{n}(z_{0})}{∑_{j=0}^{∞} |P_{j}(z_{0})|^{2}}

and P_{0}(z) ≡ 1, we even see that the radius of the circle is given by (4.6).

Corollary 4.5. If a Hamburger problem is indeterminate, then ∑_{n=0}^{∞} |a_{n}|^{−1} < ∞. In particular, if ∑_{n=0}^{∞} |a_{n}|^{−1} = ∞, (e.g., a_{n} ≡ 1), then the Hamburger problem is determinate. If

∑_{n=1}^{∞} γ_{2n}^{−1/2n} = ∞,

then the Hamburger problem is determinate. If

∑_{n=1}^{∞} γ_{n}^{−1/2n} = ∞

for a set of Stieltjes moments, that problem is both Stieltjes and Hamburger determinate.

Remarks. 1. The last pair of assertions is called Carleman’s criterion. It generalizes Proposition 1.5 and proves uniqueness in some cases where growth doesn’t imply analyticity of the Fourier transform.

2. See Corollary 5.24 for the Stieltjes analog of the first assertion.

Proof. By Proposition 4.1 and the Schwarz inequality,

∑_{n=0}^{N} |a_{n}|^{−1} ≤ 2( ∑_{n=0}^{N+1} |P_{n}(z)|^{2})^{1/2}( ∑_{n=0}^{N+1} |Q_{n}(z)|^{2})^{1/2}.

Thus, divergence of ∑_{n=0}^{∞} |a_{n}|^{−1} implies that either π or ξ (or both) fail to be ℓ_{2}, and so determinacy.

Consider Carleman’s criterion in the Hamburger case. By induction in n and (4.1) starting with P_{0}(x) = 1, we see that

P_{n}(x) = (a_{1} … a_{n})^{−1}x^{n} + lower order,

so ⟨(a_{1} … a_{n})^{−1}x^{n}, P_{n}(x)⟩ = 1 and thus, by the Schwartz inequality,

1 ≤ γ_{2n}(a_{1} … a_{n})^{−2}

hence

γ_{2n}^{−1/2n} ≤ (a_{1} … a_{n})^{−1/n}.

Therefore our result follows from the divergence criteria proven at the start of the proof and the inequality

∑_{j=1}^{n}(a_{1} … a_{j})^{−1/j} ≤ 2e ∑_{j=1}^{n} a_{j}^{−1}

(which is actually known to hold with e in place of 2e).

To prove this, note first that 1+x ≤ e^{x} so (1+n^{−1})^{n} ≤ e so using induction, n^{n} ≤ e^{n}n!, so by the geometric-arithmetic mean inequality,

(a_{1} … a_{j})^{−1/j} = (a_{1}^{−1}2a_{2}^{−1} … ja_{j}^{−1})^{1/j}(j!)^{−1/j}

≤ \frac{e}{j^{2}} ∑_{k=1}^{j} ka_{k}^{−1}.

Thus,

∑_{j=1}^{n}(a_{1} … a_{j})^{−1/j} ≤ e ∑_{k=1}^{n} a_{k}^{−1}(∑_{j=k}^{n} \frac{k}{j^{2}} ) ≤ 2e ∑_{j=1}^{n} a_{j}^{−1}

since

∑_{j=k}^{∞} \frac{k}{j^{2}} ≤ \frac{1}{k} + k ∫_{k}^{∞} \frac{dy}{y^{2}} ≤ 2.

By Proposition 1.6, Carleman’s criterion in the Hamburger case means that if a set of Stieltjes moments obeys ∑_{n=1}^{∞}(γ_{n})^{−1/2n} = ∞, then it is Stieltjes determinate. Thus by Theorem 3.3, it suffices to show that

∑_{n=1}^{∞}(γ_{n}(c))^{−1/2n} ≥ D(c) ∑_{n=1}^{∞} γ_{n}^{−1/2n}

for c > 0 and D(c) < ∞ to conclude Hamburger determinacy.

Note first that \frac{γ_{j+1}}{γ_{j}} ≥ \frac{γ_{j}}{γ_{j−1}} by the Schwarz inequality. Thus, if α = \frac{γ_{0}}{γ_{1}} , we have that

γ_{n−j} ≤ γ_{n}α^{j}.

Therefore,

γ_{n}(c) ≤ γ_{n} ∑_{j=0}^{n} \binom{n}{j}c^{j}α^{j} = (1 + cα)^{n}γ_{n}

and so

∑_{n=1}^{∞}(γ_{n}(c))^{−1/2n} ≥ (1 + cα)^{−1/2} ∑_{n=1}^{∞} γ_{n}^{−1/2n}. □

To complete the proof of Theorem 3, we need to show that if every solution of [(A^{∗} − z_{0})u]_{n} = 0 (n ≥ 1) is ℓ^{2} for a fixed z = z_{0}, the same is true for all z ∈ ℂ and that \frac{∂π}{∂x}, \frac{∂ξ}{∂x} are in ℓ^{2}. We will do this by the standard method of variation of parameters. So we write a general solution of [(A^{∗} − z)u]_{n} = 0 (n ≥ 1) in terms of P_{n}(z_{0}) and Q_{n}(z_{0}) using

\binom{u_{n+1}}{u_{n}} ≡ α_{n}\binom{P_{n+1}(z_{0})}{P_{n}(z_{0})} + β_{n}\binom{Q_{n+1}(z_{0})}{Q_{n}(z_{0})}.  (4.7)

Since W (P_{·}(z_{0}), Q_{·}(z_{0})) = −1, the two vectors on the right of (4.7) are linearly independent, and so they span ℂ^{2} and (α_{n}, β_{n}) exist and are unique. Indeed, by the Wronskian relation:

α_{n} = W (Q_{·}(z_{0}), u)(n)  (4.8a)
β_{n} = −W (P_{·}(z_{0}), u)(n).  (4.8b)

A straightforward calculation using the fact that P, Q obey (4.1) with z = z_{0}, (2.11), and (4.8) shows that (4.1) at n is equivalent to

\binom{α_{n}}{β_{n}} = [1 + (z − z_{0})S(n, z_{0})]\binom{α_{n−1}}{β_{n−1}},  (4.9)

where

S(n, z_{0}) = ( \begin{matrix}−Q_{n}(z_{0})P_{n}(z_{0}) & −Q_{n}(z_{0})Q_{n}(z_{0}) \\ P_{n}(z_{0})P_{n}(z_{0}) & P_{n}(z_{0})Q_{n}(z_{0}) \end{matrix} ) .  (4.10)

For example, using (2.11) and (4.7) for n → n − 1:

α_{n} = W (Q_{·}(z_{0}), u)(n)
= W (Q_{·}(z_{0}), u)(n − 1) + (z_{0} − z)Q_{n}(z_{0})u_{n}
= α_{n−1} − (z − z_{0})Q_{n}(z_{0}){α_{n−1}P_{n}(z_{0}) + β_{n−1}Q_{n}(z_{0})}.

Notice that

S^{2} = det(S) = Tr(S) = 0.  (4.11)

The following is obvious and explains why ℓ^{2} solutions are so natural:

If π(z_{0}) = {P_{n}(z_{0})}_{n=0}^{∞} and ξ(z_{0}) = {Q_{n}(z_{0})}_{n=0}^{∞} are both in ℓ^{2}, then

∑_{n=0}^{∞} ∥S(n, z_{0})∥ < ∞.  (4.12)

Lemma 4.6. Let A_{n} be a sequence of matrices with ∑_{n=0}^{∞} ∥A_{n}∥ < ∞. Let D_{N}(z) = (1 + zA_{N})(1 + zA_{N−1}) … (1 + zA_{0}). Then D_{∞}(z) = lim D_{N}(z) exists for each z ∈ ℂ and defines an entire function of z obeying

∥D_{∞}(z)∥ ≤ c_{ε} exp(ε|z|)

for each ε > 0.

Proof. Notice first that

∥(1 + B_{N}) … (1 + B_{0})∥ ≤ ∏_{j=0}^{N}(1 + ∥B_{j}∥) ≤ exp(∑_{j=0}^{N} ∥B_{j}∥)  (4.13)

and that

∥(1 + B_{N}) … (1 + B_{0}) − 1∥ ≤ ∏_{j=0}^{N}(1 + ∥B_{j}∥) − 1 ≤ exp(∑_{j=0}^{N} ∥B_{j}∥) − 1.  (4.14)

From (4.14), we see that

∥D_{N+j}(z) − D_{N}(z)∥ ≤ [exp(|z| ∑_{N+1}^{N+j} ∥A_{j}∥) − 1]∥D_{N}(z)∥

≤ [exp(|z| ∑_{N+1}^{∞} ∥A_{j}∥) − 1] exp(|z| ∑_{j=0}^{N} ∥A_{j}∥),

from which it follows that D_{N}(z) is Cauchy uniformly for z in balls of ℂ. Thus, D_{∞} exists and is entire in z. By (4.13),

∥D_{∞}(z)∥ ≤ ∏_{j=0}^{N}(1 + |z|∥A_{j}∥) exp(|z| ∑_{j=N+1}^{∞} ∥A_{j}∥),

so given ε, choose N so ∑_{j=N+1}^{∞} ∥A_{j}∥ ≤ \frac{ε}{2} and use the fact that the polymonial ∏_{j=0}^{N}(1+ |z|∥A_{j}∥) can be bounded by c_{ε} exp(\frac{1}{2}ε|z|). □

Given Proposition 4.4 and Theorem 2.14, the following completes the proof of Theorem 3:

Theorem 4.7. Let A be a Jacobi matrix and consider solutions of (4.1) for z ∈ ℂ. If for some z_{0} ∈ ℂ, all solutions are in ℓ^{2}, then that is true for each z ∈ ℂ. Moreover, π(z) and ξ(z) are analytic ℓ_{2}-valued functions so, in particular, \frac{∂π}{∂z} , \frac{∂ξ}{∂z} ∈ ℓ_{2} for all z. The disk D(z_{0}) varies continuously as z runs through ℂ_{+}.

Proof. Define

T (n, −1; z, z_{0}) = (1 + (z − z_{0})S(n, z_{0})) … (1 + (z − z_{0})S(0, z_{0})).

By (4.12) and Lemma 4.6,

sup_{n} ∥T (n, −1,; z, z_{0})∥ ≤ exp(|z − z_{0}| ∑_{j=0}^{∞} ∥S(j, z)∥) < ∞.

Thus for any initial \binom{α_{−1}}{β_{−1}}, if u has the form (4.7), then sup_{n} ∥\binom{α_{n}}{β_{n}}∥ ≤ sup_{n} ∥T (n, −1; z, z_{0})∥∥\binom{α_{−1}}{β_{−1}}∥ ≡ C. Thus, u_{n} = α_{n}P_{n}(z_{0}) + β_{n}Q_{n}(z_{0}) has |u_{n}|^{2} ≤ C^{2}[P_{n}(z_{0})^{2}+Q_{n}(z_{0})^{2}] which is in ℓ_{1}. Since π(z) is associated to T (n, −1; z, z_{0})\binom{1}{0} and ξ(z) is associated to T (n, −1; z, z_{0})\binom{0}{1}, the claimed analyticity holds because sup_{n} ∥T (n, −1; z, z_{0})∥ is bounded as z varies in bounded sets.

Continuity of D(z_{0}) follows from the formula for ∂D(z_{0}), viz. w ∈ ∂D(z_{0}) if and only if

∥π(z_{0})∥^{2}|w|^{2} + 2 Re[(2⟨ξ(z_{0}), π(z_{0})⟩ − \frac{i}{2 Im z_{0}} )w] + ∥ξ(z_{0})∥^{2} = 0. □

By Theorem 3 and Lemma 4.6, if the Hamburger problem is indeterminate,

T (∞, −1; z, z_{0}) ≡ lim_{n→∞} T (n, −1; z, z_{0})  (4.15)

exists. We define four functions A(z), B(z), C(z), D(z) by

T (∞, −1; z, z_{0} = 0) = ( \begin{matrix}−B(z) & −A(z) \\ D(z) & C(z) \end{matrix} ) .  (4.16)

The Nevanlinna matrix is defined by

N(z) = ( \begin{matrix}A(z) & C(z) \\ B(z) & D(z) \end{matrix} ) .  (4.17)

Theorem 4.8.

(i) Each of the functions A, B, C, D is an entire function obeying

|f(z)| ≤ c_{ε} exp(ε|z|)

for each ε > 0. (ii) For z small,

B(z) = −1 + O(z)  (4.18a)

and

D(z) = αz + O(z^{2})  (4.18b)

with α > 0. (iii) AD − BC ≡ 1.

Proof. (i) follows from Lemma 4.6. (ii) follows if we note that

T (∞, −1; z, z_{0} = 0) = 1 + z ∑_{n=0}^{∞} S(n, z_{0} = 0) + O(z^{2}).

(4.18a) is immediate and (4.18b) follows since (4.10) says that α = ∑_{n=0}^{∞} P_{n}(0)^{2} > 0. (iii) holds since by (4.11), det T = 1. □

For our purposes, the formula for A, B, C, D as matrix elements of an infinite product is sufficient, but there is an infinite sum formula connected to standard equations for perturbed solutions (see, e.g., [32, 11]) that lets us make contact with the more usual definitions [1].

38 B. SIMON

Theorem 4.9. If the Hamburger problem is indeterminate, then,

A(z) = z ∑_{n=0}^{∞} Q_{n}(0)Q_{n}(z)
B(z) = −1 + z ∑_{n=0}^{∞} Q_{n}(0)P_{n}(z)
C(z) = 1 + z ∑_{n=0}^{∞} P_{n}(0)Q_{n}(z)
D(z) = z ∑_{n=0}^{∞} P_{n}(0)P_{n}(z).

Proof.

T (n, −1; z, z_{0} = 0) = (1 + zS(n, z_{0} = 0)) … (1 + zS(0, z_{0} = 0))

= 1 + ∑_{j=0}^{n} zS(j, z_{0} = 0) ∏_{k=0}^{j−1}(1 + zS(k, z_{0} = 0))

so it suffices to show that

S(j, z_{0} = 0)(1 + zS(j − 1, z_{0} = 0)) … (1 + zS(0, z_{0} = 0))
= ( \begin{matrix}−Q_{j}(0)P_{j}(z) & −Q_{j}(0)Q_{j}(z) \\ P_{j}(0)P_{j}(z) & P_{j}(0)Q_{j}(z) \end{matrix} ).  (4.19)

By definition,

(1 + zS(j − 1, z_{0} = 0)) … (1 + zS(0, z_{0} = 0)) = ( \begin{matrix}α_{j−1}^{(1)} & α_{j−1}^{(2)} \\ β_{j−1}^{(1)} & β_{j−1}^{(2)} \end{matrix} ) ,

where

α_{j−1}^{(1)}\binom{P_{j}(0)}{P_{j−1}(0)} + β_{j−1}^{(1)}\binom{Q_{j}(0)}{Q_{j−1}(0)} = \binom{P_{j}(z)}{P_{j−1}(z)}

and

α_{j−1}^{(2)}\binom{P_{j}(0)}{P_{j−1}(0)} + β_{j−1}^{(2)}\binom{Q_{j}(0)}{Q_{j−1}(0)} = \binom{Q_{j}(z)}{Q_{j−1}(z)}.

Thus by (4.10),

LHS of (4.19) = ( \begin{matrix}−Q_{j}(0)[α_{j−1}^{(1)}P_{j}(0) + β_{j−1}^{(1)}Q_{j}(0)] & −Q_{j}(0)[α_{j−1}^{(2)}P_{j}(0) + β_{j−1}^{(2)}Q_{j}(0)] \\ P_{j}(0)[α_{j−1}^{(1)}P_{j}(0) + β_{j−1}^{(1)}Q_{j}(0)] & P_{j}(0)[α_{j−1}^{(2)}P_{j}(0) + β_{j−1}^{(2)}Q_{j}(0)] \end{matrix} )
= RHS of (4.19). □

π(0) = {P_{n}(0)} and ξ(0) = {Q_{n}(0)} span D(A^{∗})/D(A) by Proposition 4.4. Notice that A^{∗}π(0) = 0 while A^{∗}ξ(0) = δ_{0}, so ⟨π(0), A^{∗}ξ(0)⟩ = 1 and (2.7) holds. Thus, we can parametrize the self-adjoint extensions B_{t} of A by

D(B_{t}) = D(A) + {α(tπ(0) + ξ(0)) | α ∈ ℂ}  (4.20a)

if t < ∞ and

D(B_{∞}) = D(A) + {απ(0)}.  (4.20b)

This is equivalent to defining t by

t = (δ_{0}, B_{t}^{−1}δ_{0}).  (4.20c)

Theorem 4.10. For each t ∈ ℝ ∪ {∞} and z ∈ ℂ\ℝ,

(δ_{0}, (B_{t} − z)^{−1}δ_{0}) = −\frac{C(z)t + A(z)}{D(z)t + B(z)} .  (4.21)

Proof. Let us sketch the proof before giving details:

T (−1, ∞; z, z_{0} = 0) ≡ T (∞, −1; z, z_{0} = 0)^{−1}.  (4.22)

Then

\binom{α_{−1}}{β_{−1}} ≡ T (−1, ∞; z, z_{0} = 0)\binom{t}{1}  (4.23)

is such that the u_{n} associated to \binom{α_{−1}}{β_{−1}} obeys (4.1) and is asymptotically tπ(0)+ξ(0) and

so in D(B_{t}). (B_{t} − z)^{−1}δ_{0} will thus be −\frac{u_{t}}{u_{t}(−1)} and (δ_{0}, (B_{t} − z)^{−1}δ_{0}) will be −\frac{u_{t}(0)}{u_{t}(−1)}. But u_{t}(0) = α_{−1} and u_{t}(−1) = −β_{−1} so (δ_{0}, B_{t} − z)^{−1}δ_{0}) will be \frac{α_{−1}}{β_{−1}} , which is given by (4.21).

Here are the details. If φ ∈ D(A^{∗}), then by (2.14), φ ∈ D(B_{t}) if and only if

lim_{n→∞} W (φ, tπ(0) + ξ(0))(n) = 0.  (4.24)

Suppose u solves (4.1) for a given z. Then u ∈ D(A^{∗}) and u has the form (4.7), where by (4.15), lim_{n→∞} \binom{α_{n}}{β_{n}} = \binom{α_{∞}}{β_{∞}} exists. Clearly,

W (u, tπ(0) + ξ(0))(n) = −α_{n} + β_{n}t

so (4.24) holds if and only if

α_{∞} = tβ_{∞}.

Thus, if T (−1, ∞; z, z_{0} = 0) is given by (4.22), then \binom{α_{−1}}{β_{−1}} given by (4.23) is initial data for a solution u of (4.1) that has u ∈ D(B_{t}). A solution u of (4.1) has

(A^{∗} − z)u = −u_{−1}δ_{0}

and thus, if u is associated to the data in (4.23),

(B_{t} − z)u = −u_{−1}δ_{0}

and

(δ_{0}, (B_{t} − z)^{−1}δ_{0}) = −\frac{u_{0}}{u_{−1}} .

But

\binom{u_{0}}{u_{−1}} = α_{−1}\binom{P_{0}(0)}{P_{−1}(0)} + β_{−1}\binom{Q_{0}(0)}{Q_{−1}(0)} = \binom{α_{−1}}{−β_{−1}}

so

(δ_{0}, (B_{t} − z)^{−1}δ_{0}) = \frac{α_{−1}}{β_{−1}} .

Since T (∞, −1; z, z_{0} = 0) has the form (4.16) and has determinant 1, its inverse has the form

( \begin{matrix}C(z) & A(z) \\ −D(z) & −B(z) \end{matrix} )

and so α_{−1} = C(z)t + A(z), β_{−1} = −D(z)t − B(z). Thus (4.21) is proven. □

Remark. Our convention for the parameter t differs from that in Akhiezer [1], which is the traditional one! If s = −t^{−1}, then

−\frac{Ct + A}{Dt + B} = −\frac{As − C}{Bs − D} .

The parameter s is what he calls t. Our Φ(z) later and his Φ(z), which I will call Ψ(z), are related by Φ(z) = −Ψ(z)^{−1}. Since this transformation takes Herglotz functions to themselves, we both deal with Herglotz functions. See [9] for an interesting alternate reparametrization.

We turn next to half of Theorem 5. We will prove that each B_{t} has point spectrum with any two disjoint. The condition that μ_{t}({t}) > ρ({t}) for any other solution of the moment problem will wait until after we have proven Theorem 4.

Theorem 4.11 (half of Theorem 5). Suppose the Hamburger problem is indeterminate. Each B_{t} has pure point spectrum only. The eigenvalues of the different B_{t}’s are distinct and every x ∈ ℝ is an eigenvalue of some B_{t}. If x is an eigenvalue of B_{t}, then

μ_{t}(\{x\}) = \frac{1}{∑_{n=0}^{∞} |P_{n}(x)|^{2}} .  (4.25)

Moreover, if λ_{n}(B_{t}) are the eigenvalues of B_{t} (ordered in some convenient way), then for any p > 1,

∑_{n} |λ_{n}(B_{t})|^{−p} < ∞.  (4.26)

Proof. Let μ_{t} be the spectral measure for B_{t}. By Theorem 4.10,

∫ \frac{dμ_{t}(x)}{x − z} = −\frac{C(z)t + A(z)}{D(z)t + B(z)}

is a meromorphic function of z, since A, B, C, D are entire. That implies that μ_{t} is pure point only and these points are given by solutions of

D(z)t + B(z) = 0.

Since A, B, C, D are real when z = x is real, for any x, x ∈ spec(B_{t}) for t = −\frac{B(x)}{D(x)}. Since AD − BC ≡ 1, B and D have no simultaneous zeros, so the eigenvalues are distinct for different t’s. (4.26) is a standard result on the zeros of an entire function like D(z)t+B(z), which obeys (i) of Theorem 4.8.

To prove (4.25), note that if x is an eigenvalue of B_{t}, then the corresponding normalized eigenvector φ obeys A^{∗}φ = xφ. It follows that φ_{n} = φ_{0}P_{n}(x) and then by normalization, that φ_{0}^{2} = 1/ ∑_{n=0}^{∞} |P_{n}(x)|^{2}. Thus, μ_{t}({x}) = φ_{0}^{2} is given by (4.25). □

Now define the map F (z) : ℂ ∪ {∞} → ℂ ∪ {∞} by

F (z)(w) = −\frac{C(z)w + A(z)}{D(z)w + B(z)} .  (4.27)

F (z) as a fractional linear transformation is one-one and onto, and takes ℝ ∪ {∞} onto a circle or straight line. By Proposition 4.4 and (4.21), F (z)[ℝ ∪ {∞}] is precisely the circle ∂D(z) when Im z ≠ 0.

Proposition 4.12. If Im z > 0, F (z) maps the upper half-plane ℂ_{+} = {z ∈ ℂ | Im z > 0} onto the interior of the disk D(z)^{int}.

Proof. For each z ∈ ℂ\ℝ, F (z) must map ℂ_{+} onto either D(z)^{int} or ℂ\D(z) since ℝ is mapped to ∂D(z) and a fractional linear transformation maps ℂ∪{∞} to itself bijectively. Thus it suffices to show F (z)[i] ∈ D(z)^{int} for all z with Im z > 0. But F (z)[i] moves analytically, and so continuously. It can only move from D(z)^{int} to ℂ\D(z) by lying on ∂D(z), which is impossible if Im z > 0 since then ∂D(z) = F (z)[ℝ ∪ {∞}]. Thus it suffices to show F (z)[i] ∈ D(z)^{int} for z = ix with x > 0 and small.

Now, F (z)[i] ∈ D(z)^{int} if and only if F (z) takes the lower half-plane to ℂ∪{∞}\D(z), which holds if and only if F (z)^{−1}[∞] is in the lower half-plane. Thus it suffices to show for x small and positive that F (ix)^{−1}[∞] ∈ ℂ_{−}.

F (ix)^{−1}[∞] is that w with D(ix)w + B(ix) = 0, that is, w = −\frac{B(ix)}{D(ix)}. But by (4.18), B(ix) = −1 + O(x) and D(ix) = iαx + O(x^{2}) with α > 0, so w = −iα^{−1}x^{−1} + O(1) lies in ℂ_{−}, which suffices to prove the proposition. □

We now follow the classical argument of Nevanlinna [29] to obtain Theorem 4. We need to relate solutions of (1.1) to asymptotics of the Stieltjes transform of ρ following Hamburger [8] and Nevanlinna [29].

Proposition 4.13. Let M^{H}(γ) = \{ρ | ρ obeys (1.1)\}. Let G_{ρ}(z) = ∫ \frac{dρ(x)}{x−z}. Then for any N as y → ∞,

y^{N+1}[G_{ρ}(iy) + ∑_{n=0}^{N}(−i)^{n+1}y^{−n−1}γ_{n}] → 0  (4.28)

uniformly for ρ in M^{H}(γ). Conversely, if G is a function on ℂ_{+} with Im G(z) > 0 there and so that (4.28) holds for each N, then G(z) = G_{ρ}(z) for some ρ in M^{H}(γ).

Proof. By the geometric series with remainder, the left side of (4.28) is

−(−i)^{N+1}y^{N+1} ∫ \frac{dρ(x)}{x − iy} \frac{x^{N+1}}{y^{N+1}} ≡ R_{N}(ρ)  (4.29)

so, using |x − iy|^{−1} ≤ y^{−1},

|R_{N}(ρ)| ≤ y^{−1} ∫ |x|^{N+1}dρ(x)\{ \begin{matrix}= γ_{N+1}y^{−1} & if N is odd \\ ≤ \frac{1}{2}[γ_{N} + γ_{N+2}]y^{−1} & if N is even. \end{matrix}

This shows that (4.28) holds and the convergence is uniform for ρ ∈ M^{H}(γ).

For the converse, first use the Herglotz representative theorem which says that if G maps ℂ_{+} to ℂ_{+}, then for some measure dρ, some c ≥ 0 and some real d:

G(z) = cz + d + ∫ dρ(x)[\frac{1}{x − z} − \frac{x}{1 + x^{2}} ],  (4.30)

where, a priori, ρ only obeys

∫ \frac{dρ(x)}{1 + x^{2}} < ∞.

By (4.28),

y[G(iy)] → iγ_{0}.  (4.31)

If (4.30) holds, then y^{−1}G(iy) → ic, so c = 0. Since c = 0, (4.30) says

y Im G(iy) = ∫ \frac{y^{2}}{x^{2} + y^{2}} dρ(x),

so (4.31) and the monotone convergence theorem implies that ∫ dρ(x) = γ_{0}. Once this is true, (4.30) implies that as y → ∞,

Re G(iy) → d − ∫ dρ(x) \frac{x}{1 + x^{2}}  (4.32)

so (4.31) implies the right side of (4.32) is zero, and thus (4.30) becomes

G(z) = ∫ \frac{dρ(x)}{x − z} .  (4.33)

We will now prove inductively that ρ obeys (1.1), that is, ρ ∈ M^{H}(γ). Suppose that we know that (1.1) holds for n = 0, 1, …, 2M − 2. (4.28) for N = 2M then implies that

∫ \frac{(iy)^{2}x^{2M−1}}{x − iy} dρ(x) + iyγ_{2M−1} → −γ_{2M}.

Taking real and imaginary parts, we see that

γ_{2M−1} = lim_{y→∞} ∫ \frac{y^{2}x^{2M−1}}{x^{2} + y^{2}} dρ(x)  (4.34)
γ_{2M} = lim_{y→∞} ∫ \frac{y^{2}x^{2M}}{x^{2} + y^{2}} dρ(x).  (4.35)

(4.35) and monotone convergence implies that ∫ x^{2M} dρ(x) < ∞, so that dominated convergence and (4.34), (4.35) imply that (1.1) holds for n = 0, 1, 2, …, 2M. □

Let F be the set of analytic maps from ℂ_{+} to ℂ̄_{+} ∪ {∞}. By the open mapping theorem for analytic functions, Φ ∈ F either maps ℂ_{+} to ℂ_{+} or Φ has a constant value in ℝ ∪ {∞}.

Theorem 4.14 (Nevanlinna [29]). Let γ be a set of indeterminate Hamburger moments. Then there is a one-one correspondence between solutions, ρ, of the moment problem and functions Φ ∈ F given by

G_{ρ} ≡ ∫ \frac{dρ(x)}{x − z} = −\frac{C(z)Φ(z) + A(z)}{D(z)Φ(z) + B(z)} .  (4.36)

In particular, if ρ is not a von Neumann solution, G_{ρ}(z) ∈ D(z)^{int} for all z ∈ ℂ_{+}.

Remarks. 1. By (4.21), the von Neumann solutions μ_{t} correspond precisely to the case Φ(z) ≡ t ∈ ℝ ∪ {∞}.

2. We will call the Herglotz function associated to some ρ ∈ M^{H}(γ) the Nevanlinna function of ρ and denote it by Φ_{ρ}. Conversely, given Φ (and γ), we denote by ρ_{Φ} the associated solution of the moment problem.

3. In Section 6, we will discuss the ρ_{Φ}’s associated to the case where the measure μ in the Herglotz representation (1.19) is a measure with finite support.

Proof. Let Φ ∈ F and let G(z) denote the right side of (4.36). In terms of the map F (z) of (4.27), G(z) = F (z)(Φ(z)), so by Proposition 4.12, G(z) ∈ D(z). By the uniformity in Proposition 4.13, the fact that the μ_{t} ∈ M^{H}(γ) and that {G_{μ_{t}}(z) | t ∈ ℝ∪{∞}} = ∂D(z), this implies that G(z) obeys (4.28). Thus by Proposition 4.13, G(z) = G_{ρ}(z) for some ρ ∈ M^{H}(γ).

Conversely, let ρ ∈ M^{H}(γ) and consider G_{ρ}(z). By Theorem 4.3, Proposition 4.4 and Proposition 4.12, Φ(z) ≡ F (z)^{−1}(G_{ρ}(z)) ∈ ℂ̄_{+} ∪ {∞}, and so Φ ∈ F. Thus, G_{ρ}(z) = F (z)(Φ(z)), that is, (4.36) holds.

If ρ is not a von Neumann solution, then Im Φ_{ρ}(z) > 0, so G_{ρ}(z) = F (z)(Φ_{ρ}(z)) is in the interior of D(z). □

Proposition 4.15. Let ρ ∈ M^{H}(γ). If ρ is a von Neumann solution, then the polynomials are dense in L^{2}(ℝ, dρ) and for each z_{0} ∈ ℂ_{+}, there is no other μ in M^{H}(γ) with G_{μ}(z_{0}) = G_{ρ}(z_{0}). If ρ is not a von Neumann solution, then the polynomials are not dense in L^{2}(ℝ, dρ) and for each z_{0} ∈ ℂ_{+}, there are distinct μ’s in M^{H}(γ) with G_{μ}(z_{0}) = G_{ρ}(z_{0}).

Proof. If ρ is a von Neumann solution, the polynomials are dense by construction and the uniqueness result holds since G_{μ}(z_{0}) = G_{ρ}(z_{0}) if and only if Φ_{μ}(z_{0}) = Φ_{ρ}(z_{0}) and Φ_{ρ}(z_{0}) is then real.

Conversely, suppose ρ is not a von Neumann solution. Then, G_{ρ}(z_{0}) ∈ D(z)^{int}, so by the proof of Theorem 4.3, (x − z_{0})^{−1} is not in the L^{2} closure of the polynomials. Thus, the polynomials are not dense. If Φ_{ρ} is not a constant, set Ψ(z) = −|Φ_{ρ}(z_{0})|^{2}Φ_{ρ}(z)^{−1} + 2Re Φ_{ρ}(z_{0}). If Φ_{ρ}(z) is a constant, q (of necessity with Im q > 0), set Ψ(z) = cz + d with c = \frac{Im q}{Im z_{0}} (> 0) and d = Re q −c Re z_{0}. In either case, Ψ ≠ Φ, Ψ is Herglotz, and Ψ(z_{0}) = Φ(z_{0}). Thus, if μ = ρ_{Ψ}, we have G_{μ}(z_{0}) = F (z_{0})(Ψ(z_{0})) = F (z_{0})(Φ(z_{0})) = G_{ρ}(z_{0}). □

To complete the proof of Theorem 5, we need the following fact about Herglotz functions:

Theorem 4.16. Let Φ(z) be a Herglotz function so

Φ(z) = cz + d + ∫ dμ(x)[\frac{1}{x − z} − \frac{x}{1 + x^{2}} ]

with d real, c ≥ 0, and either c > 0 or dμ ≠ 0. Suppose for some x_{0}, Φ(x_{0} + iε) → t ∈ ℝ. Then either |\frac{[Φ(x_{0}+iε)−t]}{iε}| → ∞ or else ∫ \frac{dμ(x)}{(x−x_{0})^{2}} < ∞ and

lim_{ε↓0} \frac{Φ(x_{0} + iε) − t}{iε} = c + ∫ \frac{dμ(x)}{(x − x_{0})^{2}} .  (4.37)

Remark. Do not confuse the μ in the Herglotz representation for Φ and the measure ρ of (4.36). In particular, μ is allowed to have finite support even though we are suppposing that ρ does not.

Proof. Note first that

\frac{Im Φ(x_{o} + iε)}{ε} = c + ∫ \frac{dμ(x)}{(x − x_{0})^{2} + ε^{2}} .  (4.38)

On the other hand,

Im \frac{∂Φ}{∂z} (x_{0} + iε) = ∫ \frac{dμ(x) 2ε(x − x_{0})}{[(x − x_{0})^{2} + ε^{2}]^{2}} .  (4.39)

If ∫ \frac{dμ(x)}{(x−x_{0})^{2}} = ∞, then (4.38) implies Im [\frac{Φ(x_{0}+iε)−t}{ε}] → ∞ , so if the limit is finite, then

∫ \frac{dμ(x)}{(x−x_{0})^{2}} < ∞. (4.39) and the dominated convergence theorem implies that Im \frac{∂Φ}{∂z}(x_{0} +

iε) → 0 so that \frac{[Re Φ(x_{0}+iε)−t]}{iε} = \frac{1}{ε} ∫_{0}^{ε} Im \frac{∂Φ}{∂z}(x_{0} + iy) dy → 0. This and (4.38) implies (4.37). □

Theorem 4.17 (end of Theorem 5). Let γ be an indeterminate moment problem. Let ρ ∈ M^{H}(γ) correspond to a Φ which is not a constant in ℝ ∪ {∞} (so that ρ is not a von Neumann solution). Suppose α ≡ ρ({x_{0}}) > 0 for some point x_{0} ∈ ℝ. Then there is a von Neumann solution μ_{t} so that μ_{t}({x_{0}}) > α.

Proof. We will suppose that D(x_{0}) ≠ 0. If D(x_{0}) = 0, then B(x_{0}) ≠ 0 and we can use

−\frac{C(z)Φ(z) + A(z)}{D(z)Φ(z) + B(z)} = −\frac{A(z)(−Φ(z))^{−1} − C(z)}{B(z)(−Φ(z))^{−1} − D(z)}

and −Φ(z)^{−1} in place of Φ(z) to give an identical argument. Define t = −\frac{B(x_{0})}{D(x_{0})} ∈ ℝ. Since AD − BC = 1,

C(x_{0})t + A(x_{0}) = \frac{1}{D(x_{0})}  (4.40)

is non-zero and has the same sign as D.

(−iε)G_{ρ}(x_{0} + iε) = (iε)[\frac{C(x_{0} + iε)Φ(x_{0} + iε) + A(x_{0} + iε)}{D(x_{0} + iε)Φ(x_{0} + iε) + B(x_{0} + iε)}] → α.

This is only possible if G_{ρ}(x_{0} + iε) → ∞, which requires that Φ(x_{0} + iε) → t. Then

C(x_{0} + iε)Φ(x_{0} + iε) + A(x_{0} + iε) → C(x_{0})t + A(x_{0})

while

\frac{D(x_{0} + iε)Φ(x_{0} + iε) + B(x_{0} + iε)}{iε} → D′(x_{0})t + B′(x_{0}) + D(x_{0}) lim_{ε↓0} \frac{Φ(x_{0} + iε) − t}{iε} .

Thus using (4.40),

α = \frac{1}{D(x_{0})^{2} lim_{ε↓0} \frac{Φ(x_{0}+iε)−t}{iε} + D(x_{0})[D′(x_{0})t + B′(x)]} .

On the other hand, taking Φ(x) ≡ t to describe μ_{t},

μ_{t}(\{x_{0}\}) = \frac{1}{D(x_{0})[D′(x_{0})t + B′(x_{0})]}

must be positive. It follows from Lemma 4.16 that ∫ \frac{dμ(x)}{(x−x_{0})^{2}} < ∞ and that

α^{−1} > μ_{t}({x_{0}})^{−1}. □

Finally, we turn to the Nevanlinna parametrization in the indeterminate Stieltjes case. As we have seen, among the B_{t} there are two distingtuished ones, B_{F} and B_{K}. The Krein extension has ker(A^{∗}) ⊂ D(B_{K}) so π(0) ∈ D(B_{K}), which means t = ∞ in the notation of (4.20). The Friedrichs extension is B_{F} = B_{t_{0}} where t_{0} = (δ_{0}, A_{F}^{−1}δ_{0}).

Theorem 4.18 (Stieltjes case of Theorem 4). If the Stieltjes problem for γ is indeterminate, its solutions obey (4.36) for precisely those Φ that either are a constant in [t_{0}, ∞)∪{∞} or that obey Φ(z) is analytic in ℂ\[0, ∞) with Φ(x) ∈ [t_{0}, ∞) if x ∈ (−∞, 0). Here t_{0} = (δ_{0}, A_{F}^{−1}δ_{0}).

Remark. Such Φ’s are precisely those of the form

Φ(z) = d + ∫_{0}^{∞} \frac{dμ(x)}{x − z} ,  (4.41)

where d ≥ t_{0} and ∫_{0}^{∞} \frac{dμ(x)}{x+1} < ∞.

Proof. Clearly, the solutions of the Stieltjes problem are precisely the solutions ρ of the Hamburger problem with lim_{ε↓0} G_{ρ}(−y + iε) real for each y ∈ (0, ∞). Since the B_{t}’s obey t = (δ_{0}, B_{t}^{−1}δ_{0}) and B_{K} < B_{t} < B_{F} means t_{0} < (δ_{0}, B_{t}^{−1}δ_{0}) < ∞, we know that for y ∈ (−∞, 0), F (y) maps [t_{0}, ∞) into (0, ∞) (F ( ·) is given by (4.27)). It follows that if Φ obeys Φ(x) ∈ [t_{0}, ∞) for x ∈ (−∞, 0), then G_{ρ}(y + iε) has a limit in (0, ∞) on (−∞, 0), that is, Φ defines a solution of the Stieltjes moment problem. The converse follows from this argument and the next theorem. □

Theorem 4.19. Let dμ_{F} and dμ_{K} be the Friedrichs and Krein solutions of an indeterminate Stieltjes moment problem and let dρ be another solution. Then for all y ∈ (0, ∞),

∫_{0}^{∞} \frac{dμ_{F} (x)}{x + y} ≤ ∫_{0}^{∞} \frac{dρ(x)}{x + y} ≤ ∫_{0}^{∞} \frac{dμ_{K}(x)}{x + y} .  (4.42)

This will be proven below as part of Theorem 5.2.

Corollary 4.20. If ρ ∈ M^{S}(γ) and dρ ≠ dμ_{F}, then for y ∈ [0, ∞),

∫_{0}^{∞} \frac{dμ_{F} (x)}{x + y} < ∫_{0}^{∞} \frac{dρ(x)}{x + y} .  (4.43)

(Note that the inequality is strict.)

Proof. For each y ∈ (0, ∞), F (−y)( ·) is a strictly monotone map of [t_{0}, ∞) to [G_{μ_{F}} (−y), G_{μ_{K}} (−y)) by the proof of Theorem 4.18 and the fact that any fractional linear map of ℝ to ℝ is monotone in any region where it is finite. By a limiting argument, the same is true for y = 0. Thus, (4.43) follows from the relation of Nevanlinna functions Φ_{ρ}(y) > Φ_{μ_{F}} (y) ≡ t_{0}. This is immediate from (4.41). □

As a corollary of this, we see

Corollary 4.21. Let γ be a set of indeterminate Stieltjes moments and let dμ_{F} be its Friedrichs solution. Let r = ∫ x^{−1}dμ_{F} (x) and let

γ̃_{0} = 1;  γ̃_{n} = r^{−1}γ_{n−1},  n = 1, 2, ….

Then {γ̃_{j}}_{j=0}^{∞} is a determinate Hamburger moment problem.

Remark. This lets us produce Stieltjes determinate moment problems with essentially arbitrary rates of growth for γ_{n}. By using Theorem 2.13, we obtain determinate Hamburger problems with arbitrary fast growth also.

Proof. r^{−1}x^{−1}dμ_{F} ≡ dλ_{0}(x) is a Stieltjes measure whose moments are γ̃. Let dλ(x) be another such Stieltjes measure and let dρ(x) = rx dλ(x). Then the moments of dρ are γ and ∫ x^{−1}dρ(x) = r. So, by Corollary 4.20, ρ = μ_{F}. Thus, λ = λ_{0}. Thus, λ_{0} is the unique solution of the Stieltjes problem. But 0 ∉ supp(dλ_{0}). Thus by Proposition 3.1, the Hamburger problem is also determinate. □

§5. Padé Approximants, Continued Fractions, and Finite Matrix Approximations

In this section, we will primarily consider aspects of the Stieltjes moment problem. We will discuss finite matrix approximations which converge to the Friedrichs and Krein extensions of A and, as a bonus, obtain Theorem 4.19 and Theorems 5, 6, 7, and 8. We will see that the convergence of the Padé approximants (equivalently, continued fractions of Stieltjes) can be reinterpreted as strong resolvent convergence of certain finite matrix approximations to A_{F} and A_{K}.

Before we start, we note that our continued fraction expressions are related to but distinct from those of Stieltjes. Our basic object (the approximations to A_{F}) are of the form:

\frac{1}{−z + b_{0} − \frac{a_{0}^{2}}{−z + b_{1} − \frac{a_{1}^{2}}{−z + b_{2} + ⋯}}}  (5.1a)

with two terms (an affine polynomial) at each stage while Stieltjes’ are of the form

\frac{1}{c_{1}w + \frac{1}{c_{2} + \frac{1}{c_{3}w + ⋯}}}  (5.1b)

with a single term in each step and alternate constant and linear terms. This gives him twice as many continued fractions, so his alternate between the two sets of rational approximations we get. We will say more about this later.

Until Theorem 5.29 below, we will suppose {γ_{n}}_{n=0}^{∞} is a set of Stieltjes moments. Given the Jacobi matrix (1.16), we will consider two N ×N approximations. The first is obvious:

A_{F}^{[N]} = ( \begin{matrix}b_{0} & a_{0} \\ a_{0} & b_{1} \\ & & . & . & . \\ & & & & & b_{N−2} & a_{N−2} \\ & & & & & a_{N−2} & b_{N−1} \end{matrix} ) .  (5.2a)

The second differs by the value of the NN coefficient:

A_{K}^{[N]} = ( \begin{matrix}b_{0} & a_{0} \\ a_{0} & b_{1} \\ & & . & . & . \\ & & & & & b_{N−2} & a_{N−2} \\ & & & & & a_{N−2} & b_{N−1} − α_{N−1} \end{matrix} ) ,  (5.2b)

where α_{N−1} is chosen so that A_{K}^{[N]} has a zero eigenvalue. That such an α_{N−1} exists is the content of

Lemma 5.1. There is a unique α_{N−1} > 0 so that (5.2b) has a zero eigenvalue. A_{K}^{[N]} ≥ 0 and α_{N−1} obeys

(b_{N−1} − α_{N−1})P_{N−1}(0) + a_{N−2}P_{N−2}(0) = 0  (5.3)

and

α_{N−1} = −a_{N−1} \frac{P_{N}(0)}{P_{N−1}(0)} .  (5.4)

Moreover,

(b_{N} − α_{N})(α_{N−1}) − a_{N−1}^{2} = 0.  (5.5)

Proof. Any solution of A_{K}^{[N]}u = 0 must obey the eigenfunction equation (4.1) and so be a multiple of P_{j}(0), that is,

u_{j} = P_{j}(0), j = 0, 1, …, N − 1.

This obeys the condition at the lower right of the matrix if and only if (5.3) holds.

If P_{N−1}(0) were 0, then A_{F}^{[N−1]} would have a zero eigenvalue, which is impossible since A_{F}^{[N−1]} is strictly positive definite (since the form S_{N−1} is strictly positive definite). Thus, P_{N−1}(0) ≠ 0 and so (5.3) has a unique solution. Since A_{F}^{[N]} is positive definite, this solution must have α_{N−1} > 0.

The eigenfunction equation for P_{N}(0) then yields (5.4). (5.3) for N → N +1 and (5.4) yield two formulas for \frac{P_{N} (0)}{P_{N−1}(0)}. Setting these to each other yields (5.5). □

The first main result of this section, towards which we are heading, is Theorem 5.2 (includes Theorem 4.19). Let

f_{N}^{+}(z) ≡ (δ_{0}, (A_{K}^{[N]} − z)^{−1}δ_{0})  (5.6)

and

f_{N}^{−}(z) ≡ (δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}).  (5.7)

Then for x ∈ (0, ∞),

f_{N}^{+}(−x) ≥ f_{N+1}^{+}(−x) converges to (δ_{0}, (A_{K} + x)^{−1}δ_{0}) ≡ f^{+}(−x)

while

f_{N}^{−}(−x) ≤ f_{N+1}^{−}(−x) converges to (δ_{0}, (A_{F} + x)^{−1}δ_{0}) ≡ f^{−}(−x).

The convergence holds for any z ∈ ℂ\[0, ∞) and is uniform on compacts. Moreover, for any solution ρ of the moment problem and x > 0,

f^{−}(−x) ≤ ∫ \frac{dρ(y)}{x + y} ≤ f^{+}(−x).  (5.8)

Remarks. 1. Notice that (5.8) is (4.42).

2. Let A^{[N]}(β) be (5.2b) with α_{N−1} replaced by −β. Then A_{K}^{[N]} is quite a natural approximation: It is that A^{[N]}(β) which is minimal among the non-negative matrices — reasonable to capture A_{K}, the minimal non-negative self-adjoint extension. From this point of view, A_{F}^{[N]} seems less natural. One would instead want to consider lim_{β→∞} A^{[N]}(β). In fact, one can show this limit is “essentially” A_{F}^{[N−1]} in the sense that

lim_{β→∞}(δ_{0}, (A^{[N]}(β) − z)^{−1}δ_{0}) = (δ_{0}, (A_{F}^{[N−1]} − z)^{−1}δ_{0}).

Thus, the A^{[N]}(β) interpolate between A_{K}^{[N]} and A_{F}^{[N−1]}.

Proposition 5.3. Let Ã_{K}^{[N]} be the operator on ℓ^{2} which is A_{K}^{[N]} ⊕0. Then Ã_{K}^{[N]} converges to A_{K} in strong resolvent sense.

Remark. For self-adjoint operators {A_{n}}_{n=1}^{∞} and A_{∞}, we say A_{n} converges to A_{∞} in strong resolvent sense if and only if for all z ∈ ℂ\ℝ, (A_{n} − z)^{−1}φ → (A_{∞} − z)^{−1}φ for all vectors φ. It is a basic theorem (see Reed-Simon [33]) that it suffices to prove this for a single z ∈ ℂ\ℝ. The same proof shows that if all A_{n} and A_{∞} are non-negative, one then also has convergence for z ∈ (−∞, 0). Moreover, one has convergence uniformly for z in compact subsets of ℂ\[0, ∞).

Proof. Suppose first that the Hamburger problem is indeterminate. Thus, π(0) = (P_{0}(0), …) ∈ ℓ_{2}. Let P^{[n]}(0) be the vector (P_{0}(0), …, P_{n−1}(0), 0, 0, …). Then A_{K}^{[n]}P^{[n]} = 0 so (A_{K}^{[n]} − z)^{−1}P^{[n]} = −z^{−1}P^{[n]} for any z ∈ ℂ\ℝ. Since ∥(A_{K}^{[n]} − z)^{−1}∥ ≤ |Im z|^{−1} for any such z, and ∥P^{[n]} − π(0)∥ → 0, we conclude that

(A_{K}^{[n]} − z)^{−1}π(0) → −z^{−1}π(0) = (A_{K} − z)^{−1}π(0)  (5.9)

since A_{K}π(0) = 0.

Suppose φ = (A − z)η for η ∈ D(A), that is, a finite sequence. Then (A_{K} − z)η = φ = (A_{K}^{[n]} − z)η for n large, and thus lim_{n→∞}(A_{K}^{[n]} − z)^{−1}φ = η = (A_{K} − z)^{−1}φ. Thus, (A_{K}^{[n]} − z)^{−1}φ → (A − z)^{−1}φ for φ ∈ \overline{Ran(A − z)}. In the determinate case, that is all φ’s; while in the indeterminate case, we claim that \overline{Ran(A − z)} + [π(0)] is all of ℓ^{2} so (5.9) completes the proof.

That leaves the proof of our claim that \overline{Ran(A − z)} + [π(0)] is ℓ^{2}. We first note that π(0) ∉ D(Ā) by Proposition 4.4. (A_{K} −z)^{−1}( \overline{Ran(A − z)}+[π(0)]) = D(Ā)+[π(0)] since (A_{K} − z)^{−1})π(0) = −z^{−1}π(0). But D(A_{K})/D(Ā) has dim 1 and π(0) ∉ D(Ā) so (A_{K} − z)^{−1}( \overline{Ran(A − z)} + [π(0)]) = D(A_{K}) and thus, since Im z ≠ 0, \overline{Ran(A − z)} + [π(0)] = ℓ_{2}, as claimed. □

Proposition 5.4. Let ˜ Ã_{K}^{[N]} be the (N + 1) × (N + 1) matrix which is A_{K}^{[N]} ⊕ 0, that is, it has zeros in its last row and column. Then

˜Ã_{K}^{[N]} ≤ A_{K}^{[N+1]}.  (5.10)

In particular, for x > 0,

(δ_{0}, (A_{K}^{[N+1]} + x)^{−1}δ_{0}) ≤ (δ_{0}, ( ˜Ã_{K}^{[N]} + x)^{−1}δ_{0}) = (δ_{0}, (A_{K}^{[N]} + x)^{−1}δ_{0}).

Remark. Propositions 5.3 and 5.4 prove the part of Theorem 5.2 involving monotonicity of f_{N}^{+} and its convergence.

Proof. B ≡ A_{K}^{[N+1]} − ˜ Ã_{K}^{[N]} is an (N + 1) × (N + 1) matrix which has all zeros except for a 2 × 2 block in its lower right corner. This block is

( \begin{matrix}α_{N−1} & a_{N−1} \\ a_{N−1} & b_{N} − α_{N} \end{matrix} ) ,

which is symmetric with a positive trace and determinant zero (by (5.5)). Thus, the block is a positive rank one operator so B is positive, proving (5.10). □

The monotonicity of the A_{F}^{[N]}’s and their convergence is a special case of monotone convergence theorems for forms. Consider non-negative closed quadratic forms whose domain may not be dense. Each such form q_{B} with D(q_{B}) is associated to a nonnegative self-adjoint operator B on \overline{D(q_{B})}. We define (B − x)^{−1} to be 0 on D(q_{B})^{⊥}. This essentially sets B = ∞ on D(q_{B})^{⊥} consistent with the intention that D(q_{B}) is the set of φ for which q_{B}(φ) < ∞.

Given two forms q_{B} and q_{C}, we say q_{B} ≤ q_{C} if and only if D(q_{B}) ⊇ D(q_{C}) (intuitively q_{C}(φ) < ∞ implies q_{B}(φ) < ∞) and for all φ ∈ q_{C}, q_{C}(φ) ≥ q_{B}(φ). It is a fundamental result [12, 37] that if q_{B} ≤ q_{C}, then for all x ≥ 0 and all φ,

(φ, (C + x)^{−1}φ) ≤ (φ, (B + x)^{−1}φ).  (5.11)

One monotone convergence theorem for forms [12, 37] says the following: Let {q_{A_{n}}}_{n=1}^{∞} be a sequence of quadratic forms with q_{A_{n}} ≥ q_{A_{n+1}}. Then (A_{n} + x)^{−1} converges to the resolvent, (A_{∞} + x)^{−1}, of an operator A_{∞}. It is the operator associated to the form q_{A_{∞}} defined to be the closure of q̃_{A_{∞}} where D(q̃_{A_{∞}}) = ∪D(q_{A_{n}}) with q̃_{A_{∞}}(φ) = lim_{n→∞} q_{A_{n}}(φ) provided that the form q̃_{A_{∞}} is closable.

We can apply this to the A_{F}^{[N]}. Let q_{F}^{[N]} be defined to be the form with D(q_{F}^{[N]}) = {(φ_{0}, …, φ_{N−1}, 0, …, 0, …)} and q_{F}^{[N]}(φ) = (φ, Aφ). Then D(q_{F}^{[N]}) ⊂ D(q_{F}^{[N+1]}) and q_{F}^{[N]}(φ) = q_{F}^{[N+1]}(φ) if φ ∈ D(q_{F}^{[N]}). Thus, q_{F}^{[N]} ≥ q_{F}^{[N+1]} (in essence, Ã_{F}^{[N]} = A_{F}^{[N]} ⊕ ∞ on ℂ^{N+1}). The monotone convergence theorem applies. A_{∞} is just the closure of φ ↦ (φ, Aφ) on finite sequences, that is, the Friedrichs extension A_{F}, so we have the convergence and monotonicity part of Theorem 5.2 for f_{N}^{−} (using (5.11)):

Proposition 5.5. Let x > 0. Then (δ_{0}, (A_{F}^{[N+1]} + x)^{−1}δ_{0}) ≥ (δ_{0}, (A_{F}^{[N]} + x)^{−1}δ_{0}) and this converges to (δ_{0}, (A_{F} + x)^{−1}δ_{0}) as N → ∞.

To head towards the remaining part of Theorem 5.2, viz. (5.8), we need expressions for (φ, (A_{F}^{[N]} − z)^{−1}φ) and (φ, (A_{K}^{[N]} − z)^{−1}φ), which are of independent interest.

Proposition 5.6.

(δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}) = −\frac{Q_{N}(z)}{P_{N}(z)}  (5.12)

Proof. The only possible eigenfunctions of A_{F}^{[N]} are the vectors π^{[N]}(z) ≡ (P_{0}(z), …, P_{N−1}(z)). By the eigenfunction equation (4.1), this is an eigenfunction of A_{F}^{(N)} if and only if P_{N}(z) = 0. Thus, the N poles of (δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}) are precisely the N-zeros of P_{N}(z). The zeros are distinct because the eigenvalues are simple (since π^{[N]}(z) is the only potential eigenfunction, since eigenfunctions must have u_{0} ≠ 0).

Now let the B_{F}^{[N]} be the (N − 1) × (N − 1) matrix obtained by removing the top row

and left column of A_{F}^{[N]}. By Cramer’s rule, (δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}) = \frac{det(B_{F}^{[N]}−z)}{det(A_{F}^{[N]}−z)}, so the

N − 1 zeros of (δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}) are precisely the N − 1 eigenvalues of B_{F}^{[N]}. The only possible eigenfunctions of B_{F}^{[N]} are ξ(z)^{[N−1]} = (Q_{1}(z), …, Q_{N−1}(z)) (since Q_{0}(z) = 0, Q_{1}(z) = \frac{1}{a_{0}} ). Thus, the eigenvalues of B_{F}^{[N]} are precisely z’s with Q_{N}(z) = 0. It follows that

(δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}) = \frac{d_{N}Q_{N}(z)}{P_{N}(z)} ,

and we need only show that d_{N} ≡ −1.

Since (δ_{0}, (A_{F}^{[N=1]} − z)^{−1}δ_{0}) = (b_{0} − z)^{−1} and Q_{1}(z) = \frac{1}{a_{0}} , P_{1}(z) = \frac{(z−b_{0})}{a_{0}}, we see that d_{1} = −1. On the other hand, Proposition 4.1 implies that

\frac{Q_{k}(z)}{P_{k}(z)} − \frac{Q_{k−1}(z)}{P_{k−1}(z)} = \frac{1}{a_{k−1}P_{k}(z)P_{k−1}(z)} = o(\frac{1}{z}).

Moreover, (δ_{0}, (A − z)^{−1})δ_{0}) = −z^{−1} + O(z^{−2}), so d_{k} = d_{k−1}, that is, d_{N} = −1 for all N. □

Remarks. 1. The proof shows that for suitable c_{N} (indeed, an induction shows that c_{N} = (−1)^{N}(a_{0} … a_{N−1})^{−1}), we have

P_{N}(z) = c_{N} det(A_{F}^{[N]} − z)  (5.13)

and

Q_{N}(z) = −c_{N} det(B_{F}^{[N]} − z).  (5.14)

2. Since A_{F}^{[N]} is strictly positive definite, (5.13) shows once more that P_{N}(0) ≠ 0 in the Stieltjes case. Since B_{F}^{[N]} is also strictly positive definite (as a submatrix of A_{F}^{[N]}), (5.14) shows that Q_{N}(0) ≠ 0 also in the Stieltjes case.

(5.13)–(5.14) imply some facts about the zeros of P_{N} and Q_{N}:

Proposition 5.7. All the zeros of each P_{N}(z) and each Q_{N}(z) are real. Moreover, there is exactly one zero of Q_{N}(z) and one zero of P_{N−1}(z) between any pair of successive zeros of P_{N}(z).

Proof. The first assertion follows from the fact that the eigenvalues of a real-symmetric matrix are all real. The second assertion follows from the fact that if X is an N × N real-symmetric matrix and Y is an N − 1 × N − 1 submatrix, then by the min-max principle, there is exactly one eigenvalue of Y between any pair of eigenvalues of X. □

Remark. Since the Q_{N}’s are orthogonal polynomials for another moment problem (see Proposition 5.16), between any two successive zeros of Q_{N}(z), there is a single zero of Q_{N−1}(z).

Now define

M_{N}(z) = P_{N}(z) − P_{N}(0) \frac{P_{N−1}(z)}{P_{N−1}(0)}  (5.15)
N_{N}(z) = Q_{N}(z) − P_{N}(0) \frac{Q_{N−1}(z)}{P_{N−1}(0)}  (5.16)

since A_{F}^{[j]} is strictly positive, 0 is never an eigenvalue and P_{j}(0) ≠ 0 for all j, so M is well-defined. Notice M_{N}(0) = 0.

Proposition 5.8.

(δ_{0}, (A_{K}^{[N]} − z)^{−1}δ_{0}) = −\frac{N_{N}(z)}{M_{N}(z)}  (5.17)

Proof. Let c_{N} be the constant in (5.13) and let B_{K}^{[N]} be A_{K}^{[N]} with the top row and left-most column removed. We will prove that

M_{N}(z) = c_{N} det(A_{K}^{[N]} − z);  N_{N}(z) = −c_{N} det(B_{K}^{[N]} − z),  (5.18)

from which (5.17) holds by Cramer’s rule.

Clearly,

det(A_{K}^{[N]} − z) = det(A_{F}^{[N]} − z) − α_{N−1} det(A_{F}^{[N−1]} − z)

so

c_{N} det(A_{K}^{[N]} − z) = P_{N}(z) − β_{N}P_{N−1}(z),

where

β_{N} = α_{N−1} \frac{c_{N}}{c_{N−1}} .

But det(A_{K}^{[N]}) = 0 by construction. So P_{N}(0) − β_{N}P_{N−1}(0) = 0 and thus, c_{N} det(A_{K}^{[N]} − z) = M_{N}(z).

In the same way,

c_{N} det(B_{K}^{[N]} − z) = −Q_{N}(z) + β_{N}Q_{N−1}(z) = −N_{N}(z). □

Because of our convergence theorem, we have

Corollary 5.9. Let \{λ_{i}^{[N]}\}_{i=1}^{N} be the zeros of P_{N}(z) and let ν_{i}^{[N]} = −\frac{Q_{N} (λ_{i}^{[N]})}{P′_{N} (λ_{i}^{[N]})}. Then

∑_{i=1}^{N} ν_{i}^{[N]}δ(λ−λ_{i}^{[N]}) converges to dμ_{F} (λ), the Friedrichs solution of the moment problem. A similar formula holds for the Krein solution, dμ_{K}(λ), with P, Q replaced by M, N.

The following concludes the proof of Theorem 5.2. It extends the calculation behind (4.4):

Theorem 5.10. Let dρ solve the Stieltjes moment problem. Then for x > 0 and N ≥ 1,

−\frac{Q_{N}(−x)}{P_{N}(−x)} ≤ ∫ \frac{dρ(y)}{x + y} ≤ −\frac{N_{N}(−x)}{M_{N}(−x)} .  (5.19)

Proof. \frac{P_{N} (y)−P_{N} (z)}{y−z} is a polynomial in y of degree N − 1 so

∫ dρ(y)P_{N}(y)[\frac{P_{N}(y) − P_{N}(z)}{y − z}] = 0.  (5.20)

54 B. SIMON

On the other hand, (4.4) says that

∫ dρ(y) \frac{P_{N}(y)}{y − z} = P_{N}(z) ∫ \frac{dρ(y)}{y − z} + Q_{N}(z).  (5.21)

Thus for z = −x with x > 0,

0 ≤ ∫ \frac{dρ(y)P_{N}^{2}(y)}{y + x} = P_{N}(−x) ∫ \frac{dρ(y)P_{N}(y)}{y + x} = P_{N}(−x)^{2} ∫ \frac{dρ(y)}{x + y} + Q_{N}(−x)P_{N}(−x).

Dividing by P_{N}(−x)^{2}, we obtain the left-most inequality in (5.19).

Similarly, since M_{N}(0) = 0, \frac{M_{N} (z)}{z} is a polynomial of degree N −1 and so [y^{−1}M_{N}(y)− z^{−1}M_{N}(z)]/y − z is a polynomial of degree N − 2 in y, which is orthogonal to P_{N}(y) and P_{N−1}(y) and so to M_{N}(y). Thus

∫ dρ(y) \frac{M_{N}(y)}{y − z}[\frac{M_{N}(y)}{y} − \frac{M_{N}(z)}{z}] = 0.  (5.22)

Since for N ≥ 0,

Q_{N}(z) = ∫ dρ(y) \frac{P_{N}(y) − P_{N}(z)}{y − z}  (5.23)

for each z, we see that for N ≥ 1,

N_{N}(z) = ∫ dρ(y) \frac{M_{N}(y) − M_{N}(z)}{y − z} .  (5.24)

Therefore, for x > 0,

0 ≤ ∫ \frac{dρ(y)}{y + x} \frac{M_{N}(y)^{2}}{y} = \frac{M_{N}(−x)}{(−x)} ∫ dρ(y) \frac{M_{N}(y)}{y + x}  by (5.22))

= \frac{M_{N}(−x)^{2}}{(−x)} ∫ \frac{dρ(y)}{x + y} + \frac{M_{N}(−x)N_{N}(−x)}{(−x)} .

Dividing by \frac{M_{N} (−x)^{2}}{x}, we obtain the second half of (5.19). □

Next, we turn to Theorem 6 and the connection of Padé approximants to these finite matrix approximations. Given a formal power series ∑_{n=0}^{∞} κ_{n}z^{n}, we define the Padé approximants f^{[N,M]}(z) as follows (see Baker and Graves-Morris [4] for background on Padé approximants): We seek a function f^{[N,M]}(z) so that

f^{[N,M]}(z) = \frac{A^{[N,M]}(z)}{B^{[N,M]}(z)} ,  (5.25)

where A^{[N,M]}(z) is a polynomial of degree at most N, B^{[N,M]}(z) is a polynomial of degree at most M, as z → 0,

f^{[N,M]}(z) − ∑_{j=0}^{N+M} κ_{j}z^{j} = O(z^{N+M+1})  (5.26)

and

B^{[N,M]}(0) = 1.  (5.27)

There is at most one solution, f, for these equations since if Ã, B̃ are another pair of such polynomials, by (5.26), ÃB − AB̃ = O(z^{N+M+1}) must be zero since ÃB − AB̃ is a polynomial of degree N + M. Thus,

A^{[N,M]}(z)B̃^{[N,M]}(z) − Ã^{[N,M]}(z)B^{[N,M]}(z) = 0.  (5.28)

This implies A/B = Ã/B̃, showing f is uniquely determined as an analytic function. If

deg A = N,  deg B = M,  A, B relatively prime,  (5.29)

then (5.28) shows Ã = A and B̃ = B, so A and B are uniquely determined. It can be shown ([4]) that if

det((κ_{N−M+i+j−1})_{1≤i,j≤M}) ≠ 0  (5.30)

then A, B exist and obey (5.29).

There are degenerate cases where a Padé approximant exists, but A, B are not unique (e.g., for ∑_{n=0}^{∞} κ_{n}z^{n} = 1 + z, f^{[2,1]}(z) = 1 + z can be written as A(z) = (1 + z)(1 + αz), B(z) = (1 + αz) for any α). In any event, if a solution of (5.25)–(5.27) exists, we say the [N, M] Padé approximant exists and denote it by f^{[N,M]}(z).

In the context of Theorem 6, we are interested in the Padé approximants for the Taylor series of

f(z) = ∫_{0}^{∞} \frac{dρ(x)}{1 + zx} ,  (5.31)

where ρ is a measure with finite moments. Without loss, normalize ρ so ∫ dρ = 1. Recall the context of Theorem 6. We have

κ_{n} = (−1)^{n} ∫_{0}^{∞} x^{n} dρ(x)

and want to formally sum near z = 0:

f(z) ∼ ∑_{n=0}^{∞} κ_{n}z^{n}.

56 B. SIMON

If we define

γ_{n} = ∫_{0}^{∞} x^{n} dρ(x),

then near w = ∞,

G_{ρ}(w) ≡ ∫_{0}^{∞} \frac{dρ(x)}{x − w} ∼ − ∑_{n=0}^{∞} γ_{n}w^{−n−1}.  (5.32)

Thus formally,

f(z) = \frac{1}{z} G_{ρ}(−\frac{1}{z}).  (5.33)

We begin by noting: Proposition 5.11. As |w| → ∞,

−\frac{Q_{N}(w)}{P_{N}(w)} = − ∑_{j=0}^{2N−1} γ_{j}w^{−j−1} + O(w^{−2N−1})  (5.34)
−\frac{N_{N}(w)}{M_{N}(w)} = − ∑_{j=0}^{2N−2} γ_{j}w^{−j−1} + O(w^{−2N})  (5.35)

Proof. Let A be the Jacobi matrix for the moment problem {γ_{n}}_{n=0}^{∞}. Then

γ_{n} = ⟨δ_{0}, A^{n}δ_{0}⟩.  (5.36)

On the other hand, we have an expansion converging near infinity for

⟨δ_{0}, (A_{F}^{[N]} − w)^{−1}δ_{0}⟩ = − ∑_{j=0}^{∞}(δ_{0}, (A_{F}^{[N]})^{j}δ_{0})w^{−j−1},

so (5.34) follows from

⟨δ_{0}, (A_{F}^{[N]})^{j}δ_{0}⟩ = ⟨δ_{0}, A^{j}δ_{0}⟩, j = 0, 1, …, 2N − 1. (5.37)

To prove (5.37), note that

(A_{F}^{[N]})^{j}δ_{0} = A_{F}^{j} δ_{0}  (5.38)

if j = 0, …, N − 1 and for B symmetric,

⟨δ_{0}, B^{2j}δ_{0}⟩ = ∥B^{j}δ_{0}∥

and that

⟨δ_{0}, B^{2j+1}δ_{0}⟩ = ⟨B^{j}δ_{0}, B(B^{j}δ_{0})⟩.  (5.39)

To obtain (5.35), note that if A_{F}^{[N]} is replaced by A_{K}^{[N]}, (5.38) still holds for j = 0, …, N − 1, but (5.39) fails for j = N − 1 and so the analog of (5.37) only holds for j = 0, …, 2N − 2. □

Remark. While (5.37) holds for j = 0, 1, 2, …, 2N − 1, it never holds for j = 2N. For if it did, we would have for any polynomial, R, of degree 2N that

⟨δ_{0}, R(A_{F}^{[N]})δ_{0}⟩ = ⟨δ_{0}, R(A)δ_{0}⟩.

But this is false for R(x) = x^{2}P_{N−1}(x)^{2} since

⟨δ_{0}, R(A)δ_{0}⟩ = ∥Aδ_{N−1}∥^{2} = a_{N−1}^{2} + a_{N−2}^{2} + b_{N−1}^{2},

while

⟨δ_{0}, R(A_{F}^{[N]})δ_{0}⟩ = ∥A_{N}^{[F]}δ_{N−1}∥^{2} = a_{N−2}^{2} + b_{N−1}^{2}.

Proof of Theorem 6. By (5.33) and (5.34) as z → 0,

−\frac{Q_{N}(−\frac{1}{z})}{zP_{N}(−\frac{1}{z})} = ∑_{j=0}^{2N−1} κ_{j}z^{j} + O(z^{2N}).  (5.40)

Now z^{N−1}Q_{N}(−\frac{1}{z}) is a polynomial of degree N − 1 since Q_{N}(0) ≠ 0 and z^{N}P_{N}(−\frac{1}{z}) is a polynomial of degree N since P_{N}(0) ≠ 0. Moreover, lim_{z→0} z^{N}P_{N}(−\frac{1}{z}) ≠ 0 since P_{N} has degree N. Thus (5.40) identifies f^{[N−1,N]} and −z^{N−1}Q_{N}(−\frac{1}{z})/z^{N}P_{N}(−\frac{1}{z}). Similarly, noting that since M_{N}(0) = 0 (but M′_{N}(0) ≠ 0 since A_{K}^{N} has simple eigenvalues), z^{N}M_{N}(−\frac{1}{z}) is a polynomial of degree N − 1, and we identify f^{[N−1,N−1]}(z) and −z^{N−1}N_{N}(−\frac{1}{z})/z^{N}M_{N}(−\frac{1}{z}). Theorem 5.2 thus implies Theorem 6. □

Remarks. 1. Since P_{N} and Q_{N} are relatively prime (by Proposition 5.7) and similarly for M_{N}, N_{N}, we are in the situation where the numerator and denominator are uniquely determined up to a constant.

2. Suppose γ_{n} ≤ Ca^{n} so ρ is unique and supported on some interval [0, R]. Then if R is chosen as small as possible, ∑_{n=0}^{∞} κ_{n}z^{n} has radius of convergence R^{−1}, but the Padé approximants converge in the entire region ℂ\(−∞, −R^{−1}]!

Before leaving our discussion of Padé approximants for series of Stieltjes, we will consider the sequences f^{[N+j,N]}(z) for j fixed. We will show they each converge as N → ∞ with limits that are all distinct (as j varies) if the Stieltjes moment problem is indeterminate and all the same if a certain set of Stieltjes moment problems are all determinate.

Given a set of moments {γ_{j}}_{j=1}^{∞} and ℓ = 1, 2, …, define for j = 0, 1, …

γ_{j}^{(ℓ)} ≡ \frac{γ_{j+ℓ}}{γ_{ℓ}} .  (5.41)

Proposition 5.12. Suppose that {γ_{j}}_{j=0}^{∞} is a set of Stieltjes (resp. Hamburger) moments. Then {γ_{j}^{(ℓ)}}_{j=0}^{∞} is the set of moments for a Stieltjes (resp. Hamburger) problem for ℓ = 1, 2, … (resp. ℓ = 2, 4, 6, …). This problem is indeterminate if the original problem is.

Proof. We will consider the Stieltjes case. Let ρ ∈ M^{S}(γ). Then γ_{ℓ}^{−1}x^{ℓ} dρ(x) ≡ dρ^{(ℓ)}(x) solves the moment problem for γ^{(ℓ)}. If ρ_{1}, ρ_{2} ∈ M^{S}(γ), then ρ_{1}^{(ℓ)}, ρ_{2}^{(ℓ)} are both in M^{S}(γ^{(ℓ)}). If ρ_{1}^{(ℓ)} = ρ_{2}^{(ℓ)}, then ρ_{1} − ρ_{2} is supported at zero, which means it is zero since ∫ dρ_{1}(x) = γ_{0} = ∫ dρ_{2}(x). Thus, if the γ problem is indeterminate, so is the γ^{(ℓ)} problem. □

Example. As Corollary 4.21 shows, there can be determinate moment problems so that γ^{(1)} is indeterminate (in the language of that corollary, γ̃^{(1)} = γ). Thus, the converse of the last statement in Proposition 5.12 fails.

As an aside, we present a proof of a criterion for indeterminacy of Hamburger [8], especially interesting because of a way of rewriting it in terms of a single ratio of determinants (see Theorem A.7). Note this aside deals with the Hamburger problem.

Proposition 5.13. A necessary and sufficient condition for a set of Hamburger moments to be indeterminate is that

(i) ∑_{j=0}^{∞} |P_{j}(0)|^{2} < ∞ and

(ii) ∑_{j=0}^{∞} |P_{j}^{(2)}(0)|^{2} < ∞

where P_{j}^{(2)}(x) are the orthogonal polynomials for the γ^{(2)} moment problem.

Proof. If {γ_{j}}_{j=0}^{∞} is indeterminate by Proposition 5.12, so is {γ_{j}^{(2)}}_{j=0}^{∞}, and then (i), (ii) follow by Theorem 3. The converse is more involved. The intuition is the following: If dρ is an even measure, then P_{2j}^{(2)}(x) = \sqrt{γ_{2}} \frac{P_{2j+1}(x)}{x}, so P_{m}^{(2)}(0) = \sqrt{γ_{2}} \frac{∂P_{m+1}}{∂x} (0) (both sides are zero if m is odd). So (ii) is equivalent to ∑_{j=0}^{∞} |\frac{∂P_{j}(0)}{∂x}|^{2} < ∞ and condition (v) of Theorem 3 says that (i), (ii) imply indeterminacy. Our goal is to prove in general that when (i) holds, P_{j}^{(2)}(0) and \frac{∂P_{j+1}}{∂x} (0) are essentially equivalent.

Let S_{n}(x) = P_{n}^{(2)}(x), η_{n} = \frac{∂P_{n}}{∂x} (0), and define

α_{n} = ∫ xS_{n}(x) dρ(x)  (5.42)
β_{n} = ∫ [xS_{n}(x)][P_{n+1}(x)] dρ(x),  (5.43)

where ρ is any solution of the γ moment problem. Since S_{n} is an orthogonal polynomial for dρ^{(2)}, we have

∫ [xS_{n}(x)]x^{j} xdρ(x) = 0, j = 0, 1, …, n − 1

and thus

∫ xS_{n}(x)P_{j}(x) dρ(x) = α_{n}P_{j}(0), j = 0, 1, …, n

so the orthogonal expansion for xS_{n}(x) is

xS_{n}(x) = β_{n}P_{n+1}(x) + α_{n} ∑_{j=0}^{n} P_{j}(0)P_{j}(x).  (5.44)

Since xS_{n}(x) vanishes at zero,

β_{n}P_{n+1}(0) + α_{n} ∑_{j=0}^{n} |P_{j}(0)|^{2} = 0.  (5.45)

Since

∫ |xS_{n}(x)|^{2} dρ(x) = γ_{2} ∫ S_{n}(x)^{2} dρ^{(2)}(x) = γ_{2},

we have by (5.44) that

γ_{2} = β_{n}^{2} + α_{n}^{2} ∑_{j=0}^{n} |P_{j}(0)|^{2}.  (5.46)

Moreover, taking derivatives of (5.44) at x = 0,

S_{n}(0) = β_{n}η_{n+1} + α_{n} ∑_{j=0}^{n} P_{j}(0)η_{j}.  (5.47)

By (5.46), β_{n} is bounded, so by (5.45),

|α_{n}| ≤ C_{1}|P_{n+1}(0)|.  (5.48)

By hypothesis, ∑_{j=0}^{∞} |P_{j}(0)|^{2} < ∞, so P_{n+1}(0) → 0 and thus by (5.46), β_{n} → \sqrt{γ_{2}} as n → ∞. Since β_{n} > 0 for all n (this follows from the fact that both xS_{n}(x) and P_{n+1}(x) have positive leading coefficient multiplying x^{n+1}), we see that β_{n}^{−1} is bounded. Using (5.48), the Schwarz inequality on the sum in (5.47), and ∑_{j=0}^{∞} |P_{j}(0)|^{2} < ∞ again, we see that

|η_{n+1}|^{2} ≤ C(|S_{n}(0)|^{2} + |P_{n+1}(0)|^{2} ∑_{j=0}^{n} |η_{j}|^{2})

≤ C(|S_{n}(0)|^{2} + |P_{n+1}(0)|^{2})(1 + ∑_{j=0}^{n} |η_{j}|^{2}).

It follows that

[1 + ∑_{j=0}^{n+1} |η_{j}|^{2}] ≤ [1 + C(|S_{n}(0)|^{2} + |P_{n+1}(0)|^{2})][1 + ∑_{j=0}^{n} |η_{j}|^{2}]

and so by induction that

sup_{1≤n<∞} ∑_{j=0}^{n+1} |η_{j}|^{2} ≤ sup_{1≤n<∞} ∏_{j=1}^{n+1}[1 + C(|S_{j−1}(0)|^{2} + |P_{j}(0)|^{2})] < ∞

since ∑_{j=0}^{∞} |P_{j}(0)|^{2}+∑_{j=0}^{∞} |S_{j}(0)|^{2} < ∞ by hypothesis. Thus, if (i) and (ii) hold, η_{n} ∈ ℓ_{2} and thus, the problem is indeterminate by Theorem 3. □

Theorem 5.14. Let {γ_{n}}_{n=0}^{∞} be a set of Stieltjes moments. Fix ℓ ≥ 1 and let P_{N}^{(ℓ)}(z), Q_{N}^{(ℓ)}(z), M_{N}^{(ℓ)}(z), and N_{N}^{(ℓ)}(z) be the orthogonal polynomials and other associated polynomials for the γ^{(ℓ)} moment problem. Let f^{[N,M]}(z) be the Padé approximants for the series of Stieltjes ∑_{j=0}^{∞} κ_{j}z^{j} where κ_{j} = (−1)^{j}γ_{j}. Then

f^{[N+ℓ−1,N]}(z) = ∑_{j=0}^{ℓ−1}(−1)^{j}γ_{j}z^{j} + (−1)^{ℓ}γ_{ℓ}z^{ℓ}[\frac{Q_{N}^{(ℓ)}(−\frac{1}{z})}{zP_{N}^{(ℓ)}(−\frac{1}{z})}]  (5.49)

and

f^{[N+ℓ−1,N−1]}(z) = ∑_{j=0}^{ℓ−1}(−1)^{j}γ_{j}z^{j} + (−1)^{ℓ}γ_{ℓ}z^{ℓ}[\frac{N_{N}^{(ℓ)}(−\frac{1}{z})}{zM_{N}^{(ℓ)}(−\frac{1}{z})}].  (5.50)

In particular:

(1) For each ℓ, (−1)^{ℓ}f^{[N+ℓ−1,N]}(x) is monotone increasing to a finite limit for all x ∈ (0, ∞).

(2) For all z ∈ ℂ\(−∞, 0], lim_{N→∞} f^{[N+ℓ−1,N]}(z) ≡ f_{ℓ}(z) exists.

(3) The γ^{(ℓ)} moment problem is (Stieltjes) determinate if and only if f_{ℓ}(z) = f_{ℓ+1}(z).

(4) If the γ^{(ℓ)} moment problem is determinate, then f_{0}(z) = f_{1}(z) = ⋯ = f_{ℓ+1}(z).

(5) If the γ^{(ℓ)} moment problem is indeterminate, then as x → ∞,

f_{ℓ+1}(x) = (−1)^{ℓ}γ_{ℓ}μ_{K}^{(ℓ)}({0})x^{ℓ} + O(x^{ℓ−1}),  (5.51)

where μ_{K}^{(ℓ)} is the Krein solution of the γ^{(ℓ)} problem. In particular if ℓ > 1, −f_{ℓ+1}(x) is not a Herglotz function (as it is if the problem is determinate).

Remarks. 1. Thus, it can happen (e.g., if γ is determinate but γ^{(1)} is not (cf. Corollary 4.21) that f^{[N−1,N]} and f^{[N,N]} have the same limit, but that f^{[N+1,N]} has a different limit.

2. If the {γ_{j}}_{j=0}^{∞} obey a condition that implies uniqueness and which is invariant under γ_{j} → γ_{j+1} (e.g., the condition of (1.12b) of Proposition 1.5), then all γ^{(ℓ)} are determinate; see Theorem 5.19 below.

3. Even for ℓ = 1 and consideration of f^{[N−1,N]} and f^{[N,N]}, we have something new — the remarkable fact that xdμ_{K}(x)/γ_{1}^{−1} is dμ_{F}^{(1)}(x). Multiplication by x kills the point measure at x = 0 and produces a measure supported on some set [R, ∞) with R > 0. We also see that monotonicity for f_{N}^{−} implies monotonicity of f_{N}^{+}. The direction of monotonicity flips because of the minus sign ((−1)^{ℓ} = −1) in (5.49).

4. In particular, we have P_{N}^{(1)}(x) = c_{N}x^{−1}M_{N+1}(x), that is, for any ℓ ≠ m, ∫ x^{−1}M_{m}(x)M_{ℓ}(x) dρ(x) = 0, something that can be checked directly.

5. This theorem implies that all the f^{[N,M]} with N ≥ M − 1 exist and their denominators obey a three-term recursion relation.

6. The connection in Remark 3 extended to xdμ_{K}^{(ℓ)}(x) = c_{ℓ}dμ_{F}^{(ℓ+1)}(x) implies that if any γ^{(ℓ)} is indeterminate, then dμ_{K}(x) is a point measure. Thus if κ_{n} = (−1)^{n} ∫_{0}^{∞} x^{n} dρ(x) where ρ is associated to a determinate Stieltjes problem and dρ is not a point measure, then all the f_{ℓ}(z)’s, ℓ = 0, 1, 2, …, are equal.

Proof. It is easy to check that the right side of (5.49) and (5.50) are ratios of polynomials whose degrees are at most of the right size. (Because P_{N}^{(ℓ)}(0) ≠ 0 ≠ M_{N}^{(ℓ)′}(0), it is easy that the denominators are always precisely of the right size.) Since M_{N}^{(ℓ)}(z) and P_{N}^{(ℓ)}(z) are of degree n, the rationalized denominators do not vanish at \frac{1}{z} = 0. By Proposition 5.11 for the γ^{(ℓ)} problem, they have the proper asymptotic series to be f^{[N+ℓ−1,N]}(z) and f^{[N+ℓ−1,N−1]}(z), respectively. That proves (5.49) and (5.50).

Assertions (1), (2), (3) are then just Theorem 6 for the γ^{(ℓ)} moments. To prove assertion (4), note that if γ^{(ℓ)} is determined by Proposition 5.12, so are γ^{(ℓ−1)}, γ^{(ℓ−2)}, …, γ^{(1)}, γ, and so by (3), we have f_{ℓ+1} = f_{ℓ} = ⋯ = f_{0}.

To prove assertion (5), note that if the γ^{(ℓ)} moment problem is indeterminate, dμ_{K}^{(ℓ)}({0}) > 0. Thus,

lim_{x→∞} ∫ \frac{dμ_{K}^{(ℓ)}(y)}{1 + xy} = μ_{K}^{(ℓ)}\{0\} > 0.  (5.52)

By (5.50),

f_{ℓ+1}(x) = ∑_{j=0}^{ℓ−1}(−1)^{j}γ_{j}x^{j} + (−1)^{ℓ}γ_{ℓ}x^{ℓ} ∫_{0}^{∞} \frac{dμ_{K}^{(ℓ)}(y)}{1 + xy} ,

so (5.52) implies (5.51). □

To deal with the sequences f^{[N−1+ℓ,N]}(z) with ℓ < 0, we need to introduce yet another modified moment problem associated to a set of Hamburger (or Stieltjes) moments. To motivate what we are looking for, let ρ ∈ M^{H}(γ) and let G_{ρ}(z) be the associated Stieltjes transform. Then −G_{ρ}(z)^{−1} is a Herglotz function which has an asymptotic series to all orders as z → i∞. Since G_{ρ}(z) ∼ −\frac{1}{z}(1 + γ_{1}z^{−1} + γ_{2}z^{−2} + O(z^{−3})),

−G_{ρ}(z)^{−1} ∼ z − γ_{1} − (γ_{2} − γ_{1}^{2})z^{−1} + O(z^{−2})

so (by the proof of Proposition 4.13) the Herglotz representation of −G_{ρ}(z)^{−1} has the form

−G_{ρ}(z)^{−1} = z − γ_{1} + (γ_{2} − γ_{1}^{2}) ∫ \frac{dρ̃(x)}{x − z} ,  (5.53)

where dρ̃ is a normalized measure. Its moments, which we will call γ_{j}^{(0)}, only depend on the asymptotic series for G_{ρ}(z), and so only on the original moments. We could do everything abstractly using (5.53), but we will be able to explicitly describe the relation between the moment problems.

The key is the following formula, a Ricatti-type equation well known to practitioners of the inverse problem [6] (which we will, in essence, prove below):

−m_{0}(z)^{−1} = z − b_{0} + a_{0}^{2}m_{1}(z),  (5.54)

where m_{0}(z) = ⟨δ_{0}, (A − z)^{−1}δ_{0}⟩ is just G_{ρ}(z) for a spectral measure, and m_{1}(z) = ⟨δ_{0}, (A^{[1]} −z)^{−1}δ_{0}⟩, where A^{[1]} is obtained from A by removing the top row and left-most column. ( 5.54) is just (5.53) if we note that b_{0} = γ_{1} and a_{0}^{2} = γ_{2} − γ_{1}^{2}. Thus, we are led to define γ_{j}^{(0)} as the moments associated to the Jacobi matrix A^{[1]} obtained by removing the top row and left-most column, that is,

A^{[1]} = ( \begin{matrix}b_{1} & a_{1} & 0 & … & … \\ a_{1} & b_{2} & a_{2} & … & … \\ 0 & a_{2} & b_{3} & a_{3} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \end{matrix} ) .

Proposition 5.15. Let {γ_{j}}_{j=0}^{∞} be a set of Hamburger moments. Then the {γ_{j}^{(0)}}_{j=0}^{∞} Hamburger problem is determinate if and only if {γ_{j}}_{j=0}^{∞} is.

Proof. Let Ã^{[1]} be A^{[1]} with a row of zeros added at the top and column of zeros on the left. Since Ã^{[1]} = 0⊕A^{[1]}, Ã^{[1]} is essentially self-adjoint if and only if A^{[1]} is. A−Ã^{[1]} is a matrix with three non-zero elements and essential self-adjointness is provided by bounded perturbations. Thus, A is essentially self-adjoint if and only if A^{[1]} is. By Theorem 2, we have the equivalence of determinacy for the Hamburger problem. □

Remark. As we will see shortly, this result is not true in the Stieltjes case.

Let P_{N}^{(0)}(x), Q_{N}^{(0)}(x), and f_{N}^{(0)}(z) = −\frac{Q_{N}^{(0)}(z)}{P_{N}^{(0)}(z)} be the polynomials and finite matrix

approximation for the γ^{(0)} problem. Then

Proposition 5.16.

(i)  P_{N}^{(0)}(x) = a_{0}Q_{N+1}(x)  (5.56)
(ii)  Q_{N}^{(0)}(x) = −\frac{1}{a_{0}} P_{N+1}(x) + \frac{x − b_{0}}{a_{0}} Q_{N+1}(x)  (5.57)
(iii)  − f_{N+1}^{−}(z)^{−1} = z − b_{0} + a_{0}^{2}f_{N}^{(0)}(z)  (5.58)

Remarks. 1. (i) implies that the Q_{N}’s are orthogonal polynomials for some measure. 2. (5.58) is (5.54).

Proof. For each fixed x, P_{N}^{(0)}(x), Q_{N}^{(0)}(x) obey the same difference equation as P_{N+1}(x), Q_{N+1}(x) so we need (5.56), (5.57) at the initial points N = −1, 0 where it is required that

P_{−1}^{(0)}(x) = 0, P_{0}^{(0)}(x) = 1; Q_{−1}^{(0)}(x) = −\frac{1}{a_{0}} , Q_{0}^{(0)}(x) = 0

(Q_{−1}^{(0)}(x) = −\frac{1}{a_{0}} since we don’t make the convention a_{−1}^{(0)} = 1, but rather a_{−1}^{(0)} = a_{0}

consistent with a_{n}^{(0)} = a_{n+1}). This is consistent with (5.56) if we note that

Q_{0}(x) = 0, Q_{1}(x) = \frac{1}{a_{0}} ; P_{0}(x) = 1, P_{1}(x) = \frac{(x − b_{0})}{a_{0}} .

This proves (i), (ii).

They in turn imply

a_{0}^{2} \frac{Q_{N}^{(0)}(x)}{P_{N}^{(0)}(x)} = x − b_{0} − \frac{P_{N+1}(x)}{Q_{N+1}(x)} ,

which, by (5.12), proves (iii). □

Proposition 5.17. Let {γ_{j}}_{j=0}^{∞} be a set of Stieltjes moments. If the Hamburger problem is indeterminate, then the γ^{(0)} Stieltjes problem is indeterminate (even if the γ Stieltjes problem is determinate).

Proof. As we will see below (Proposition 5.22 ≡ Theorem 8), a set of Stieltjes moments which is Hamburger indeterminate is Stieltjes determinate if and only if L ≡ lim_{N→∞}[−\frac{Q_{N} (0)}{P_{N} (0)}] is infinite. By (5.58),

a_{0}^{2}L^{(0)} = b_{0} − L^{−1}.

Since L > 0, L^{(0)} is never infinite. □

Remarks. 1. Thus, if γ is Hamburger indeterminate but Stieltjes determinate, γ and γ^{(0)} have opposite Stieltjes determinacy.

2. The spectral way to understand this result is to note that if γ is Hamburger indeterminate, the Friedrichs extensions of A and A^{[1]} have interlacing eigenvalues. Given that A ≥ 0, A^{[1]} must be strictly positive.

For ℓ < 0 and integral, we let

γ_{j}^{(ℓ)} = [γ^{(0)}]_{j}^{(−ℓ)} = \frac{γ_{j−ℓ}^{(0)}}{γ_{−ℓ}^{(0)}} .

Then:

Theorem 5.18. Let {γ_{n}}_{n=0}^{∞} be a set of Stieltjes moments. Fix ℓ ≤ 0 and let P_{N}^{(ℓ)}(z), Q_{N}^{(ℓ)}(z), M_{N}^{(ℓ)}(z), and N_{N}^{(ℓ)}(z) be the orthogonal polynomials and other associated polynomials for the γ^{(ℓ)} moment problem. Let f^{[N,M]}(z) be the Padé aproximants for the series of Stieltjes ∑_{j=0}^{∞} κ_{j}z^{j} where κ_{j} = (−1)^{j}γ_{j}. Then

f^{[N+ℓ−1,N]}(z) = \{1 − γ_{1}z − ∑_{j=0}^{−ℓ−1}(−1)^{j}γ_{j}^{(0)}z^{j+2} + (−1)^{ℓ+1}γ_{−ℓ}^{(0)}z^{−ℓ+2}[\frac{Q_{N+ℓ−1}^{(ℓ)}(−\frac{1}{z})}{zP_{N+ℓ−1}^{(ℓ)}(−\frac{1}{z})}]\}^{−1}

(5.59)

and

f^{[N+ℓ−1,N+1]}(z) = \{1−γ_{1}z− ∑_{j=0}^{−ℓ−1}(−1)^{j}γ_{j}^{(0)}z^{j+2}+(−1)^{ℓ+1}γ_{−ℓ}^{(0)}z^{−ℓ+2}[\frac{N_{N+ℓ−1}^{(ℓ)}(−\frac{1}{z})}{zM_{N+ℓ}^{(ℓ)}(−\frac{1}{z})}]\}^{−1}.

(5.60)

In particular,

(1) For each ℓ, (−1)^{ℓ}f^{[N+ℓ−1,N]}(x) is monotone increasing to a finite limit for all x ∈ (0, ∞).

(2) For all z ∈ ℂ\(−∞, 0], lim_{N→∞} f^{[N+ℓ−1,N]}(z) ≡ f_{ℓ}(z) exists.

(3) The γ^{(ℓ)} moment problem is Stieltjes determinate if and only if f_{ℓ}(z) = f_{ℓ−1}(z).

(4) If the γ^{(ℓ)} moment problem is determinate, then f_{0}(z) = f_{−1}(z) = ⋯ = f_{ℓ}(z) = f_{ℓ−1}(z).

(5) If the γ^{(ℓ)} moment problem is indeterminate, then as x → ∞,

f_{ℓ−1}(x) = (−1)^{ℓ+1}(γ_{−ℓ}^{(0)})^{−1}x^{ℓ−2}μ_{K}^{(ℓ)}({0})^{−1} + O(x^{ℓ−3}),

where μ_{K}^{(ℓ)} is the Krein solution of the γ^{(ℓ)} problem. In particular, −f_{ℓ−1}(x) is not a Herglotz function (as it is if the problem is determinate).

Proof. By (5.58), we have a relation between formal power series:

(∑_{j=0}^{∞}(−1)^{j}γ_{j}z^{j})^{−1} = 1 + γ_{1}z − (γ_{2} − γ_{1}^{2})z^{2}[∑_{j=0}^{∞}(−1)^{j}γ_{j}^{(0)}z^{j}],  (5.61)

from which one obtains

f^{[M,N+2]}(z) = (1 + γ_{1}z − (γ_{2} − γ_{1}^{2})z^{2}f^{(0)[N,M]}(z))^{−1}.

Thus, (5.59) is just (5.49) and (5.60) is (5.50). The consequence (1)–(5) follows as in the proof of Theorem 5.14. □

We summarize and extend in the following:

Theorem 5.19. Let ∑_{j=0}^{∞} κ_{j}z^{j} be a series of Stieltjes. Then for each ℓ = 0, ±1, ±2, …, f_{ℓ}(z) = lim_{N→∞} f^{[N+ℓ,N]}(z) exists uniformly for z in compact subsets of ℂ\(−∞, 0]. Moreover,

(1) If any two f_{ℓ}’s are different, then all of them are meromorphic.

(2) If the |κ_{j}| = γ_{j} moment problem is determinate and the measure solving the moment problem is not a discrete point measure, then all f_{ℓ}’s are equal.

(3) If |κ_{j}| ≤ C^{j}(2j)!, then all f_{ℓ}’s are equal.

Proof. The first assertion is Theorem 5.14 for ℓ ≥ 1 and Theorem 5.18 for ℓ ≤ 0. If some f_{ℓ} ≠ f_{ℓ+1}, then some γ^{(ℓ)} are indeterminate, so the corresponding dρ_{F} ’s are pure point and f_{ℓ}’s meromorphic. This proves (1). Under the hypothesis of (2), f_{0} is not meromorphic, so by (1), all f’s are equal. To prove (3), an induction using (5.61) proves |γ_{j}^{(0)}| ≤ C̃^{j}(2j)!, so by Proposition 1.5, all the γ^{(ℓ)} moment problems are determinate. □

This completes our discusion of Padé approximants for series of Stieltjes. We return to consequences of Theorem 5.2 for the study of the Stieltjes moment problem.

As we have seen for any x > 0, −\frac{Q_{N} (−x)}{P_{N} (−x)} is monotone increasing and −\frac{xN_{N} (−x)}{M_{N} (−x)}

is monotone decreasing. By taking limits (recall M_{N}(0) = 0), we see that −\frac{Q_{N} (0)}{P_{N} (0)} is

monotone increasing and \frac{N_{N} (0)}{M′_{N} (0)} is decreasing.

Proposition 5.20.

(i)  lim_{N→∞} −\frac{Q_{N}(0)}{P_{N}(0)} = ∫ y^{−1} dμ_{F} (y)
(ii)  lim_{N→∞} \frac{N_{N}(0)}{M′_{N}(0)} = μ_{K}(\{0\}),

where μ_{F} (resp. μ_{K}) is the Friedrichs (resp. Krein) solution.

Proof. By Theorem 5.2 and Proposition 5.6, for x > 0,

−\frac{Q_{N}(−x)}{P_{N}(−x)} ≤ ∫ (y + x)^{−1} dμ_{F} (y),

so taking x to zero and N to infinity, we see that

lim_{N→∞} −\frac{Q_{N}(0)}{P_{N}(0)} ≤ ∫ y^{−1} dμ_{F} (y).

On the other hand, since −\frac{Q_{N} (−x)}{P_{N} (−x)} = (δ_{0}, (A_{F}^{[N]} + x)^{−1}δ_{0}) is monotone increasing as x decreases, we have for each N and x > 0,

−\frac{Q_{N}(−x)}{P_{N}(−x)} ≤ lim_{N→∞} −\frac{Q_{N}(0)}{P_{N}(0)} ,

66 B. SIMON

so taking N to infinity for fixed x > 0 and using Theorem 5.2,

∫ (x + y)^{−1} dμ_{F} (y) ≤ lim_{N→∞} −\frac{Q_{N}(0)}{P_{N}(0)} .

Taking x to zero, we see that

∫ y^{−1} dμ_{F} (y) ≤ lim_{N→∞} −\frac{Q_{N}(0)}{P_{N}(0)} ,

so (i) is proven.

The proof of (ii) is similar. By Theorem 5.2 and Proposition 5.8,

−\frac{xN_{N}(−x)}{M_{N}(−x)} ≥ ∫ \frac{x}{y + x} dμ_{K}(y) ≥ μ_{K}(\{0\}),

so taking x to zero and N to infinity,

lim_{N→∞} \frac{N_{N}(0)}{M′_{N}(0)} ≥ μ_{K}(\{0\}).

On the other hand, since −\frac{xN_{N} (−x)}{M_{N} (−x)} = (δ_{0}, x(A_{K}^{[N]} + x)^{−1}δ_{0}) is monotone decreasing as x decreases, we have for each N and x > 0,

−\frac{xN_{N}(−x)}{M_{N}(−x)} ≥ lim_{N→∞} \frac{N_{N}(0)}{M′_{N}(0)} ,

so taking N to infinity for fixed x > 0 and using Theorem 5.2,

∫ \frac{x}{x + y} dμ_{K}(y) ≥ lim_{N→∞} \frac{N_{N}(0)}{M′_{N}(0)} .

Taking x to zero, we see that

μ_{K}(\{0\}) ≥ lim_{N→∞} \frac{N_{N}(0)}{M′_{N}(0)} ,

so (ii) is proven. □

This leads us to define (note that we use \frac{M′_{N} (0)}{N_{N} (0)}, not \frac{N_{N} (0)}{M′_{N} (0)} so M, L ∈ (0, ∞) ∪ \{∞\}):

L = lim_{N→∞} −\frac{Q_{N}(0)}{P_{N}(0)}  (5.62)
M = lim_{N→∞} \frac{M′_{N}(0)}{N_{N}(0)},  (5.63)

so Proposition 5.20 says that

L = ∫ y^{−1} dμ_{F} (y)
M^{−1} = μ_{K}({0}).

By (4.25), μ_{K}({0}) = 1/ ∑_{n=0}^{∞} |P_{N}(0)|^{2}, so

M = ∑_{n=0}^{∞} |P_{n}(0)|^{2}.  (5.64)

Note. (4.25) was only proven in the indeterminate case, but the argument applies in the determinate case also. If x is an eigenvalue of a solution μ of the moment problem associated to a self-adjoint extension, then μ({x}) = 1/ ∑_{n=0}^{∞} |P_{n}(x)|^{2}.

Theorem 5.21 (≡ Theorem 7). Let {γ_{n}}_{n=0}^{∞} be the moments of a Stieltjes problem. Then the problem is indeterminate if and only if

L < ∞  and  M < ∞.

Equivalently, the problem is determinate if and only if

L = ∞  or  M = ∞.

Proof. If L < ∞ and M < ∞, then ∫ y^{−1}dμ_{F} (y) < ∞, while μ_{K}({0}) > 0, so clearly, μ_{F} ≠ μ_{K} and the problem is indeterminate. Conversely, if the problem is indeterminate, then by Proposition 3.1, α = inf spec(A_{F}) > 0, so ∫ y^{−1} dμ_{F} (y) ≤ α^{−1} < ∞ and L < ∞. Moreover, since the Stieltjes problem is indeterminate, so is the Hamburger problem; and thus by (5.64), M < ∞. □

Theorem 5.22 (≡ Theorem 8). Let {γ_{n}}_{n=0}^{∞} be a set of Stieltjes moments. Then the Stieltjes problem is determinate while the Hamburger problem is indeterminate if and only if

∑_{n=0}^{∞} |Q_{n}(0)|^{2} < ∞  (5.65)

and L = ∞.

Proof. For a set of Stieltjes moments −\frac{Q_{n}(0)}{P_{n}(0)} is positive and monotone increasing, so

|Q_{n}(0)| ≥ \frac{|Q_{1}(0)P_{n}(0)|}{|P_{1}(0)|}. Since Q_{1}(0) ≠ 0, we see that (5.65) implies also that M < ∞. Thus, (5.65) is equivalent to the Hamburger problem being indeterminate. Given that M < ∞, Theorem 5.21 says that determinacy of the Stieltjes problem is equivalent to L = ∞. □

As our next topic, we will further examine the conditions M < ∞ and L < ∞ to see they are conditions of Stieltjes and Krein in a different form. Define

ℓ_{n} = −\frac{Q_{n}(0)}{P_{n}(0)} + \frac{Q_{n−1}(0)}{P_{n−1}(0)} ,  n ≥ 1  (5.66)

and

m_{n} = \frac{M′_{n}(0)}{N_{n}(0)} − \frac{M′_{n−1}(0)}{N_{n−1}(0)} ,  n ≥ 2  (5.67a)
m_{1} = \frac{M′_{1}(0)}{N_{1}(0)} .  (5.67b)

By the monotonicity properties of this section, ℓ_{n} > 0, m_{n} > 0. By definition and Q_{0}(0) = 0,

−\frac{Q_{N}(0)}{P_{N}(0)} = ∑_{n=1}^{N} ℓ_{n},  L = ∑_{n=1}^{∞} ℓ_{n}

\frac{M′_{N}(0)}{N_{N}(0)} = ∑_{n=1}^{N} m_{n},  M = ∑_{n=1}^{∞} m_{n}.

Proposition 5.23. For N ≥ 1,

m_{N} = |P_{N−1}(0)|^{2}  (5.68)
ℓ_{N} = −[a_{N−1}P_{N}(0)P_{N−1}(0)]^{−1}.  (5.69)

Remark. Since A_{F}^{[N]} > 0, P_{N}(z) has no zeros on (−∞, 0] and thus, P_{N}(z) has the same sign near −∞ and 0. But P_{N}(z) = c_{N}z^{N} + lower order with c_{N} > 0, so (−1)^{N}P_{N}(0) > 0 (cf. (5.13)). Thus, P_{N}(0)P_{N−1}(0) < 0 and the minus sign in (5.69) is just what is needed to ensure that ℓ_{N} > 0.

Proof. A_{K}^{[N]} has {P_{0}(0), …, P_{N−1}(0)} as its eigenfunction with eigenvalue zero. Thus, lim_{x→0} x(δ_{0}, (A_{K}^{[N]} + x)^{−1}δ_{0}) = |P_{0}(0)|^{2}/ ∑_{j=0}^{N−1} |P_{j}(0)|^{2}. Since P_{0}(0) = 1, we see that

\frac{M′_{N}(0)}{N_{N}(0)} = ∑_{j=0}^{N−1} |P_{j}(0)|^{2},  (5.70)

which implies (5.68). (5.69) follows immediately from Proposition 4.1. □

Corollary 5.24. If A given by (1.16) is the Jacobi matrix associated to an indeterminate Stieltjes moment problem, then

∑_{n=0}^{∞} a_{n}^{−1/2} < ∞.

In particular, if ∑_{n=0}^{∞} a_{n}^{−1/2} = ∞ for the Jacobi matrix associated to a Stieltjes problem, then the problem is determinate.

Proof. If the problem is indeterminate, by Proposition 5.23 and Theorem 5.21, a_{n}^{−1/2}|P_{n}(0)P_{n−1}(0)|^{−1/2} and |P_{n}(0)P_{n−1}(0)|^{1/2} both lie in ℓ^{2}, so their product lies in ℓ^{1}. □

Now we will define, following Stieltjes and Krein, functions

U_{n}(x) = \frac{P_{n}(−x)}{P_{n}(0)} ,  n ≥ 0  (5.71)
V_{n}(x) = −\frac{Q_{n}(−x)}{P_{n}(0)} ,  n ≥ 0  (5.72)
G_{n}(x) = −a_{n−1}M_{n}(−x)P_{n−1}(0),  n ≥ 1  (5.73)
H_{n}(x) = a_{n−1}N_{n}(−x)P_{n−1}(0),  n ≥ 1.  (5.74)

We claim that

Proposition 5.25.

(i) U_{n}(0) = 1 (ii) G_{n}(0) = 0 (iii) H_{n}(0) = 1 (iv) U_{n}(x)H_{n}(x) − V_{n}(x)G_{n}(x) = 1 for n ≥ 1

(v) f_{n}^{+}(−x) − f_{n}^{−}(−x) = \frac{1}{U_{n}(x)G_{n}(x)} for n ≥ 1

Proof. (i) is immediate from the definition (5.71), (ii) from M_{n}(0) = 0, and (iii) follows from Proposition 4.1 and the definition (5.16) of N_{N} (and explains why we multiply M and N by a_{n−1}P_{n−1}(0)). To prove (iv), we note that for some constants α_{n}, β_{n}, γ_{n},

U_{n}(x) = α_{n}P_{n}(−x),  V_{n}(x) = −α_{n}Q_{n}(−x)
G_{n}(x) = β_{n}P_{n}(−x) + γ_{n}P_{n−1}(−x),  H_{n}(x) = −β_{n}Q_{n}(−x) − γ_{n}Q_{n−1}(−x),

so

U_{n}(x)H_{n}(x) − V_{n}(x)G_{n}(x) = −α_{n}γ_{n}[P_{n}(−x)Q_{n−1}(−x) − Q_{n}(−x)P_{n−1}(−x)]

is constant as x is varied for fixed n. But at x = 0, by (i)–(iii) this combination is 1. (v) follows from (iv) and the definitions. □

The following says that for x > 0, U, G can be associated with the equation of motion of a string of beads (see [1]). This is the starting point of deep work of Krein [17].

Theorem 5.26.

(i) U_{n}(x) − U_{n−1}(x) = ℓ_{n}G_{n}(x), n ≥ 1 (ii) G_{n+1}(x) − G_{n}(x) = m_{n+1}xU_{n}(x), n ≥ 1 (iii) G_{1}(x) = m_{1}x U_{0}(x) = 1

Proof. (i) By definition of U_{n},

U_{n}(x) − U_{n−1}(x) = \frac{M_{n}(−x)}{P_{n}(0)} = ℓ_{n}G_{n}(x)

by (5.69) and (5.73).

(ii) We have for n ≥ 1,

a_{n}P_{n+1}(−x) + b_{n}P_{n}(−x) + a_{n−1}P_{n−1}(−x) = −xP_{n}(−x).  (5.75)

(5.75) for x = 0 implies that

b_{n} = −a_{n}P_{n+1}(0)P_{n}(0)^{−1} − a_{n−1}P_{n−1}(0)P_{n}(0)^{−1},  (5.76)

which we can substitute into (5.75) to obtain

a_{n}M_{n+1}(−x) − a_{n−1}P_{n−1}(0)P_{n}(0)^{−1}M_{n}(−x) = −xP_{n}(x).

Multiplying by −P_{n}(0) and using the definitions (5.71)/(5.73), we see

G_{n+1}(x) − G_{n}(x) = xP_{n}(0)^{2}U_{n}(x)

so (5.68) implies the result.

(iii) U_{0}(x) is a constant so U_{0}(0) = 1 implies U_{0}(x) ≡ 1. G_{1} is a linear polynomial with G_{1}(0) = 0, so G_{1}(x) = G′_{1}(0)x. For any n, G′_{n}(0) = lim_{x↓0} \frac{x^{−1}G_{n}(x)}{H_{n}(x)} = ∑_{j=1}^{n} m_{j}. □

We can use these equations to get further insight into Theorem 5.2. First, we obtain explicit bounds on the rate of convergence of f_{N}^{+}(−x)−f_{N}^{−}(−x) to zero in the determinate case, where either ∑ m_{j} or ∑ ℓ_{j} or both diverge.

Theorem 5.27. For x > 0,

f_{N}^{+}(−x) − f_{N}^{−}(−x) ≤ [m_{1}x^{2}(∑_{j=1}^{n} m_{j})(∑_{j=1}^{n} ℓ_{j})]^{−1}.

Proof. It follows from Theorem 5.26 and induction that U_{n}(x) and G_{n}(x) are nonnegative for x > 0 and then that they are monotone in n. In particular, U_{n}(x) ≥ 1, G_{n}(x) ≥ m_{1}x from which, by (i), (ii),

U_{n}(x) ≥ m_{1}(∑_{j=1}^{n} ℓ_{j})x

G_{n}(x) ≥ (∑_{j=1}^{n} m_{j})x.

(v) of Proposition 5.25 then completes the proof. □

And we obtain directly that in the indeterminate case, f^{±} are meromorphic functions.

Theorem 5.28. (i) For any z ∈ ℂ, we have that for n ≥ 1,

|U_{n}(z)| ≤ ∏_{j=1}^{n}(1 + ℓ_{j}) ∏_{j=1}^{n}(1 + m_{j}|z|)

|G_{n}(z)| ≤ ∏_{j=1}^{n−1}(1 + ℓ_{j}) ∏_{j=1}^{n}(1 + m_{j}|z|).

(ii) If M < ∞ and L < ∞, then for all z ∈ ℂ, U_{n}(z), V_{n}(z), G_{n}(z), H_{n}(z) converge to functions U_{∞}(z), V_{∞}(z), G_{∞}(z), H_{∞}(z), which are entire functions obeying

|f(z)| ≤ C_{ε} exp(ε(z))

for each ε > 0.

(iii) f^{−}(z) = \frac{V_{∞}(−z)}{U_{∞}(−z)}, f^{+}(z) = \frac{H_{∞}(−z)}{G_{∞}(−z)} .

(iv) f^{+}(z) − f^{−}(x) = \frac{1}{U_{∞}(−z)G_{∞}(−z)} .

Remark. In terms of the Nevanlinna functions A, B, C, D, one can see (using the fact that the Friedrichs solution is associated to B_{t} with t = L) that G_{∞}(−z) = −D(z), H_{∞}(−z) = C(z), U_{∞}(−z) = −B(z) − LD(z), and V_{∞}(−z) = A(z) + LC(z), where L = ∑_{j=1}^{∞} ℓ_{j}.

Proof. (i) follows by an elementary induction from Theorem 5.26. Similarly, one follows the proof of that theorem to show that V_{n}, H_{n} obey

V_{n}(x) − V_{n−1}(x) = ℓ_{n}H_{n}(x),  n ≥ 1
H_{n+1}(x) − H_{n}(x) = m_{n+1}xV_{n}(x),  n ≥ 1
H_{1}(x) = 1 V_{0}(x) = 0

and obtains inductively that

|V_{n}(z)| ≤ ∏_{j=1}^{n}(1 + ℓ_{j}) ∏_{j=2}^{n}(1 + m_{j}|z|)

|H_{n}(z)| ≤ ∏_{j=1}^{n−1}(1 + ℓ_{j}) ∏_{j=2}^{n}(1 + m_{j}|z|).

Thus, if L, M < ∞, we first see that U, V, H, G are bounded and then, since ∑_{j≥n} ℓ_{j} → 0, ∑_{j≥n} m_{j} → 0, that each sequence is Cauchy as n → ∞. The C_{ε} bound is easy for products ∏_{j=1}^{∞}(1 + m_{j}|z|) with ∑_{1}^{∞} m_{j} < ∞. (iii), (iv) are then immediate from the definitions. □

To make the link to Stieltjes’ continued fractions, we need to note the relation between a_{n}, b_{n}, ℓ_{n}, and m_{n}. Immediately from Proposition 5.23, we have

a_{n} = (ℓ_{n+1}\sqrt{m_{n+1}m_{n+2}} )^{−1},  n ≥ 0  (5.77)

and by (5.76),

b_{n} = m_{n+1}^{−1}(ℓ_{n}^{−1} + ℓ_{n+1}^{−1}),  n ≥ 1  (5.78)
b_{0} = m_{1}^{−1}ℓ_{1}^{−1}.  (5.79)

(Parenthetically, we note that these equations can be used inductively to define m_{j}, ℓ_{j} given a_{j}, b_{j}. We have m_{1} = 1, so (5.79) gives ℓ_{1}. Given ℓ_{1}, …, ℓ_{j} and m_{1}, …, m_{j}, we can use a_{j−1} and (5.77) to find m_{j+1} and then b_{j} and (5.78) to get ℓ_{j+1}.)

Stieltjes’ continued fractions are of the form (5.1b). Let

d_{0} = c_{1}^{−1},  d_{n} = (c_{n}c_{n+1})^{−1}  (5.80)

for n ≥ 1 so (5.1b) becomes

\frac{d_{0}}{w + \frac{d_{1}}{1 + \frac{d_{2}}{w + …}}}

Now use the identity

w + \frac{β_{1}}{1 + \frac{β_{2}}{f(w)}} = w + β_{1} − \frac{β_{1}β_{2}}{β_{2} + f(w)}

to see that (5.1b) has the form (5.1a) if w = −z, d_{0} = 1, and

b_{0} = d_{1}  (5.81a)
b_{n} = d_{2n+1} + d_{2n}  (5.81b)
a_{n}^{2} = d_{2n+1}d_{2n+2}.  (5.81c)

Thus, (5.80) and (5.81) are consistent with (5.77)–(5.79) if and only if

c_{2j−1} = m_{j},  c_{2j} = ℓ_{j}.  (5.82)

Thus, we have seen that (5.1a) is (5.1b) if w = −z and the c’s are given by (5.82). Stieltjes’ criterion that depends on whether or not ∑_{j=1}^{∞} c_{j} < ∞ is equivalent to L < ∞ and M < ∞.

The connection (5.82) is not surprising. In (5.1b) if w = 0, the continued fraction formally reduces to c_{2} + c_{4} + ⋯ consistent with ∑_{j=1}^{∞} ℓ_{j} = − lim_{N→∞} \frac{Q_{N} (0)}{P_{N} (0)}. On the other hand, if we multiply by w, the continued fraction formally becomes

\frac{1}{c_{1} + \frac{1}{c_{2}w + \frac{1}{c_{3} + ⋯}}}

which, formally at w = 0, is (c_{1} +c_{3} +⋯)^{−1}, consistent with ∑_{j=1}^{∞} m_{j} = lim_{N→∞} lim_{x↓0} [−\frac{xN_{N} (−x)}{M_{N} (−x)}].

We conclude this section by discussing Padé approximants for the Hamburger analog of series of Stieltjes. A series of Hamburger is a formal power series ∑_{j=0}^{∞} κ_{j}z^{j} with

κ_{j} = (−1)^{j} ∫_{−∞}^{∞} x^{j} dρ(x)

for some measure ρ. We will begin with the “principal” Padé approximants f^{[N−1,N]}(z):

Theorem 5.29. Let ∑_{j=0}^{∞} κ_{j}x^{j} be a series of Hamburger. Then:

(i) The f^{[N−1,N]}(z) Padé approximants always exist and are given by

f^{[N−1,N]}(z) = (δ_{0}, (1 + zA_{F}^{[N]})^{−1}δ_{0}) = −\frac{z^{N−1}Q_{N}(−\frac{1}{z})}{z^{N}P_{N}(\frac{1}{z})} .  (5.83)

(ii) If the associated Hamburger moment problem is determinate, then for any z with Im z ≠ 0,

lim_{N→∞} f^{[N−1,N]}(z) = −\frac{z^{N−1}Q_{N}(−\frac{1}{z})}{z^{N}P_{N}(\frac{1}{z})}  (5.84)

exists and equals ∫ \frac{dρ(x)}{1+zx} for the unique solution, ρ, of the moment problem. (iii) The sequence f^{[N−1,N]}(z) is pre-compact in the family of functions analytic in

ℂ_{+} (in the topology of uniform convergence on compact sets). (iv) Any limit of f^{[N−1,N]}(z) is of the form ∫ \frac{dρ(x)}{1+zx} where ρ is a von Neumann solution

of the Hamburger problem.

Proof. (i) The proof that (5.12) holds (Proposition 5.6) is the same as in the Stieltjes case, and from there we get (5.34) as in the Stieltjes case. Since lim_{z→0} z^{N}P_{N}(\frac{1}{z}) ≠ 0, we see f^{[N−1,N]}(z) exists and is given by (5.83).

(iii) Since

|zf^{[N−1,N]}(z)| = |(δ_{0}, (A_{F}^{[N]} + z^{−1})^{−1}δ_{0})| ≤ |Im(z^{−1})|^{−1},

we see that

|f^{[N−1,N]}(z)| ≤ |z| |Im(z)|^{−1}

is bounded on compacts. By the Weierstrass-Vitali theorem, such functions are precompact.

(ii), (iv) By the proof of Proposition 4.13, if dρ is any measure whose first M + 2 moments are γ_{0}, γ_{1}, …, γ_{M+2}, then

|y^{M+1}[G_{ρ}(iy) + ∑_{n=0}^{M}(−i)^{n+1}y^{−n−1}γ_{n}]| ≤ ay^{−1}  (5.85)

with a only depending on \{γ\} (either a = γ_{M+1} or a = \frac{1}{2}[γ_{M}+γ_{M+2}]). Since ⟨δ_{0}, (A_{F}^{[N]})^{j}δ_{0}⟩ = γ_{j} for j ≤ 2N − 1, we see that for M fixed, (5.85) holds uniformly for N large if G_{N}(z) = −z^{−1}f^{[N−1,N]}(−\frac{1}{z}), so any limit point of the G’s obeys (5.85) and is Herglotz. Thus by Proposition 4.13, it is of the form G_{ρ}(z) with ρ ∈ M^{H}(γ). Thus the limit for f^{[N−1,N]}(z) as N → ∞, call it f(z), has the form

f(z) = ∫ \frac{dρ(x)}{1 + xz}

with ρ ∈ M^{H}(γ). In the determinate case, ρ must be the unique solution, so all limit points are equal. Thus compactness implies convergence, and we have proven (ii).

In the indeterminate case, we note that δ_{j} = P_{j}(A_{F}^{[N]})δ_{0} for j = 0, …, N − 1 so

(A_{F}^{[N]} − z)^{−1}δ_{0} = ∑_{j=0}^{N−1}⟨P_{j}(A_{F}^{[N]})δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}⟩δ_{j}.

As in the proof of Theorem 4.3, we conclude that (with G_{N}(z) = ⟨δ_{0}, (A_{F}^{[N]} − z)^{−1}δ_{0}⟩),

∑_{j=0}^{N−1} |Q_{j}(z) + G_{N}(z)P_{j}(z)|^{2} = \frac{Im G_{N}(z)}{Im z} .

Thus, since Q, P ∈ ℓ_{2} if G_{N}(z) → G(z) through a subsequence, the limit ζ = G(z) obeys equality in (4.5). By Theorem 4.14, any such solution of the moment problem is a von Neumann solution. □

We want to show that in many cases, lim f^{[N−1,N]}(z) will not exist. Consider a set of Stieltjes moments, {γ_{n}}_{n=0}^{∞} and the associated even Hamburger problem with moments

Γ_{2n} = γ_{n},  Γ_{2n+1} = 0.  (5.86)

Given ρ ∈ M^{S}(γ), let ρ̃ ∈ M^{H}(Γ) be given by

dρ̃(x) = \frac{1}{2}(χ_{[0,∞)}(x) dρ(x^{2}) + χ_{(−∞,0]}(x) dρ(x^{2}))  (5.87)

so that Theorem 2.12 says that this sets up a one-one correspondence between even integrals in M^{H}(Γ) and M^{S}(γ). Let P̃_{n}, Q̃_{n} be the orthogonal polynomials for Γ and P_{n}, Q_{n}, M_{n}, N_{n} for γ. Define d_{n} by d_{n}^{2} ∫ M_{n}(x)^{2}x^{−1} dρ(x) = 1. Then we claim that

Gρ̃(z) = zG_{ρ}(z^{2})  (5.88)
P̃_{2n}(x) = P_{n}(x^{2})  (5.89)
P̃_{2n−1}(x) = \frac{d_{n}M_{n}(x^{2})}{x}  (5.90)
Q̃_{2n}(x) = xQ_{n}(x^{2})  (5.91)
Q̃_{2n−1}(x) = d_{n}N_{n}(x^{2}).  (5.92)

To prove (5.89) and (5.90), we note that the right sides have the correct degree, and since P_{n}(x^{2}) is even and \frac{M_{n}(x^{2})}{x} is odd, they are dρ̃ orthogonal to each other. Easy calculations prove the dρ̃ orthogonality of two P_{n}(x^{2})’s and two \frac{M_{n}(x^{2})}{x}’s. The formulae for Q̃ and Gρ̃ follow by direct calculations using Theorem 4.2 for Q.

Theorem 5.30. Let γ, Γ be related by (5.86) and let dρ̃^{(F)}, dρ̃^{(K)} be the images of dρ_{F} , dρ_{K} (for γ) under the map ρ ↦ ρ̃ given by (5.87). Let f̃^{[N,M]}(z) be the Padé approximants for ∑_{n=0}^{∞}(−1)^{n}Γ_{n}z^{n}. Then

lim_{N→∞} f̃^{[2N−1,2N]}(z) = ∫ \frac{dρ̃^{(F)}(x)}{1 + xz}  (5.93)
lim_{N→∞} f̃^{[2N,2N+1]}(z) = ∫ \frac{dρ̃^{(K)}(x)}{1 + xz} .  (5.94)

In particular, if the Γ problem is indeterminate, lim f^{[N−1,N]}(z) does not exist.

Proof. This follows directly from Theorem 6 for the γ problem and (5.88)–(5.92). □

Of course, in the special case, the non-convergence of f^{[N−1,N]}(z) is replaced by something almost as nice since the even and odd subsequences separately converge. But there is no reason to expect this to persist if Γ is not even and indeed, simple numerical examples ([14]) illustrate non-convergence is the rule. One can understand why this happens. In the indeterminate case, there is a one-parameter family of extensions, each with discrete spectrum. There is, in general, nothing to tack down A_{F}^{[N]} so it wanders. In the even case, two self-adjoint extensions are even, one with an eigenvalue at 0 and one without, and the A_{F}^{[N]} for odd N and even N track these. This suggests that A_{K}^{[N]}, which is tacked down to have a zero eigenvalue, should converge in strong resolvent sense and the corresponding f^{[N,N]}’s should converge. This is true and is not a previously known result:

Theorem 5.31. Let {γ_{n}}_{n=0}^{∞} be a set of Hamburger moments and let ∑_{n=0}^{∞}(−1)^{n}γ_{n}z^{n} be the associated series of Hamburger. Then the f^{[N,N]}(z) Padé approximants exist if and only if

P_{N}(0) ≠ 0.  (5.95)

Moreover, (5.95) holds if and only if there exists α_{N} so that the matrix A_{K}^{[N+1]} given by (5.2b) has a zero eigenvalue, and then

f^{[N,N]}(z) = (δ_{0}, (1 + zA_{K}^{[N+1]})^{−1}δ_{0}).  (5.96)

If N → ∞ through the sequence of all N’s for which (5.95) holds, then

lim_{\begin{matrix}N→∞ \\ N:P_{N} (0)≠0 \end{matrix}} f^{[N,N]}(z) = ∫ \frac{dρ(x)}{1 + xz} ,  (5.97)

where dρ is the unique solution of the moment problem in case it is determinate and the unique von Neumann solution with μ({0}) > 0 if the moment problem is indeterminate.

Remarks. 1. If P_{N}(0) = 0, then, by the second-order equation P_{N}(0) obeys, we have that P_{N+1}(0) ≠ 0 ≠ P_{N−1}(0), so (5.95) can fail at most half the time. Typically, of course, it always holds (e.g., under a small generic perturbation of the Jacobi matrix, one can prove that it will hold even if initially it failed).

2. If the γ’s are a set of even moments, then (5.95) holds exactly for N even and

f^{[2M,2M]}(z) = \frac{z^{2M}Q_{2M+1}(−\frac{1}{z})}{z^{2M+1}P_{2M+1}(−\frac{1}{z})} ,

so we get convergence of those measures we called dρ̃ above.

3. The non-existence of f^{[N,N]} is using the Baker definition that we have made. There is a classical definition which would define f^{[N,N]}(z) even if P_{N}(0) = 0, (but (5.26) and (5.27) would both fail!). Under this definition, f^{[N,N]}(z) would be f^{[N−1,N−1]}(z) when P_{N}(0) = 0 and (5.97) would hold using all N rather than just those N with P_{N}(0) ≠ 0.

Proof. Define

M̃_{N}(z) = P_{N−1}(0)P_{N}(z) − P_{N}(0)P_{N−1}(z)  (5.98)
Ñ_{N}(z) = P_{N−1}(0)Q_{N}(z) − P_{N}(0)P_{N−1}(z).  (5.99)

Then by the Wronskian calculations (see Proposition 4.1); also see (5.70)),

Ñ_{N}(0) = \frac{1}{a_{N−1}} ≠ 0  (5.100)

M̃′_{N}(0) = ∑_{j=0}^{N−1} \frac{P_{j}(0)^{2}}{a_{N−1}} ≠ 0  (5.101)

and, of course,

M̃_{N}(0) = 0.  (5.102)

Define

A_{N}(z) = z^{N} Ñ_{N+1}(−\frac{1}{z})  (5.103)
B_{N}(z) = z^{N+1}M̃_{N+1}(−\frac{1}{z}).  (5.104)

By (5.100), deg A_{N} = N and by (5.101/5.102), deg B_{N} = N. Moreover, by the definition (5.98),

B_{N}(0) = 0 ⟺ P_{N}(0) = 0  (5.105)

and

B_{N}(0) = 0 ⟹ P′_{N}(0) ≠ 0  (5.106)

since P_{N}(0) = 0 ⇒ P_{N−1}(0) ≠ 0.

Moreover, we claim that

A_{N}(z) − B_{N}(z) ∑_{j=0}^{2N} γ_{j}z^{j} = O(z^{2N+1}).  (5.107)

This can be seen either by algebraic manipulation as in the remark following the proof of Theorem A.5 or by noting that we have proven it in (5.39) for Stieltjes measures. So if ρ ∈ M^{H}(γ) and μ_{0} is a Stieltjes measure and ρ_{t} = (1 − t)μ_{0} + tρ, then B_{N}(−z), A_{N}(−z) are real analytic functions of t and (5.107) is true for all small t in (0, 1) since A_{F}^{[N+1]}(t) > 0 for small t (and fixed N) and that is all the proof of (5.106) depends on.

If P_{N}(0) ≠ 0, B_{N}(0) ≠ 0, so B^{[N,N]}(z) = \frac{B_{N} (z)}{B_{N} (0)} and A^{[N,N]}(z) = \frac{A_{N} (z)}{B_{N} (0)} yield denominator and numerator of f^{[N,N]}(t); so f^{[N,N]} exists.

Moreover, the proof of Lemma 5.1 shows that an α_{N} exists giving A_{K}^{[N+1]} a zero eigenvalue if and only if P_{N}(0) ≠ 0. In that case, the proof of Proposition 5.8 is valid, and we have that (5.17) holds. Noting that \frac{N_{N}}{M_{N}} = \frac{Ñ_{N}}{M̃_{N}}, we see that (5.96) is then valid.

On the other hand, suppose that P_{N}(0) = 0 and that f^{[N,N]}(z) exists. Then by (5.107), we have that

A_{N}(z)B^{[N,N]} − B_{N}(z)A^{[N,N]}(z) = O(z^{2N+1})

and so it is zero since the left side of this is of degree 2N.

Since P_{N}(0) = 0, M̃_{N+1}(z) = −P_{N+1}(0)P_{N}(z) = cM̃_{N}(z) where c = −P_{N+1}(0) P_{N−1}(0)^{−1}, and similarly, Ñ_{N=1}(z) = cÑ_{N}(z). It follows that

A_{N}(z) = czA_{N−1}(z), B_{N}(z) = czB_{N−1}(z).

Thus, if P_{N}(0) = 0 but f^{[N,N]}(z) exists, then

f^{[N,N]}(z) = f^{[N−1,N−1]}(z),

that is, the [N − 1, N − 1] Padé approximant would need to have an error of order O(z^{2N+1}).

Since P_{N}(0) = 0, A_{K}^{[N]} = A_{F}^{[N]} and so we would have that ⟨δ_{0}, (A_{F}^{[N]})^{j}δ_{0}⟩ = γ_{j} for j = 0, 1, …, 2N. But as noted in the remark following the proof of Proposition 5.11, this can never happen. We conclude that if P_{N}(0) = 0, f^{[N,N]} must fail to exist.

To complete the proof, define A_{K} to be Ā if the problem is determinate and the unique von Neumann extension with eigenvalue zero if the problem is indeterminate (so A_{K}π(0) = 0). Then the proof of Proposition 5.3 goes through without change and shows that A_{K}^{[N]} converges to A_{K} in strong resolvent sense. This implies (5.97). □

As for other f^{[N+ℓ−1,N]}(z) for series of Hamburger, we can only define γ^{(j)} moment problems for j an even integer (positive or negative). This means that we can use (5.49)/(5.59) if ℓ is an even integer and (5.50)/(5.60) if ℓ is an odd integer. The result is the following:

Theorem 5.32. Let {γ_{n}}_{n=0}^{∞} be a set of Hamburger moments and let ∑_{n=0}^{∞}(−1)^{n}γ_{n}z^{n} be the associated series of Hamburger. Then:

(i) If ℓ is any odd integer, the Padé approximants f^{[N+ℓ−1,N]}(z) exist if and only if P_{M}^{(j)}(0) ≠ 0 where for ℓ > 0, j = ℓ − 1, and M = N and if ℓ < 0, j = ℓ + 1 and M = N + ℓ − 1. The functions that exist converge to a finite limit f_{ℓ}(z).

(ii) If ℓ is any even integer, the Padé approximants f^{[N+ℓ−1,N]}(z) all exist and lie in a compact subset in the topology of uniform convergence on compact subsets of ℂ_{+}. Any limit point, f_{ℓ}(z), is associated to a von Neumann solution, ρ^{(ℓ)}, of the γ^{(ℓ)} moment problem via

f_{ℓ}(z) = ∑_{j=0}^{ℓ−1}(−1)^{j}γ_{j}z^{j} + (−1)^{ℓ}γ_{ℓ}z^{ℓ} ∫_{−∞}^{∞} \frac{dρ^{(ℓ)}(x)}{1 + xz}

if ℓ > 0 and

f_{ℓ}(z) = \{1 − γ_{1}z − ∑_{j=0}^{−ℓ−1}(−1)^{j}γ_{j}^{(0)}z^{j+2} + (−1)^{ℓ+1}γ_{−ℓ}^{(0)}z^{−ℓ+2} ∫_{−∞}^{∞} \frac{dρ^{(ℓ)}(x)}{1 + xz}\}^{−1}

if ℓ < 0. In particular, if the γ^{(ℓ)} problem is determinate, f^{[N+ℓ−1,N]}(z) is convergent.

(iii) Let ℓ be an even integer. If the γ^{(ℓ)} problem is determinate (as a Hamburger problem), then for ℓ ≥ 0, f_{ℓ+1}(z) = f_{ℓ}(z) = ⋯ = f_{0}(z) = f_{−1}(z), and for ℓ ≤ 0, f_{ℓ−1}(z) = f_{ℓ}(z) = ⋯ = f_{0}(z) = f_{1}(z). In particular, if |γ_{j}| ≤ C^{j+1}j!, then all f^{[N+ℓ−1,N]}(z) converges to the same ℓ-independent limits as N → ∞.

§6. Solutions of Finite Order

In this section, we will continue the analysis of the indeterminate Hamburger moment problem. Throughout, {γ_{n}}_{n=0}^{∞} will be a set of indeterminate Hamburger moments and ρ will be a solution of (1.1), the moment problem for γ.

We have picked out the von Neumann solutions as coming from self-adjoint extensions of the Jacobi matrix, A. But in a certain sense, every ρ does come from a self-adjoint extension of A! For let H_{ρ} = L^{2}(ℝ, dρ), let D(B) = {f ∈ H_{ρ} | ∫ x^{2}|f(x)|^{2} dρ(x) < ∞}, and let

(Bf)(x) = xf(x).

Let ℂ[X] be the set of polynomials in x which lies in H_{ρ} since all moments are finite, and let H_{0} = \overline{ℂ[X]}. Corollary 4.15 says that H_{0} = H_{ρ} if and only if ρ is a von Neumann solution. Let A_{ρ} = B ↾ ℂ[X]. Then to say ρ is a solution of the Hamburger moment problem is precisely to say that A_{ρ} is unitarily equivalent to the Jacobi matrix, A, associated to γ under the natural map (so we will drop the subscript ρ). In that sense, B is a self-adjoint extension of A, but not in the von Neumann sense.

B is minimal in the sense that there is no subspace of H_{ρ} containing H_{0} and left invariant by all bounded functions of B (equivalently, left invariant by all (B − z)^{−1}, z ∈ ℂ\ℝ or by all e^{iBs}, s ∈ ℝ). So (if ρ is not a von Neumann extension) we are in the strange situation where D(B) ∩ H_{0} is dense in H_{0}, B[D(B) ∩ H_{0}] ⊂ H_{0}, but H_{0} is not invariant for the resolvents of B!

It is not hard to see that the set of all solutions of the moment problem is precisely the set of all self-adjoint “extensions” of A in this extended sense, which are minimal (H_{0} is cyclic for B) and modulo a natural unitary equivalence. This is a point of view originally exposed by Naimark [24, 25, 26, 27] and developed by Livsic [21], Gil de Lamadrid [7], and Langer [20]. See the discussion of Naimark’s theory in Appendix 1 of [2].

In the language of Cayley transforms, the extensions associated to ρ’s that we will call of order at most n below are parametrized by unitary maps, U, from K_{+} ⊕ ℂ^{n} to K_{−} ⊕ ℂ^{n} with U_{1}, U_{2} equivalent if and only if there is a unitary map V : ℂ^{n} → ℂ^{n} so U_{1}(1 ⊕ V) = (1 ⊕ V)U_{2}. From this point of view, these extensions are then parametrized by the variety of conjugacy classes of U(n + 1) modulo a U(n) subgroup. This has dimension (n + 1)^{2} − n^{2} = 2n + 1. Our parametrization below in terms of a ratio of two real polynomials of degree at most n also is a variety of dimension 2n + 1, showing the consistency of the parametrization. (The ρ’s of exact order n will be a manifold.)

Given any ρ, we define the order of ρ,

ord(ρ) = dim(H_{ρ}/H_{0})

so the von Neumann solutions are exactly the solutions of order 0. Our main result in this section will describe all solutions of finite order which turn out to correspond precisely to those ρ’s whose Nevanlinna function Φ_{ρ} are ratios of real polynomials.

They will also be ρ’s for which dρ/ ∏_{i=1}^{n} |x − z_{i}|^{2} is the measure of a determinate moment problem. We will therefore need some preliminaries about such problems. Given z_{1}, …, z_{n} ∈ ℂ_{+}, by using partial fractions we can write for x real,

\frac{x^{m}}{∏_{i=1}^{n} |x − z_{i}|^{2}} = P_{m−2n}(x; z_{1}, …, z_{n}) + ∑_{i=1}^{n}[\frac{z_{i}^{m}}{z_{i} − z̄_{i}} \frac{1}{x − z_{i}} + \frac{z̄_{i}^{m}}{z̄_{i} − z_{i}} \frac{1}{x − z̄_{i}} ] (6.1)

where P_{m−2n} is a polynomial in x of degree m − 2n. To see (6.1), write ∏_{i=1}^{n} |x − z_{i}|^{2} = ∏_{i=1}^{n}(x − z_{i})(x − z̄_{i}) and analytically continue. The sum in (6.1) comes from computing the residues of the poles of these functions. Define

Γ_{m}^{(0)}(z_{1}, …, z_{n}) = E_{x}^{γ}(P_{m−2n}(x; z_{1}, …, z_{n})),  (6.2)

where, as usual, E_{x}^{γ} is the expectation with respect to any solution of the moment problem. For ζ_{1}, …, ζ_{n} ∈ ℂ_{+}, let

Γ_{m}(z_{1}, …, z_{n}; ζ_{1}, …, ζ_{n}) = Γ_{m}^{(0)}(z_{1}, …, z_{n}) + ∑_{i=1}^{n}[\frac{z_{i}^{m}}{z_{i} − z̄_{i}} ζ_{i} + \frac{z̄_{i}^{m}}{z̄_{i} − z_{i}} ζ̄_{i}].  (6.3)

Theorem 6.1. Fix z_{1}, …, z_{n} and ζ_{i}, …, ζ_{n} in ℂ_{+}. There is a one-one correspondence between ν ∈ M^{H}(Γ(z_{1}, …, z_{n}; ζ_{1}, …, ζ_{n})) and those ρ in M^{H}(γ) which obey

∫ \frac{dρ(x)}{x − z_{i}} = ζ_{i},  i = 1, …, n  (6.4)

under the association

dρ(x) ↔ dν(x) = ∏_{i=1}^{n} |x − z_{i}|^{−2} dρ(x).  (6.5)

That is, ρ has moments {γ_{m}}_{m=0}^{∞} with subsidiary conditions (6.4) if and only if ν given by (6.5) has the moments {Γ_{m}(z_{1}, …, z_{n}; ζ_{1}, …, ζ_{n})}_{m=0}^{∞}.

Proof. This is pure algebra. The functions \{x_{m}\}_{m=0}^{∞}, \{\frac{1}{x−z_{i}} \}_{i=1}^{n} are linearly independent as analytic functions (finite sums only). (6.1) says that each function x^{m}/ ∏_{i=1}^{n}(x − z_{i})(x − z̄_{i}) ≡ Q_{m}(x) is in the span, S, of these functions. On the other hand, since ∏_{i=1}^{n}(x − z_{i})(x − z̄_{i}) ≡ L(x) is a polynomial in x, and x^{ℓ}Q_{m}(x) = Q_{ℓ+m}(x), L(x)Q_{m}(x) is in the span of the \{Q_{m}(x)\}. Similarly, since \frac{L(x)}{(x−z_{i})} is a polynomial, L(x)Q_{0}(x)\frac{1}{(x−z_{i})} lies in that span.

Thus, {Q_{m}(x)} is also a basis of S. It follows that there is a one-one correspondence between assignments of numbers Q_{m}(x) ↦ Γ_{m} and of assignments x^{m} ↦ γ_{m}, \frac{1}{x−z_{i}} ↦ ζ_{i}, \frac{1}{x−z̄_{i}} ↦ λ_{i} since each defines a linear functional on S. For the linear functional to be real in Γ_{m} representation, all Γ_{m} must be real. For reality in (γ, ζ, λ) representation, we must have γ_{m} real and λ_{i} = ζ̄_{i}. (6.3) is just an explicit realization of the map from γ, ζ, ζ̄ to Γ. Thus, a real dρ has moments γ_{m} with ζ_{i} conditions if and only if ∫ x^{m} ∏_{i=1}^{n} |x − z_{i}|^{−2} dρ(x) = Γ_{m} for all m. Since (6.5) shows ν is positive if and only if ρ is, we see that the claimed equivalence holds. □

As an aside of the main theme of this section, we can construct determinate Hamburger moments quite close to indeterminate ones. The result is related to Corollary 4.21.

Theorem 6.2. Let {γ_{n}}_{n=0}^{∞} be a set of Hamburger moments. Then for any a ∈ (0, ∞), there is a c_{a} so that the moment problem with moments

γ̃_{2n+1} = 0
γ̃_{2n} = γ_{2n} − a^{2}γ_{2n−2} + a^{4}γ_{2n−2} + ⋯ + (−1)^{n}a^{2n}γ_{0} + (−1)^{n+1}c_{a}a^{2n+2} (6.6)

is a determinate moment problem.

Proof. Let ρ_{1} be any solution for the Hamburger problem. Since \frac{1}{2}[dρ_{1}(x)+dρ_{1}(−x)] = dρ̃ has moments

∫ x^{m} dρ̃(x) = \{ \begin{matrix}0 & m odd \\ γ_{2m} & m even \end{matrix} ≡ γ′_{2m},

we can suppose γ_{2m+1} = 0. Let ρ be a von Neumann solution to the γ′_{m} moment problem which is invariant under x → −x. If the γ′_{m} problem is determinate, then the unique solution is invariant. Otherwise, this follows from the proof of Theorem 2.13. Let dμ(x) = \frac{dρ(x)}{1+a^{2}x^{2}} . By Corollary 4.15 and Theorem 6.1, ρ is the unique solution of the γ′ moment problem with ∫ \frac{dρ(x)}{x−ia^{−1}} ≡ ζ_{a^{−1}}. Thus, the Γ_{m}(ia^{−1}, ζ_{a^{−1}}) problem is determinate. If c_{a} = a^{−1} Im ζ_{a^{−1}}, then a calculation using the geometric series with remainder shows that Γ_{m}(−ia^{−1}, ζ_{a^{−1}}) = γ̃_{m} given by (6.6). □

Returning to the main theme of this section, we next examine when a Herglotz function is a real rational function. Define for z ∈ ℂ_{+},

φ_{z}(x) ≡ \frac{1 + xz}{x − z}  (6.7a)

viewed as a continuous function on ℝ ∪ {∞} with

φ_{z}(∞) = z.  (6.7b)

If dσ(x) = [\frac{dμ(x)}{1+x^{2}}] + cδ_{∞} as a finite measure on ℝ ∪ \{∞\}, then (1.19) can be rewritten as

Φ(z) = d + ∫_{ℝ∪{∞}} φ_{z}(x) dσ(x).  (6.8)

If σ({∞}) = 0 and ∫ |x| dσ(x) < ∞, define d̃= d − ∫ x dσ(x).

Proposition 6.3. A Herglotz function is a ratio of two real polynomials if and only if the representation (6.8) has a σ with finite support. If σ has exactly N points in its support, there are three possibilities for the degrees of P, Q in Φ(z) = \frac{P (z)}{Q(z)} with P, Q relatively prime polynomials:

(i) (σ({∞}) = 0, d̃= 0), deg P = N − 1, deg Q = N (ii) (σ({∞}) = 0, d̃≠ 0), deg P = deg Q = N (iii) (σ({∞}) ≠ 0), deg P = N, deg Q = N − 1 In all cases, we say N = max(deg(P), deg(Q)) is the degree of Φ.

Proof. Elementary. □

Remark. The set of σ’s with exactly N pure points is a manifold of dimension 2N (N points, N weights). d ∈ ℝ is another parameter so the set of such Φ’s is a manifold of dimension 2N + 1.

As a final preliminary, for any z ∈ ℂ\σ(B) ≡ ℂ\ supp(dρ) and n = 1, 2, …, we introduce the functions e_{n}(z) on supp(dρ),

e_{n}(z)(x) = \frac{1}{(x − z)^{n}} ,

thought of as elements of H_{ρ} = L^{2}(ℝ, dρ).

The main result of this section is:

Theorem 6.4. Let γ be a set of indeterminate Hamburger moments and let ρ_{0} ∈ M^{H}(γ). Fix N ∈ {0, 1, … }. Then the following are equivalent:

(1) ρ_{0} has order at most N.

(2) For some set of distinct z_{1}, …, z_{N} ∈ ℂ\ℝ, H_{0} ∪ {e_{1}(z_{j})}_{j=1}^{N} span H_{ρ_{0}}.

(3) For any set of distinct z_{1}, …, z_{N} ∈ ℂ\ℝ, H_{0} ∪ {e_{1}(z_{j})}_{j=1}^{N} span H_{ρ_{0}}.

(4) For some z_{0} ∈ ℂ\ℝ, H_{0} ∪ {e_{j}(z_{0})}_{j=1}^{N} span H_{ρ_{0}}.

(5) For any z_{0} ∈ ℂ\ℝ, H_{0} ∪ {e_{j}(z_{0})}_{j=1}^{N} span H_{ρ_{0}}.

(6) For some set of z_{0}, z_{1}, …, z_{N} ∈ ℂ_{+} and ζ_{j} = G_{ρ_{0}}(z_{j}), there is no other ρ ∈ M^{H}(γ) with G_{ρ}(z_{j}) = ζ_{j}.

(7) For all sets of z_{0}, z_{1}, …, z_{N} ∈ ℂ_{+} and ζ_{j} = G_{ρ_{0}}(z_{j}), there is no other ρ ∈ M^{H}(γ) with G_{ρ}(z_{j}) = ζ_{j}.

(8) For some z_{0}, z_{1}, …, z_{N} ∈ ℂ_{+} and ζ_{j} = G_{ρ_{0}}(z_{j}), the Γ(z_{0}, …, z_{N}; ζ_{0}, …, ζ_{N}) moment problem is determinate.

(9) For any z_{0}, z_{1}, …, z_{N} ∈ ℂ_{+} and ζ_{j} = G_{ρ_{0}}(z_{j}), the Γ(z_{0}, …, z_{N}; ζ_{0}, …, ζ_{N}) moment problem is determinate.

(10) For some z_{1}, …, z_{N} ∈ ℂ_{+} and ζ_{j} = G_{ρ_{0}}(z_{j}), ρ_{0} is a von Neumann solution of the Γ(z_{1}, …, z_{N}; ζ_{1}, …, ζ_{N}) moment problem.

(11) For any z_{1}, …, z_{N} ∈ ℂ_{+} and ζ_{j} = G_{ρ_{0}}(z_{j}), ρ_{0} is a von Neumann solution of the Γ(z_{1}, …, z_{N}; ζ_{1}, …, ζ_{N}) moment problem.

(12) The Nevanlinna function Φ of ρ_{0} is a rational function of degree at most N. In particular, the Nevanlinna function Φ of ρ_{0} has degree N if and only if ρ_{0} has

order N. Moreover, if one and hence all those conditions hold, ρ_{0} is an extreme point in M^{H}(γ) and is also a pure point measure.

Remark. The measures obeying (1)–(12) are what Akhiezer calls canonical solutions of order N.

Lemma 6.5. If φ ∈ H_{0} and z ∉ σ(B), then

(B − z)^{−1}φ ∈ H_{0} + [e_{1}(z)].

Proof. Since (B − z)^{−1} is bounded and H_{0} + [e_{1}(t)] is closed, it suffices to prove this for φ(x) = P (x), a polynomial in x. But then

(B − z)^{−1}φ = (x − z)^{−1}P (x) = P (z)(x − z)^{−1} + \frac{(P (x) − P (z))}{x − z}
= P (z)e_{1}(z) + R(x)

for a polynomial R, that is, (B − z)^{−1}φ ∈ H_{0} + e_{1}(z). □

Lemma 6.6. Supose that z_{0}, z_{1}, …, z_{ℓ} ∈ ℂ\σ(B) are distinct. If e_{m}(z_{0}) is in the span of H_{0} ∪ span[{e_{j}(z_{0})}_{j=1}^{m−1}] ∪ span[{e_{1}(z_{k})}_{k=1}^{ℓ}], then so is e_{m+1}(z_{0}).

Remark. ℓ can be zero.

Proof. By hypothesis, for some φ ∈ H_{0} and {α_{j}}_{j=1}^{m−1} and {β_{k}}_{k=1}^{ℓ} in ℂ,

e_{m}(z_{0}) = φ + ∑_{j=1}^{m−1} α_{j}e_{j}(z_{0}) + ∑_{k=1}^{ℓ} β_{k}e_{1}(z_{k}).

Apply (B − z_{0})^{−1} to this using (B − z_{0})^{−1}e_{j}(z_{0}) = e_{j+1}(z_{0}), Lemma 6.5, and (B − z_{0})^{−1}e_{1}(z_{k}) = (z_{0} − z_{k})^{−1}[e_{1}(z_{0}) − e_{1}(z_{k})] and we see that e_{m+1}(z_{0}) is in the requisite span. □

Proposition 6.7. Let ℓ = ord(ρ) < ∞. Then for any set {z_{i}}_{i=1}^{ℓ} of ℓ distinct points in ℂ\σ(B), H_{0} ∪ {e_{1}(z_{i})}_{i=1}^{ℓ} span H_{ρ}.

Proof. By the Stone-Weierstrass theorem, linear combinations of {(x − z)^{−1} | z ∈ ℂ\σ(B)} are dense in the continuous functions on σ(B) vanishing at infinity, and so they are dense in L^{2}(ℝ, dρ). It follows that we can find {z_{i}}_{i=1}^{ℓ} so that H_{0} ∪ {e_{1}(z_{i})}_{i=1}^{ℓ} span H_{ρ}.

Pick z_{0} ∈ ℂ\σ(B) and let W = (B − z_{ℓ})(B − z_{0})^{−1}. Then W is bounded with bounded inverse, so it maps dense subspaces into dense subspaces. By Lemma 6.5, W [H_{0}] ⊂ H_{0} + [e_{1}(z_{0})]. For i = 1, …, ℓ − 1, W e_{1}(z_{i}) = e_{1}(z_{i}) + (\frac{z_{0}−z_{ℓ}}{z_{0}−z_{i}})[e_{1}(z_{0}) − e_{1}(z_{i})] and W e_{1}(z_{ℓ}) = e_{1}(z_{0}). Thus, W maps the span of {e_{1}(z_{i})}_{i=1}^{ℓ} into {e_{1}(z_{i})}_{i=0}^{ℓ−1}. So H_{0}∪{e_{1}(z_{i})}_{i=0}^{ℓ−1} span H_{ρ}. By successive replacement, we can move the z_{i}’s to an arbitrary set of distinct points. □

Remark. Proposition 6.7 shows (1)–(3) of Theorem 6.4 are equivalent.

Proposition 6.8. Suppose that ℓ = ord(ρ) < ∞. Then for any z_{0} ∈ ℂ\σ(B), H_{0} ∪ {e_{j}(z_{0})}_{j=1}^{ℓ} span H_{ρ}.

Proof. By hypothesis, there must be some dependency relation

φ + ∑_{j=1}^{ℓ+1} α_{j}e_{j}(z_{0}) = 0  (6.10)

for φ ∈ H_{0} and some (α_{1}, …, α_{ℓ+1}) ≠ 0. Let k + 1 = max{j | α_{j} ≠ 0}. Then solving (6.10) for e_{k+1}(z_{0}), we see that e_{k+1}(z_{0}) lies in the span of H_{0} ∪ {e_{j}(z_{0})}_{j=1}^{k}; and so by induction and Lemma 6.6, all e_{m}(z_{0}) lie in this span and so in the span of H_{0}∪{e_{j}(z_{0})}_{j=1}^{ℓ}. e_{1}(z) is an analytic function in ℂ\ℝ with Taylor coefficients at z_{0} equal to e_{n}(z_{0}), so e_{1}(z) lies in the span of H_{0} ∪{e_{j}(z_{0})}_{j=1}^{ℓ} for z in the same half plane as z_{0}. By Proposition 6.7, these e_{1}’s together with H_{0} span H_{ρ}. □

Remark. Proposition 6.8 shows that (1), (4), (5) of Theorem 6.4 are equivalent.

Proposition 6.9. Fix distinct points z_{1}, …, z_{ℓ} ∈ ℂ\ℝ. Let

dμ(x) = ∏_{i=1}^{ℓ} |x − z_{i}|^{−2} dρ(x).

Then the polynomials are dense in L^{2}(ℝ, dμ) if and only if H_{0} ∪ {e_{1}(z_{j})}_{j=1}^{ℓ} span H_{ρ}.

Proof. Let U be the unitary map from H_{ρ} to H_{μ} given by

(Uf)(x) = ∏_{i=1}^{ℓ}(x − z_{i})f(x).

Then U maps the span of ℂ[X]∪{e_{1}(z_{i})}_{i=1}^{ℓ} onto ℂ[X] for P (x) is a polynomial of degree m if and only if ∏_{i=1}^{ℓ}(x − z)^{−1}P (x) is a linear combination of a polynomial of degree max(0, m − ℓ) and {(x − z_{i})^{−1}}_{i=1}^{ℓ}. Thus, the density assertions are equivalent. □

Remark. Proposition 6.9 and Corollary 4.15 show the equivalence of (2), (3), (10), and (11) of Theorem 6.4 (given that we already know that (2) and (3) are equivalent).

Proposition 6.10. If ord(ρ_{0}) is finite, then ρ_{0} is a pure point measure with discrete support.

Proof. Let ℓ = ord(ρ_{0}). Pick distinct {z_{i}}_{i=1}^{ℓ} in ℂ_{+}. Then by Propositions 6.7 and 6.9, the polynomials are dense in L^{2}(ℝ, dμ_{ℓ}) where for k = 0, 1, …, ℓ, we set

dμ_{k}(x) = ∏_{j=1}^{k} |x − z_{i}|^{−2} dρ(x).

Let k_{0} be the smallest k for which the polynomials are dense in L^{2}(ℝ, dμ_{k}). If k_{0} = 0, then dρ is a von Neumann solution of the γ problem. If k_{0} > 0, then dμ_{k_{0}−1} is not a von Neumann solution of its moment problem. So by Proposition 4.15 and Theorem 6.1, the moment problem for Γ_{m}^{(0)} ≡ Γ_{m}(z_{1}, …, z_{k_{0}−1}, z_{k_{0}}; ζ_{1}, …, ζ_{k_{0}−1}, ζ_{k}) is indeterminate. But since the polynomials are dense in L^{2}(ℝ, dμ_{k}), dμ_{k_{0}} is a von Neumann solution of an indeterminate problem.

Either way, dμ_{k_{0}} is a von Neumann solution of an indeterminate problem. It follows Theorem 4.11 that dμ_{k_{0}} and so dρ is a discrete point measure. □

As a final preliminary to the proof of Theorem 6.4, we need a known result about interpolation of Herglotz functions. Given z_{1}, …, z_{n}; w_{1}, …, w_{n} ∈ C_{+} with the z’s distinct, define the n × n matrix D by

D_{ij}(z_{1}, …, z_{n}; w_{1}, …, w_{n}) = \frac{w_{i} − w̄_{j}}{z_{i} − z̄_{j}} .  (6.11)

Theorem 6.11. Pick z_{1}, …, z_{n} ∈ ℂ_{+} with n ≥ 1. There exists a Herglotz function Φ with

Φ(z_{i}) = w_{i},  i = 1, …, n  (6.12)

if and only if D(z_{1}, …, z_{n}; w_{1}, …, w_{n}) is a (not necessarily strictly) positive definite matrix. Moreover, the following are equivalent, given such a Φ:

(1) det D(z_{1}, …, z_{n}; w_{1}, …, w_{n}) = 0.

(2) There is a unique Herglotz function Φ obeying (6.12).

(3) Φ is a real rational polynomial of degree at most n − 1.

Remark. Since (3) is independent of the choice of {z_{i}}_{i=1}^{n}, so are (1), (2).

We will sketch a proof of this result (fleshing out some arguments in [1]) below.

Proof of Theorem 6.4. As already noted, Propositions 6.7, 6.8, and 6.9 show that (1), (2), (3), (4), (5), (10), and (11) are equivalent. Theorem 6.1 and Corollary 4.15 prove the equivalence of (10) and (11) with (6), (7), (8), and (9). By Theorems 4.14 and 6.11,

(6) and (12) are equivalent.

Since ord(ρ) = N is equivalent to ord(ρ) ≥ N and the negation of ord(ρ) ≤ N − 1, we see that Φ_{ρ_{0}} has degree precisely equal to ord(ρ_{0}). That ρ_{0} is then an extreme point is proven in Appendix B. □

We conclude this section by proving Theorem 6.11.

Proposition 6.12. Let Φ be a Herglotz function and let w_{i} = Φ(z_{i}), i = 1, …, n for distinct z_{1}, …, z_{n} ∈ ℂ_{+}. Let D be given by (6.11). Then D(z_{1}, …, z_{n}; w_{1}, …, w_{n}) is (not necessarily strictly) positive definite and det D = 0 if and only if Φ is a real ratio of polynomials of degree n − 1 or less.

Proof. Using the Herglotz representation (1.19), we get a representation of D_{ij},

\frac{w_{i} − w̄_{j}}{z_{i} − z̄_{j}} = c + ∫ dμ(x) \frac{1}{x − z_{i}} \frac{1}{x − z̄_{j}} ,

from which we obtain for α ∈ ℂ^{N}:

∑_{i,j=1}^{n} ᾱ_{i}α_{j}D_{ij} = c |∑_{i=1}^{n} α_{j}|^{2} + ∫_{−∞}^{∞} dμ(x)|∑_{i=1}^{n} \frac{α_{i}}{x_{i} − z_{i}} |^{2},  (6.13)

proving the positivity.

Suppose det(D) = 0. Then there is a non-zero α ∈ ℂ^{n} so that the right side of (6.13) is 0. Note that ∑_{i=1}^{n} \frac{α_{i}}{(x−z_{i})} = ∏_{i=1}^{n}(x − z_{i})^{−1}Q(x), where Q is a polynomial of degree n − 1 if ∑_{i=1}^{n} α_{i} ≠ 0 and of degree at most n − 2 if ∑_{i=1}^{n} α_{i} = 0. Thus, for the right side of (6.13) to vanish for some non-zero α in ℂ^{n}, either c ≠ 0 and μ is supported at n − 2 or fewer points, or else c = 0 and μ is supported at n − 1 or fewer points. Either way, by Proposition 6.3, Φ is a real rational function of degree at most n − 1.

Conversely, suppose Φ is a real rational function of degree precisely n−1. Then either μ is supported at n − 1 points (say, x_{1}, …, x_{n−1}) or c ≠ 0 and μ is supported at n − 2 points (say, x_{1}, …, x_{n−2}). The map ψ : ℂ^{n} → ℂ^{n−1} given by

ψ_{j}(α) = ∑_{i=1}^{n} \frac{α_{i}}{x_{j} − z_{i}}

(with ψ_{n−1}(α) = ∑_{i=1}^{n} α_{i} if c ≠ 0) has a non-zero kernel by dimension counting. Thus, the right side of (6.13) is zero for some α ∈ ℂ^{N}. If degree Φ is smaller than n − 1, we can find (α_{1}, …, α_{ℓ}) ∈ ℂ^{ℓ} so that ∑_{i,j=1}^{ℓ} ᾱ_{i}α_{j}D_{ij} = 0, which still implies that det(D) = 0. □

Proposition 6.13. Let z_{1}, …, z_{n}; w_{1}, …, w_{n} lie in ℂ_{+} with the z_{i} distinct. Suppose the matrix D_{ij} given by (6.11) is (not necessarily strictly) positive definite. Then there exists a Herglotz function with Φ(z_{i}) = w_{i}, i = 1, …, n.

Proof. This is a fairly standard use of the Hahn-Banach theorem. We will use the representation (6.8) for Φ and the function φ_{z} of (6.7). Begin by noting that without loss of generality, we can suppose z_{n} = i (by mapping ℂ_{+} → ℂ_{+} with a linear map that takes the original z_{n} to i) and that Re w_{n} = 0 (by adjusting the constant d in (6.8)). Indeed, since

φ_{i}(x) ≡ i,  (6.14)

this choice Re Φ(i) = 0 is equivalent to d = 0 in the representation (6.8).

Let V be the 2n − 1-dimensional subspace of C(ℝ ∪ {∞}), the real-valued continuous functions on ℝ ∪ {∞} spanned by 1, {Re φ_{z_{j}}( ·)}_{j=1}^{n−1} and {Im φ_{z_{j}}( ·)}_{j=1}^{n−1}. Any f ∈ V can be written:

f = A_{n} + ∑_{j=1}^{n−1} A_{j}φ_{j}(x) + Ā_{j} \overline{φ_{j}(x)}  (6.15)

with A_{1}, …, A_{n−1} ∈ ℂ and A_{n} ∈ ℝ. Define a linear functional f : V → ℝ by

ℓ(f) = A_{n} Im w_{n} + ∑_{j=1}^{n−1} A_{j}w_{j} + Ā_{j}w̄_{j}  (6.16)

if f has the form (6.15).

We will prove shortly that

f(x) > 0  for all x ⇒ ℓ(f) > 0.  (6.17)

Assuming this, we let X = {f ∈ C(ℝ ∪ {∞} | f(x) > 0 for all x} and Y = {f ∈ V | ℓ(f) = 0}. Since X is open and Y closed, by the separating hyperplane version of the Hahn-Banach theorem (see [33]), there is a linear functional L on ℂ(ℝ ∪ {∞}) so L > 0 on X and L ≤ 0 on Y. If L is normalized so L(1) = Im w_{n}, it is easy to see that L extends ℓ, and so defines a measure σ on ℝ ∪ {∞} with

Φ(z_{i}) ≡ ∫ φ_{z_{i}}(x) dσ(x) = w_{i}, i = 1, …, n.

Thus, if (6.17) holds, we have the required Φ.

Since f(x) > 0 on the compact set ℝ ∪ {∞} implies f ≡ ε + g with g ≥ 0 and ε > 0 (by hypothesis, 1 ∈ V) and Im w_{n} > 0, we need only show f ≥ 0 implies ℓ(f) ≥ 0.

If f has the form (6.15), we can write

f(x) = \frac{Q(x)}{∏_{j=1}^{n−1} |x − z_{j}|^{2}} ,

where Q is a polynomial. f ≥ 0 implies Q ≥ 0. Any real non-negative polynomial has roots either in complex conjugate pairs or double real roots, so Q must have even degree and have the form

Q(x) = c^{2} ∏_{k=1}^{ℓ} |x − y_{k}|^{2}

for suitable y_{1}, …, y_{ℓ} ∈ ℝ ∪ ℂ_{+}. Thus, with R(x) = c ∏_{k=1}^{ℓ}(x − y_{k}),

f = |h|^{2}  with  h(x) = \frac{R(x)}{∏_{j=1}^{n−1}(x − z_{j})} .

Since f is bounded on ℝ ∪ {∞}, deg(R) ≤ n − 1, and thus h has the form

h(x) = β_{n} + ∑_{j=1}^{n−1} β_{j}(x − z_{j})^{−1}.  (6.18)

Since (x − z_{j})^{−1} = (z_{j} − i)^{−1}[\frac{(x−i)}{(x−z_{j})} − 1], we can rewrite (6.18) as

h(x) = ∑_{j=1}^{n} α_{j} \frac{x − i}{x − z_{j}}

for real x. Thus,

f(x) = |h(x)|^{2} = ∑_{j=1}^{n} ᾱ_{i}α_{j} \frac{x^{2} + 1}{(x − z̄_{j})(x − z_{j})} = ∑_{j=1}^{n} ᾱ_{i}α_{j}[\frac{φ_{z_{j}}(x) − \overline{φ_{z_{i}}(x)}}{z_{j} − z̄_{i}}]

so that

ℓ(f) = ∑_{i,j=1}^{n} ᾱ_{i}α_{j}D_{ij},

which is non-negative by hypothesis. □

Proof of Theorem 6.11. The first assertion is a direct consequence of Propositions 6.12 and 6.13. Proposition 6.12 shows the equivalence of (1) and (3).

To prove (1) ⇒ (2), suppose (1) holds and, as in the last proof, we can suppose that z_{n} = i and Re w_{n} = 0. Then by our proof of Proposition 6.12, the set of points where the measure σ of (6.8) is supported is determined by the α with ∑_{i} D_{ij}α_{j} = 0 as an n − 1 point set x_{1}, …, x_{n−1} (with x_{n−1} = ∞ allowed). Thus, any such Φ is of the form P/Q where deg(P) ≤ n − 1 and Q(z) = ∏_{j=1}^{n−1}(z − x_{j}) (if x_{n−1} = ∞, the product only goes from 1 to n − 2), If Φ_{1} and Φ_{2} are two solutions of (6.12), they have the same Q but they could be distinct P ’s, say P_{1} and P_{2}. But by (6.12), P_{1}(z) − P_{2}(z) vanishes at z_{1}, …, z_{n}. Since P_{1} − P_{2} is a polynomial of degree n − 1, this is only possible if P_{1} − P_{2} = 0, that is, Φ_{1} = Φ_{2}. Thus, (1) ⇒ (2).

For the converse, suppose the determinant

D(z_{1}, …, z_{n}; w_{1}, …, w_{n}) > 0.  (6.19)

If n = 1, since z, w ∈ ℂ_{+}, it is easy to see there are multiple Φ’s obeying (6.12). (For example, as usual we can consider the case z = w = i and then that Φ(z) = z or Φ(z) = −z^{−1}.) So we suppose n ≥ 2.

Consider the function g(w) = det[D(z_{1}, …, z_{n}; w_{1}, …, w_{n−1}, w)] where we vary w_{n}. g(w) is of the form c|w|^{2} + dw + d̄w̄ + e with c < 0, e real, and d complex. (c is strictly negative since it is − det(D(z_{1}, …, z_{n−1}; w_{1}, …, w_{n−1}) and we are supposing (6.19) and n ≥ 2.) Thus, the set g(w) ≥ 0 is a disk and since g(w_{n}) > 0, w_{n} is in its interior. Let w_{0} be the center of this disk, R its radius, and let w(θ) = w_{0} + Re^{iθ}. Since we have proven (1) ⇒ (2), there is a unique Φ with

Φ_{θ}(z_{i}) = w_{i}, i = 1, …, n − 1;  Φ_{θ}(z_{n}) = w(θ)  (6.20)

and it is a rational function of degree at most n − 1.

As usual, we can suppose z_{1} = w_{1} = i so each Φ_{θ}(z) has the form

Φ_{θ}(z) = ∫ \frac{1 + xz}{x − z} dσ_{θ}(x),  (6.21)

where dσ_{θ} is a probability measure on ℝ ∪ {∞}. If θ_{k} → θ_{∞}, and dσ_{θ_{k}} → dρ, then the Herglotz function associated to dρ obeys (6.20) for θ_{∞} and so it must be Φ_{θ_{∞}}. It follows (since the probability measures are compact) that dρ = dσ_{θ_{∞}} and thus {dσ_{θ}} is closed and dσ_{θ} ↦ Φ_{θ} is continuous. Since a continuous bijection between compact sets has a continuous inverse, θ ↦ dσ_{θ} is continuous.

Since Φ_{θ} is unique, dσ_{θ} is a pure point measure with at most n − 1 pure points. Since the function is not determined by (z_{1}, …, z_{n−1}), there must be exactly n − 1 points. It follows that the points in the support must vary continuously in θ.

Note next that for any θ, we can find θ′ and t_{θ} ∈ (0, 1) so that

w_{n} = t_{θ}w(θ) + (1 − t_{θ})w(θ′).

Suppose that there is a unique Φ obeying (6.12). It follows that for any θ,

Φ(z) = t_{θ}Φ_{θ}(z) + (1 − t_{θ})Φ_{θ′}(z)  (6.22)

and thus Φ(z) has a representation of the form (6.21) with dσ a point measure with at most 2n − 2 pure points. By (6.22) again, each dσ_{θ} must be supported in that 2n − 2 point set, and then by continuity, in a fixed n − 1 point set. But if dσ_{θ} is supported in a fixed θ-independent n − 1 point set, so is dσ, and thus (6.19) fails. We conclude that there must be multiple Φ’s obeying (6.12). □

Appendix A: The Theory of Moments and Determinantal Formulae

The theory of moments has a variety of distinct objects constructed in principle from the moments {γ_{n}}_{n=0}^{∞}: the orthogonal polynomials {P_{n}(x)}_{n=0}^{∞}, the associated polynomials {Q_{n}(x)}_{n=0}^{∞}, the sums ∑_{j=0}^{n} P_{j}(0)^{2} and ∑_{j=0}^{n} Q_{j}(0)^{2}, the Jacobi matrix coefficients, and the approximations −\frac{Q_{n}(x)}{P_{n}(x)} and −\frac{N_{n}(x)}{M_{n}(x)}. It turns out most of these objects can be expressed as determinants. These formulae are compact and elegant, but for some numerical applications, they suffer from numerical round-off errors in large determinants.

We have already seen two sets of determinants in Theorem 1. Namely, let H_{N} be the N × N matrix,

H_{N} = ( \begin{matrix}γ_{0} & γ_{1} & … & γ_{N−1} \\ γ_{1} & γ_{2} & … & γ_{N} \\ ⋮ & ⋮ & & ⋮ \\ γ_{N−1} & γ_{N} & … & γ_{2N−2} \end{matrix} )  (A.1a)

and S_{N}, the matrix

S_{N} = ( \begin{matrix}γ_{1} & γ_{2} & … & γ_{N} \\ γ_{2} & γ_{3} & … & γ_{N+1} \\ ⋮ & ⋮ & & ⋮ \\ γ_{N} & γ_{N+1} & … & γ_{2N−1} \end{matrix} )  (A.1b)

and define

h_{N} = det(H_{N}),  s_{N} = det(S_{n})  (A.2)

(for comparison, D_{N} from [1] is our h_{N+1}). We will use h_{N}(γ) if we need to emphasize what moments {γ_{n}}_{n=0}^{∞} are involved. Thus, Theorem 1 says h_{N} > 0 for all N is equivalent to solubility of the Hamburger problem and h_{N} > 0, s_{N} > 0 for all N is equivalent to solubility of the Stieltjes problem.

There is an interesting use of determinants to rewrite h_{N} and s_{N} in terms of the moment problem that makes their positivity properties evident. Suppose dρ obeys (1.1). Then

h_{N} = ∫ det((x_{a}^{a+b})_{0≤a,b≤N−1}) ∏_{a=0}^{N−1} dρ(x_{a})

= ∫ [∏_{a=1}^{n} x_{a}^{a}] det((x_{a}^{b}))_{0≤a,b≤N−1} ∏_{a=0}^{N−1} dρ(x_{a}).

Permuting over indices, we see that in ∏_{a=1}^{n}(x_{a}^{a}) det(x_{a}^{b}), we can replace x_{a} by x_{π(a)} for any permutation, π. Since det(x_{π(a)}^{b}) = (−1)^{π} det(x_{a}^{b}), we see that

h_{N} = (N!)^{−1} ∫ [det((x_{a}^{b})_{0≤a,b≤N−1})]^{2} ∏_{a=0}^{N−1} dρ(x_{a}).

Recognizing the Vandermonde determinant, we have

h_{N} = (N!)^{−1} ∫ ∏_{0≤a<b≤N−1}(x_{a} − x_{b})^{2} ∏_{a=0}^{N−1} dρ(x_{a}).  (A.3a)

Similarly,

s_{N} = (N!)^{−1} ∫ x_{0} … x_{N−1} ∏_{0≤a<b≤N−1}(x_{a} − x_{b})^{2} ∏_{a=0}^{N−1} dρ(x_{a}).  (A.3b)

The most basic formula is:

Theorem A.1. P_{n}(x) is given by

P_{n}(x) = \frac{1}{\sqrt{h_{n}h_{n+1}}} det ( \begin{matrix}γ_{0} & γ_{1} & … & γ_{n} \\ γ_{1} & γ_{2} & … & γ_{n+1} \\ ⋮ & ⋮ & & ⋮ \\ γ_{n−1} & γ_{n} & … & γ_{2n−1} \\ 1 & x & … & x^{n} \end{matrix} ) .  (A.4)

Proof. Let S_{n}(x) be the determinant on the right side of (A.4). Then for any solution ρ of the moment problem, we have that ∫ S_{n}(x)x^{j} dρ(x) is given by the same determinant, but with the last row replaced by γ_{j} γ_{j+1} … γ_{j+n}. It follows that

∫ x^{j}S_{n}(x) dρ(x) = 0,  j = 0, 1, …, n − 1  (A.5)
= h_{n+1}  j = n.

THE CLASSICAL MOMENT PROBLEM 91

In particular, since

S_{n}(x) = h_{n}x^{n} + lower order,  (A.6)

we see that

∫ S_{n}(x)^{2} dρ(x) = h_{n}h_{n+1}.  (A.7)

From (A.5)–(A.7), it follows that P_{n}(x) = \frac{S_{n}(x)}{\sqrt{h_{n}h_{n+1}}} . □

From (A.4), we can deduce formulae for the coefficients a_{n}, b_{n} of a Jacobi matrix associated to the moments {γ_{n}}_{n=0}^{∞}. Let h̃_{n} be the n × n determinant obtained by changing the last column in H_{N}:

h̃_{n} = det ( \begin{matrix}γ_{0} & γ_{1} & … & γ_{n−2} & γ_{n} \\ ⋮ & ⋮ & & ⋮ & ⋮ \\ γ_{n} & γ_{n+1} & … & γ_{2n−3} & γ_{2n−1} \end{matrix} ) .

Thus (A.4) implies that

P_{n}(x) = \sqrt{\frac{h_{n}}{h_{n+1}}} [x^{n} − \frac{h̃_{n}}{h_{n}} x^{n−1} + lower order].  (A.8)

Theorem A.2. For n ≥ 0,

(i)  a_{n} = (\frac{h_{n}h_{n+2}}{h_{n+1}^{2}} )^{1/2}
(ii)  ∑_{j=0}^{n} b_{j} = \frac{h̃_{n+1}}{h_{n+1}} .

Proof. By the definition of a_{n} and b_{n}, we have that

xP_{n}(x) = a_{n}P_{n+1}(x) + b_{n}P_{n}(x) + a_{n−1}P_{n−1}(x).  (A.9)

Identifying the x^{n+1} and x^{n} terms in (A.9) using (A.8), we see that

\sqrt{\frac{h_{n}}{h_{n+1}}} = a_{n}\sqrt{\frac{h_{n+1}}{h_{n+2}}}  (A.10)
\sqrt{\frac{h_{n}}{h_{n+1}}} (−\frac{h̃_{n}}{h_{n}} ) = a_{n}\sqrt{\frac{h_{n+1}}{h_{n+2}}} (−\frac{h̃_{n+1}}{h_{n+1}} ) + b_{n}\sqrt{\frac{h_{n}}{h_{n+1}}} .  (A.11)

(A.10) implies (i) immediately, and given (A.10), (A.11) becomes

b_{n} = \frac{h̃_{n+1}}{h_{n+1}} − \frac{h̃_{n}}{h_{n}} ,

which implies (ii) by induction if we note the starting point comes from looking at the constant term in xP_{0}(x) = x = a_{0}P_{1}(x) + b_{0}P_{0}(x), which implies that b_{0} = \frac{h̃_{1}}{h_{1}} = \frac{γ_{1}}{γ_{0}} . □

Remark. (ii) has an alternate interpretation. (A.8) says that \frac{h̃_{n+1}}{h_{n+1}} is the sum of the n+1 roots of P_{n+1}(x). But P_{n+1}(x) is a multiple of the determinant of the Jacobi matrix A_{F}^{[n+1]}. So the sum of the roots is just the trace of A_{F}^{[n+1]}, that is, ∑_{j=0}^{n} b_{j}.

From (A.4) and Theorem 4.2, (i.e., Q_{n}(x) = E_{X}(\frac{P_{n}(X)−P_{n}(Y)}{X−Y})), we immediately get

Theorem A.3.

Q_{n}(x) = \frac{1}{\sqrt{h_{n}h_{n+1}}} det ( \begin{matrix}γ_{0} & γ_{1} & … & γ_{n} \\ γ_{1} & γ_{2} & … & γ_{n+1} \\ ⋮ & ⋮ & & ⋮ \\ γ_{n−1} & γ_{n} & … & γ_{2n−1} \\ R_{n,0}(x) & R_{n,1}(x) & … & R_{n,n}(x) \end{matrix} )  (A.12)

where

R_{n,j}(x) = ∑_{k=0}^{j−1} γ_{j−1−k}x^{k},  j ≥ 1  (A.13a)
R_{n,j=0}(x) = 0.  (A.13b)

Proof. \frac{x^{j}−y^{j}}{x−y} = ∑_{k=0}^{j−1} y^{j−1−k}x^{k} so E_{X}(\frac{x^{j}−y^{j}}{x−y}) = R_{n,j}(x). □

Remark. By (A.4), P_{n}(x) ∑_{k=0}^{2n−1} γ_{k}x^{−k−1} has the same form as (A.4) but with the bottom row replaced by S_{n,0}(x) … S_{n,n}(x) where

S_{n,j}(x) = ∑_{k=0}^{2n−1} γ_{k}x^{j−k−1} = ∑_{ℓ=−(2n−j)}^{j−1} γ_{j−1−ℓ}x^{ℓ}
= R_{n,j}(x) + x^{−1}γ_{j} + x^{−2}γ_{j+1} + ⋯ + x^{−n}γ_{j+n−1} + O(x^{−n−1}).

Recognizing x^{−1}γ_{j} +x^{−2}γ_{j+1}+⋯+x^{−n−1}γ_{j+n−1} as x^{−1} times the first row of the matrix in (A.4) plus x^{−2} times the second row plus …, we conclude that

P_{n}(x)( ∑_{k=0}^{2n−1} γ_{k}x^{−k−1}) = Q_{n}(x) + O(x^{−n−1})

and thus

−\frac{Q_{n}(x)}{P_{n}(x)} = − ∑_{k=0}^{2n−1} γ_{k}x^{−k−1} + O(x^{−2n−1})

consistent with the f^{[N−1,N]} Padé formula (5.28).

We saw the quantity L of (5.29) is important. Here is a formula for it.

Theorem A.4. L = lim_{n→∞} −\frac{Q_{n}(0)}{P_{n}(0)} and

−\frac{Q_{n}(0)}{P_{n}(0)} = \frac{t_{n}}{s_{n}} ,  (A.14a)

where s_{n} = det(S_{N}) and

t_{n} = − det ( \begin{matrix}0 & γ_{0} & γ_{1} & … & γ_{n} \\ γ_{0} & γ_{1} & γ_{2} & … & γ_{2n+1} \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ γ_{n} & γ_{n+1} & γ_{n+2} & … & γ_{2n−1} \end{matrix} ) .  (A.14b)

Proof. Follows by putting x = 0 in our formula for P_{n}(x) and Q_{n}(x). □

Next we have explicit formulae for M_{n}(x) and N_{n}(x), the polynomials introduced for Section 5 (see (5.15) and (5.16)).

Theorem A.5.

M_{n}(x) = \frac{1}{s_{n−1}} \sqrt{\frac{h_{n}}{h_{n+1}}} ( \begin{matrix}γ_{1} & γ_{2} & … & γ_{n} \\ γ_{2} & γ_{3} & … & γ_{n+1} \\ ⋮ & ⋮ & & ⋮ \\ γ_{n−1} & γ_{n−2} & … & γ_{2n−2} \\ x & x^{2} & … & x^{n} \end{matrix} )  (A.15)

N_{n}(x) = \frac{1}{s_{n−1}} \sqrt{\frac{h_{n}}{h_{n+1}}} ( \begin{matrix}γ_{1} & γ_{2} & … & γ_{n} \\ γ_{2} & γ_{3} & … & γ_{n+1} \\ ⋮ & ⋮ & & ⋮ \\ γ_{n−1} & γ_{n−2} & … & γ_{2n−2} \\ R_{n,1}(x) & R_{n,2}(x) & … & R_{n,n}(x) \end{matrix} )  (A.16)

where R_{i,j}(x) is given by (A.13).

Remarks. 1. P_{n} is given by a (n + 1) × (n + 1) matrix. To get M_{n}, which has an n × n matrix, we drop the first column and next to last row.

2. It is interesting that (A.15) does not obviously follow from the basic definition (5.15).

Proof. The right side of (A.15) (call it M̃_{n}(x)) has the following properties: (i) It is a polynomial of degree n.

(2) It obeys M̃_{n}(0) = 0.

(3) It obeys E_{x}(x^{j}M̃_{n}(x)) = 0 for j = 0, 1, …, n − 2 since the corresponding matrix has two equal rows.

(4) It obeys M̃_{n}(x) = \sqrt{h_{n}/h_{n+1}} x^{n} + lower order so it has the same highest degree term as P_{n}(x).

These properties uniquely determine M_{n}(x) so (A.15) is proven. (A.16) then follows from (5.24). □

Remark. By mimicking the argument following Theorem A.3, one finds that

M_{n}(x) ∑_{k=0}^{2n−2} γ_{k}x^{−k−1} = N_{n}(x) + O(x^{−n})

so that

−\frac{N_{n}(x)}{M_{n}(x)} = − ∑_{k=0}^{2n−2} γ_{k}x^{−k−1} + O(x^{−2n}),

consistent with the f^{[N−1,N−1]} Padé formula used in the proof of Theorem 6 (at the end of Section 5).

Remarkably, there are also single determinant formulae for ∑_{j=0}^{n} P_{j}(0)^{2} and ∑_{j=0}^{n} Q_{j}(0)^{2}.

Theorem A.6. We have that

∑_{j=0}^{n} P_{j}(0)^{2} = \frac{v_{n}}{h_{n+1}}  (A.17)

∑_{j=0}^{n} Q_{j}(0)^{2} = −\frac{w_{n+2}}{h_{n+1}} ,  (A.18)

where v_{n} is given by the n × n determinant,

v_{n} = det ( \begin{matrix}γ_{2} & … & γ_{n+1} \\ γ_{3} & … & γ_{n+2} \\ ⋮ & & ⋮ \\ γ_{n+1} & … & γ_{2n} \end{matrix} )  (A.19)

and w_{n+2} by the (n + 2) × (n + 2) determinant,

w_{n} = det ( \begin{matrix}0 & 0 & γ_{0} & γ_{1} & … & γ_{n−1} \\ 0 & γ_{0} & γ_{1} & γ_{2} & … & γ_{n} \\ γ_{0} & γ_{1} & γ_{2} & γ_{3} & … & γ_{n+1} \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ \\ γ_{n−1} & γ_{n} & γ_{n+1} & γ_{n+2} & … & γ_{2n} \end{matrix} )  (A.20)

Remark. Thus by Theorems 3 and 7, indeterminacy for both the Hamburger and Stieltjes problems can be expressed in terms of limits of ratios of determinants. In the Hamburger case, we need (A.17) and (A.18) to have finite limits in order that the problem be indeterminate. In the Stieltjes case, (A.17) and (A.14) must have finite limits for indeterminacy to hold.

Proof. We actually prove a stronger formula. Let

B_{N}(x, y) = ∑_{n=0}^{N} P_{n}(x)P_{n}(y).  (A.21)

We will show that

B_{n}(x, y) = −h_{n+1}^{−1} det ( \begin{matrix}0 & 1 & x & … & x^{N} \\ 1 & γ_{0} & γ_{1} & … & γ_{N} \\ y & γ_{1} & γ_{2} & … & γ_{N+1} \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ y^{N} & γ_{N} & γ_{N+1} & … & γ_{2N} \end{matrix} )  (A.22)

(A.17) then follows by setting x = y = 0 and noting that if C = {C_{ij}}_{1≤i≤j≤N} is an N × N matrix with C_{11} = 0, C_{1j} = δ_{2j}, and C_{i1} = δ_{i2}, then

det((C_{ij})_{1≤i,j≤N}) = − det((C_{ij})_{3≤i,j≤N}).

(A.18) then follows from Q_{j}(z) = E_{x}(\frac{[P_{j}(x)−P_{j}(z)]}{(x−z)}), since

∑_{n=0}^{N} Q_{n}(0)^{2} = E_{x}E_{y}([B_{N}(x, y) − B_{N}(x, 0) − B_{N}(0, y) + B_{N}(0, 0)]x^{−1}y^{−1}).

Thus we need only prove (A.22). To do this, let B̃_{N}(x, y) be the right side of (A.22). Consider E_{x}(B̃_{N}(x, y)x^{j}) for 0 ≤ j ≤ N. This replaces the top row in the determinant by (0 γ_{j} γ_{j+1} … γ_{N+j}). The determinant is unchanged if we subtract the row (y^{j} γ_{j} … γ_{N+j}) from this row. That is, E_{x}(B̃_{N}(x, y)x^{j}) is given by the determinant with top row (−y^{j} 0 0 … 0). Thus,

E_{x}(B̃_{N}(x, y)x^{j}) = −h_{n+1}^{−1}(−y^{j})h_{n+1} = y^{j}.

So, B̃_{N}(x, y) is a reproducing kernel. For any polynomial P (x) of degree N or smaller, E_{x}(B̃_{N}(x, y)P (x)) = P (y). But

B̃_{N}(x, y) = ∑_{j=0}^{N} E_{x}(B̃_{N}(x, y)P_{j}(x))P_{j}(x) = B_{N}(x, y)

since {P_{j}(x)}_{j=0}^{N} is an orthonormal basis in the polynomials of degree N or less. Thus (A.22) is proven. □

In terms of the γ^{(ℓ)} moment problems (with moments γ_{j}^{(ℓ)} = \frac{γ_{ℓ+j}}{γ_{ℓ}}), we recognize v_{n} as (γ_{2})^{n−1}h_{n−1}(γ^{(2)}). By (A.17) for the γ^{(2)} problem,

∑_{j=0}^{n−1} P_{j}^{(2)}(0) = \frac{v_{n−1}(γ^{(2)})}{h_{n}(γ^{(2)})} .

But h_{n}(γ^{(2)}) = v_{n}(γ)(γ_{2})^{−n} and v_{n−1}(γ^{(2)}) = γ_{2}^{1−n}y_{n−1}(γ) where

y_{n−1} = ( \begin{matrix}γ_{4} & … & γ_{n+2} \\ γ_{5} & … & γ_{n+3} \\ ⋮ & & ⋮ \\ γ_{n+1} & … & γ_{2n} \end{matrix} ) .  (A.23)

Thus, using Proposition 5.13:

Theorem A.7.

γ_{2}^{−1}(∑_{j=0}^{n} P_{j}(0)^{2})(∑_{j=0}^{n−1} P_{j}^{(2)}(0)^{2}) = \frac{y_{n−1}}{h_{n+1}} .

In particular, the Hamburger problem is determinate if and only if lim_{n→∞} \frac{y_{n−1}}{h_{n+1}} = ∞.

So we have a simple ratio of determinants to determine determinacy.

Appendix B: The Set of Solutions of the Moment Problem as a Compact Convex Set

In this appendix, we will prove (following [1]) that M^{H}(γ) and M^{S}(γ) are compact convex sets whose extreme points are dense. Each set is a subset of M_{+}(ℝ ∪ {∞}), the set of measures on the compact set ℝ∪{∞}. This set, with the condition ∫ dρ(x) = γ_{0}, is a compact space in the topology of weak convergence (i.e., ∫ f(x) dρ_{n}(x) → ∫ f(x) dρ(x) for continuous functions f on ℝ ∪ {∞}).

Theorem B.1. M^{H}(γ) and M^{S}(γ) are closed in the weak topology and so are compact convex sets.

Remark. Since the x^{n}’s are unbounded, this does not follow from the definition of the topology without some additional argument.

Proof. By Propositions 4.4 and 4.13, μ ∈ M^{H}(γ) if and only if

∫ (x − z_{0})^{−1} dμ(x) ∈ D(z_{0})  (B.1)

for all z_{0} ∈ ℂ. The set of μ’s that obey (B.1) for a fixed z_{0} is closed since D(z_{0}) is closed and (x − z_{0})^{−1} is in C(ℝ ∪ {∞}). Thus, the intersection over all z_{0} is closed.

M^{S}(γ) = M^{H}(γ) ∩ {μ | ∫ f(x) dμ(x) = 0 if supp f ⊂ (−∞, 0)} is an intersection of closed sets. □

Remark. To get compact sets of measures, we need to consider ℝ ∪ {∞} rather than ℝ. Thus a priori, the integral in (B.1) could have a point at infinity giving a constant term as z_{0} = iy with y → ∞. Since D(z_{0}) → {0} as z_{0} = iy with y → ∞, this term is absent.

Theorem B.2 (Naimark). μ ∈ M^{H}(γ) (resp. M^{S}(γ)) is an extreme point if and only if the polynomials are dense in L^{1}(ℝ, dρ).

Remark. Compare with density in L^{2}(ℝ, dρ) which picks out the von Neumann solutions.

Proof. This is a simple use of duality theory. The polynomials fail to be dense if and only if there exists a non-zero F ∈ L^{∞}(ℝ, dρ) so that

∫ x^{n}F (x) dρ(x) = 0  (B.2)

for all n. We can suppose that ∥F ∥_{∞} = 1, in which case dρ_{±} = (1 ± F) dρ both lie in M^{H}(γ) with ρ = \frac{1}{2}(ρ_{+} + ρ_{−}), so ρ is not extreme.

Conversely, suppose ρ = \frac{1}{2}(ρ_{+} + ρ_{−}) with ρ_{+} ≠ ρ_{−} and both in M^{H}(γ). Then ρ_{±} ≤ 2ρ so ρ_{+} is ρ-absolutely continuous, and the Radon-Nikodyn derivative, \frac{dρ_{+}}{dρ}, obeys

∥\frac{dρ_{+}}{dρ} ∥_{∞} ≤ 2. Let F = 1 − \frac{dρ_{+}}{dρ}, so ∥F ∥_{∞} ≤ 1 and F ≠ 0 since ρ_{+} ≠ ρ_{−}. Then

∫ F (x)x^{n} dρ = ∫ x^{n} dρ − ∫ x^{n} dρ_{+} = 0.

Thus, we have proven the result for M^{H}(γ). Since M^{S}(γ) = {ρ ∈ M^{H}(γ) | ∫ f(x) dρ(x) = 0 for f in C(ℝ ∪ {∞}) with support in (−∞, 0)}, M^{S}(γ) is a face of M^{H}(γ), so extreme points of M^{S}(γ) are exactly those extreme points of M^{H}(γ) that lie in M^{S}(γ). □

Theorem B.3. Let ρ ∈ M^{H}(γ) have ord(ρ) < ∞. Then ρ is an extreme point.

Proof. By Theorem 6.4, for some z_{1}, …, z_{n} ∈ ℂ_{+}, we have that the polynomials are dense in L^{2}(ℝ, ∏_{j+1}^{n} |x − z_{j}|^{−2}dρ). For any polynomially bounded continuous function

∫ |f(x)| dρ(x) ≤ (∫ |f(x)|^{2} ∏_{j=1}^{n} |x − z_{j}|^{−2} dρ(x))^{1/2}(∫ ∏_{j=1}^{n}(x − z_{j})^{2} dρ(x))^{1/2}

by the Schwarz inequality. It follows that the identification map is continuous from L^{2}(ℝ, ∏_{j=1}^{n}(x − z)^{−2} dρ) into L^{1}(ℝ, dρ), so the polynomials are dense in L^{1}-norm in the continuous functions, and so in L^{1}(ℝ, dρ). By Theorem B.2, ρ is an extreme point. □

Theorem B.4. The extreme points are dense in M^{H}(γ) (and in M^{S}(γ)).

Proof. The finite point measures are dense in the finite measures on ℝ ∪ {∞}. Thus, by the Herglotz representation theorem in form (6.8), if Φ is a Herglotz function, there exist real rational Herglotz functions Φ_{n} so that Φ_{n}(z) → Φ(z) for each z ∈ ℂ_{+}.

Now let ρ ∈ M^{H}(γ) and let Φ_{ρ} be the Nevanlinna function of ρ. Let Φ_{n} be as above and ρ_{n} = ρ_{Φ_{n}}. Then by (4.35), G_{ρ_{n}}(z) → G_{ρ}(z) for each z ∈ ℂ_{+} and so ρ_{n} → ρ weakly. By Theorems 6.4 and B.3, each such ρ_{n} is an extreme point.

For the Stieltjes case, we need only note that by Theorem 4.18 and the remark after it, if ρ ∈ M^{S}(γ), then the approximating Φ_{n}’s can be chosen so that ρ_{Φ_{n}} ∈ M^{S}(γ). □

Theorem B.5. For any indeterminate set of Hamburger moments {γ_{n}}_{n=0}^{∞}, M^{H}(γ) has extreme points ρ with ord(ρ) = ∞.

Proof. We first pick positive α_{j} strictly decreasing so that for any ρ ∈ M^{H}(γ) we have that

sup_{n} ∫ ∏_{j=1}^{n}(1 + α_{j}^{2}x^{2})^{2} dρ(x) ≤ 2.  (B.3)

We can certainly do this as follows: Since the integral only depends on the moments γ_{j}, we need only do it for some fixed ρ_{0} ∈ M^{H}(γ). Since

lim_{α_{1}↓0} ∫ (1 + α_{1}^{2}x^{2})^{2} dρ(x) = 1,

we can pick α_{1} > 0 so

∫ (1 + α_{1}^{2}x^{2})^{2} dρ(x) < 2.

We then pick α_{2}, α_{3}, … inductively so the integral is strictly less than 2. Then the sup in (B.3) is bounded by 2. The product must be finite for a.e. x w.r.t. dρ, so

∑_{j=1}^{∞} α_{j}^{2} < ∞.  (B.4)

So if

g_{n}(x) = ∏_{j=1}^{n}(1 + α_{j}^{2}x^{2}),

then for any real x,

g(x) = lim_{n→∞} g_{n}(x)

exists, and by (B.3),

∫ g(x)^{2} dρ(x) < ∞

for any ρ ∈ M^{H}(γ).

Now pick ζ_{1}, ζ_{2}, … in ℂ_{+} and ρ_{1}, ρ_{2}, … in M^{H}(γ) inductively so ρ_{k} obeys

∫ (x − iα_{j}^{−1})^{−1} dρ_{k}(x) = ζ_{j},  j = 1, …, k  (B.5)

and then ζ_{k+1} is the middle of the disk of allowed values for ∫ (x − iα_{j+1}^{−1})^{−1} dρ(x) for those ρ in M^{H}(γ) which obey (B.5). Pick a subsequence of the ρ_{j}’s which converges to some ρ_{∞} ∈ M^{H}(γ). Then ρ_{∞} obeys (B.5) for all k.

Let

M_{k} = {ρ ∈ M^{H}(γ) | ρ obeys (B.5)}

and

M_{∞} = ∩M_{k}.

Define for all n, m,

Γ_{m}^{(n)} = ∫ x^{m} ∏_{j=1}^{n}(1 + α_{j}^{2}x^{2})^{−1} dρ_{n}(x).

By Theorem 6.1, μ ∈ M^{H}(Γ^{(n)}) if and only if dρ ≡ g_{n} dμ lies in M_{k}. Γ_{m}^{(n)} is decreasing to Γ_{m}^{(∞)} and

Γ_{m}^{(∞)} = ∫ x^{m}g(x)^{−1} dρ(x)

by a simple use of the monotone convergence theorem.

We claim that μ ∈ M^{H}(Γ^{(∞)}) if and only if dρ = g dμ lies in M_{∞}. For

γ_{n} = lim_{n→∞} ∫ x^{m}g(x)^{−1}g_{n}(x) dρ(x)

and the right side only depends on the moments Γ_{m}^{(∞)} and similarly for calculation of ∫ (x − iα_{j})^{−1} dρ(x).

Now let μ be a von Neumann solution of the Γ^{(∞)} moment problem and dρ = g dμ. Since ∫ g^{2} dρ < ∞ by construction, the proof of Theorem B.3 shows that ρ is an extreme point. On the other hand, since ρ obeys (B.5) and ζ_{j} is not in the boundary of its allowed circle, ord(ρ) = ∞. □

Appendix C: Summary of Notation and Constructions

Since there are so many objects and constructions associated to the moment problem, I am providing the reader with this summary of them and where they are discussed in this paper.

C1 Structure of the Set of Moments.

{γ_{n}}_{n=0}^{∞} is called a set of Hamburger moments if there is a positive measure ρ on (−∞, ∞) with γ_{n} = ∫ x^{n} dρ(x) and a set of Stieltjes moments if there is a ρ with support in [0, ∞). We take γ_{0} = 1 and demand supp(ρ) is not a finite set. Theorem 1 gives necessary and sufficient conditions for existence. If there is a unique ρ, the problem is called determinate. If there are multiple ρ’s, the problem is called indeterminate. If a set of Stieltjes moments is Hamburger determinate, it is a fortiori Stieltjes determinate. But the converse can be false (see the end of Section 3).

M^{H}(γ) (resp. M^{S}(γ)) denotes the set of all solutions of the Hamburger problem (resp. all ρ ∈ M^{H}(γ) supported on [0, ∞)). They are compact convex sets (Theorem B.1) whose extreme points are dense (Theorem B.4). Indeterminate Hamburger problems have a distinguished class of solutions we have called von Neumann solutions (called Nextremal in [1] and extremal in [35]). They are characterized as those ρ ∈ M^{H}(γ) with the polynomials dense in L^{2}(ℝ, dρ). They are associated to self-adjoint extensions B_{t} of the Jacobi matrix associated to γ (see C4 below for the definition of this Jacobi matrix). The parameter t lies in ℝ ∪ {∞} and is related to B_{t} and the solution μ_{t} ∈ M^{H}(γ) by

t = (δ_{0}, B_{t}^{−1}δ_{0}) = ∫ x^{−1}dμ_{t}(x).  (C.1)

In the indeterminate Stieltjes case, there are two distinguished solutions: μ_{F}, the Friedrichs solution, and μ_{K}, the Krein solution. Both are von Neumann solutions of the associated Hamburger problem and are characterized by inf[supp(μ_{F})] > inf[supp ρ] for any other ρ ∈ M^{H}(γ) and by μ_{K}({0}) > 0 and μ_{K}({0}) > ρ({0}) for any other ρ ∈ M^{H}(γ) (see Theorems 3.2 and 4.11 and 4.17). All solutions of the Stieltjes problem lie between μ_{F} and μ_{K} in the sense of (4.42) (see Theorem 5.10).

We define

ℂ_{+} = {z ∈ ℂ | Im z > 0}.

For any probability measure ρ, we define its Stieltjes transform as a function of ℂ\supp(ρ) by

G_{ρ}(z) = ∫ \frac{dρ(x)}{x − z} .

These functions map ℂ_{+} to ℂ_{+} and have an asymptotic series

G_{ρ}(z) ∼ −z^{−1}(∑_{j=0}^{∞} γ_{j}z^{−j})  (C.2)

as |z| → ∞ with min(|Arg(z)|, |Arg(−z)|) > ε. (C.2) characterizes M^{H}(γ) and the asymptotics are uniform over M^{H}(γ) (Proposition 4.13).

For an indeterminate moment problem, we defined ord(ρ), the order of ρ, to be the codimension of the closure of the polynomials in L^{2}(ℝ, dρ). The solutions of finite order are dense in M^{H}(γ) and in M^{S}(γ), and each is an extreme point. Solutions of finite order (and, in particular, the von Neumann solutions) are discrete pure point measures; equivalently, their Stieltjes transforms are meromorphic functions (Theorems 4.11 and 6.4).

C2 Nevanlinna Parametrization.

We used F to denote the analytic maps of ℂ_{+} to ℂ̄_{+} where the closure is in the Riemann sphere. The open mapping theorem implies that if Φ ∈ F, either Φ(z) ≡ t for some t ∈ ℝ ∪ {∞} or else Φ is a Herglotz function which has a representation of the form (1.19).

Associated to any indeterminate moment problem are four entire functions, A(z), B(z), C(z), D(z), obeying growth condition of the form |f(z)| ≤ c_{ε} exp(ε|z|). We defined them via the transfer matrix relation (4.16), but they also have explicit formulae in terms of the orthogonal polynomials, P and Q (defined in C4 below)—these are given by Theorem 4.9.

We define a fractional linear transformation F (z) : ℂ ∪ {∞} → ℂ ∪ {∞} by

F (z)(w) = −\frac{C(z)w + A(z)}{D(z)w + B(z)} .  (C.3)

For indeterminate Hamburger problems, there is a one-one correspondence between ρ ∈ M^{H}(γ) and Φ ∈ F given by

G_{ρ}(z) = F (z)(Φ(z)) = −\frac{C(z)Φ(z) + A(z)}{D(z)Φ(z) + B(z)}  (C.4)

for z ∈ ℂ_{+} (Theorem 4.14). Φ is called the Nevanlinna function of ρ, denoted by Φ_{ρ}. The von Neumann solutions correspond precisely to Φ(z) = t where t is given by (C.1) (Theorem 4.10). The solutions of finite order correspond precisely to the case where Φ is a ratio of real polynomials (Theorem 6.4).

Given a set of Stieltjes moments, the ρ ∈ M^{S}(γ) are precisely those ρ ∈ M^{H}(γ) whose Nevanlinna function has the form (Theorem 4.18)

Φ(z) = d_{0} + ∫_{0}^{∞} \frac{dμ(x)}{x − z}

with ∫ dμ(x) < ∞ and d_{0} ≥ t_{F} = ∫ dρ_{F} (x) = (δ_{0}, A_{F}^{−1}δ_{0}).

For z ∈ ℂ_{+}, the image of ℂ̄_{+} ∪ {∞} under F (z) is a closed disk denoted by D(z). Von Neumann solutions ρ have G_{ρ}(z) ∈ ∂D(z) for all z while other solutions have G_{ρ}(z) ∈ D(z)^{int} for all z (Theorem 4.3, Proposition 4.4, and Theorem 4.14).

C3 Derived Moment Problems.

Any set of moments has many families of associated moments. For each real c, γ(c) is defined by

γ_{n}(c) = ∑_{j=0}^{n} \binom{n}{j}c^{j}γ_{n−j}.  (C.5)

There is a simple map (ρ ↦ ρ( · − c)) that sets up a bijection between M^{H}(γ) and M^{H}(γ(c)) and, in particular, γ is Hamburger determinate if and only if γ(c) is Hamburger determinate. But the analog is not true for the Stieljes problem. Indeed, (end of Section 3), if γ is a set of indeterminate Stieltjes moments and c_{F} = − inf supp(dμ_{F}), then γ(c) is a set of Stieltjes moments if and only if c ≥ c_{F}, and it is Stieltjes determinate if and only if c = c_{F}. The orthogonal polynomials for γ(c) and γ are related via P_{N}^{(γ(c))}(z) = P_{N}(z − c).

Given a set of Stieltjes moments {γ_{n}}_{n=0}^{∞}, one defines Hamburger moments {Γ_{n}}_{n=0}^{∞} by

Γ_{2m} = γ_{m},  Γ_{2m+1} = 0.  (C.6)

There is a simple map (dρ ↦ dμ(x) = \frac{1}{2}χ_{[0,∞)}(x) dρ(x^{2}) + \frac{1}{2}χ_{(−∞,0]}(x) dρ(x^{2})) that sets up a bijection between M^{H}(γ) and those μ ∈ M^{H}(Γ) with dμ(−x) = dμ(x). Then γ is Stieltjes determinate if and only if the Γ problem is Hamburger determinate (Theorem 2.13). There are simple relations between the orthogonal polynomials for γ and those for Γ ((5.89)–(5.92)).

For ℓ = 1, 2, …, define

γ_{j}^{(ℓ)} = \frac{γ_{ℓ+j}}{γ_{ℓ}} .  (C.7)

If γ is a set of Stieltjes moments, so are γ^{(ℓ)} for all ℓ, and if γ is a set of Hamburger moments, so are γ^{(ℓ)} for ℓ = 2, 4, . . .. The map ρ ↦ dρ^{(ℓ)} = x^{ℓ} \frac{dρ(x)}{γ_{ℓ}} is a map from M^{H}(γ) to M^{H}(γ^{(ℓ)}) which is injective. But in the indeterminate case, it is not, in general, surjective. If γ is indeterminate, so is γ^{(ℓ)}, but the converse may be false for either the Hamburger or Stieltjes problems (Corollary 4.21). For general ℓ, the connection between the orthogonal polynomials for γ and γ^{(ℓ)} is complicated (see the proof of Proposition 5.13 for ℓ = 2), but (see the remarks following Theorem 5.14)

P_{N}^{(1)}(z) = \frac{d_{N}[P_{N}(0)P_{N+1}(z) − P_{N+1}(0)P_{N}(z)]}{z} .

Given a set of Hamburger moments {γ_{j}}_{j=0}^{∞}, one defines a new set {γ_{j}^{(0)}}_{j=0}^{∞} by the following formal power series relation:

[∑_{n=0}^{∞}(−1)^{n}γ_{n}z^{n}]^{−1} = 1 − γ_{1}z + (γ_{2} − γ_{1}^{2})z^{2}(∑_{n=0}^{∞}(−1)^{n}γ_{n}^{(0)}z^{n}).  (C.8)

This complicated formula is simple at the level of Jacobi matrices. If {a_{n}}_{a=0}^{∞}, {b_{n}}_{n=0}^{∞} (resp. {a_{n}^{(0)}}_{n=0}^{∞}, {b_{n}^{(0)}}_{n=0}^{∞}) are the Jacobi matrix coefficients associated to γ (resp. γ^{(0)}), then

a_{n}^{(0)} = a_{n+1}, b_{n}^{(0)} = b_{n+1}.

The map ρ ↦ ρ̃ given by (5.53) sets up a bijection between M^{H}(γ) and M^{H}(γ^{(0)}), and, in particular, γ is a Hamburger determinate if and only if γ^{(0)} is (Proposition 5.15). This is not, in general, true in the Stieltjes case (see Proposition 5.17). There is a simple relation between the orthogonal polynomials for γ and for γ^{(0)}; see Proposition 5.16. In particular,

P_{N}^{(0)}(z) = a_{0}Q_{N+1}(z).  (C.9)

For ℓ = −1, −2, …, we defined

γ^{(ℓ)} = (γ^{(0)})^{(−ℓ)} = \frac{γ_{j−ℓ}^{(0)}}{γ_{−ℓ}^{(0)}} .  (C.10)

C4 Associated Polynomials.

Given a set of Hamburger moments, the fundamental orthogonal polynomials P_{N}(z) are defined by requiring for any ρ ∈ M^{H}(γ) that

∫ P_{j}(x)P_{ℓ}(x) dρ(x) = δ_{jℓ}  (C.11)

and P_{N}(x) = c_{NN}x^{N} + lower order terms with c_{NN} > 0. They obey a three-term recursion relation

xP_{N}(x) = a_{N}P_{N+1}(x) + b_{N}P_{N}(x) + a_{N−1}P_{N−1}(x).  (C.12)

The Jacobi matrix A associated to γ is the tridiagonal matrix given by (1.16).

The associated polynomials Q_{N}(x) are of degree N − 1 and defined to obey (C.12) but with the starting conditions, Q_{0}(x) = 0, Q_{1}(x) = \frac{1}{a_{0}} . They are related to P_{N} by (Theorem 4.2)

Q_{N}(x) = ∫ \frac{P_{N}(x) − P_{N}(y)}{x − y} dρ(y)  (C.13)

for any ρ ∈ M^{H}(γ). As noted in (C.9), they are up to constants, orthogonal polynomials for another moment problem, namely, γ^{(0)}.

In the Stieltjes case, we defined polynomials M_{N} by

M_{N}(x) = P_{N}(x) − \frac{P_{N}(0)P_{N−1}(x)}{P_{N−1}(0)} .  (C.14)

They vanish at x = 0 so \frac{M_{N} (x)}{x} is also a polynomial. It is of degree N − 1 and is up to a constant, the principal orthogonal polynomial of the moment problem γ^{(1)}. N_{N}(x) is defined analogously to (C.13) with P_{N} replaced by M_{N}. In Section 5, it was useful to change the normalizations and define multiples of P_{N}(−x), Q_{N}(−x), M_{N}(−x), and N_{N}(−x) as functions U_{N}(x), V_{N}(x), G_{N}(x), H_{N}(x) given by (5.71)–(5.74).

C5 Padé Approximants and Finite Matrix Approximations.

If {γ_{n}}_{n=0}^{∞} is a set of Hamburger (resp. Stieltjes) moments, the formal power series ∑_{n=0}^{∞}(−1)^{n}γ_{n}z^{n} is called a series of Hamburger (resp. a series of Stieltjes). Formally, it sums to

∫ \frac{dρ(x)}{1 + xz}

if ρ ∈ M^{H}(γ). The Padé approximants f^{[N,M]}(z) to these series, if they exist, are defined by (5.25)–(5.27) as the rational function which is a ratio of a polynomial of degree N to a polynomial of degree M, whose first N + M + 1 Taylor coefficients are {(−1)^{n}γ_{n}}_{n=0}^{N+M}.

For a series of Stieltjes, for each fixed ℓ = 0, ±1, …,

lim_{N→∞} f^{[N+ℓ−1,N]}(z) ≡ f_{ℓ}(z)

exists for all z ∈ ℂ\(−∞, 0) and defines a function analytic there. Indeed, for x ∈ [0, ∞), (−1)^{ℓ}f^{[N+ℓ−1,N]}(x) is monotone increasing (Theorem 6, Theorem 5.14, and Theorem 5.18). Moreover,

f_{0}(z) = ∫ \frac{dρ_{F} (x)}{1 + xz} , f_{1}(z) = ∫ \frac{dρ_{K}(x)}{1 + xz} ,

where ρ_{F}, ρ_{K} are the Friedrichs and Krein solutions of the moment problem. In particular, the moment problem is Stieltjes determinate if and only if f_{0} = f_{1}.

There is a connection between f^{[N+ℓ−1,N]}(z) and the moment problems γ^{(ℓ)} and γ^{(ℓ−1)} (if ℓ < 0, γ^{(ℓ)} and γ^{(ℓ+1)}) given in Theorem 5.14 and Theorem 5.18. In particular, if ℓ > 0 and γ^{(ℓ)} is Stieltjes determinate, then f_{−1}(z) = f_{0}(z) = ⋯ = f_{ℓ+1}(z), and if ℓ < 0 and γ^{(ℓ)} is Stieltjes determinate, then f_{1}(z) = f_{0}(z) = ⋯ = f_{ℓ}(z) = f_{ℓ−1}(z).

The situation for series of Hamburger is more complicated, but, in general, f^{[N+ℓ−1,N]}(z) converges for z ∈ ℂ_{+} and ℓ = ±1, ±3, … (Theorem 5.31 and Theorem 5.32; see Theorem 5.31 for issues of existence of the Padé approximant).

There are connections between the Padé approximants and the polynomials P, Q, M, N as well as two finite matrix approximations A_{F}^{[N]} and A_{K}^{[N]} to the Jacobi matrix, A. A_{F}^{[N]} is the upper right N × N piece of A. A_{K}^{[N]} differs by adjusting the NN matrix element of A_{F}^{[N]} so det(A_{K}^{[N]}) = 0. Then

f^{[N−1,N]}(z) = ⟨δ_{0}, (1 + zA_{F}^{[N]})^{−1}δ_{0}⟩ = −\frac{z^{N−1}Q_{N}(−\frac{1}{z})}{z^{N}P_{N}(−\frac{1}{z})}

f^{[N,N]}(z) = ⟨δ_{0}, (1 + zA_{K}^{[N+1]})^{−1}δ_{0}⟩ = −\frac{z^{N}N_{N+1}(−\frac{1}{z})}{z^{N+1}M_{N+1}(−\frac{1}{z})} .

The connection to the continued fractions of Stieltjes is discussed after Theorem 5.28.

C6 Criteria for Determinacy.

Criteria for when a Hamburger problem is determinate can be found in Proposition 1.5, Theorem 3, Corollary 4.5, Proposition 5.13, Theorem A.6, and Theorem A.7.

Criteria for when a Stieltjes problem is determinate can be found in Proposition 1.5, Theorem 7, Corollary 4.5, Theorem 5.21, Corollary 5.24, Theorem A.4, and Theorem A.6. In particular, if one defines

c_{2j} = ℓ_{j} = −[a_{j−1}P_{j}(0)P_{j−1}(0)]^{−1}
c_{2j−1} = m_{j} = |P_{j−1}(0)|^{2}

(the c’s are coefficients of Stieltjes continued fractions; the m’s and ℓ’s are natural parameters in Krein’s theory), then the Stieltjes problem is determinate if and only if (Theorem 5.21 and Proposition 5.23) ∑_{j=0}^{∞} c_{j} = ∞.

Acknowledgments. I would like to thank George Baker, Percy Deift, Rafael del Rio, Fritz Gesztesy, Mourad Ismail, Uri Keich, and Sasha Kiselev for valuable comments.

References

1. N.I. Akhiezer, The Classical Moment Problem, Hafner Publishing Co., New York, 1965.

2. N.I. Akhiezer and I.M. Glazman, Theory of Linear Operators in Hilbert Space, Vol. 2, Frederick Ungar, New York, 1961.

3. A. Alonso and B. Simon, The Birman-Krein-Vishik theory of self-adjoint extensions of semibounded operators, J. Oper. Th. 4 (1980), 251–270.

4. G. Baker and P. Graves-Morris, Padé Approximants, 2nd ed., Cambridge University Press, New York, 1996.

5. N. Dunford and J. Schwartz, Linear Operators, II. Spectral Theory, Interscience Publishers, New York, 1963.

6. F. Gesztesy and B. Simon, m-functions and inverse spectral analysis for finite and semi-infinite Jacobi matrices, to appear in J. d’Anal. Math.

7. J. Gil de Lamadrid, Determinacy theory for the Livšic moments problem, J. Math. Anal. Appl. 34 (1971), 429–444.

8. H. Hamburger, Über eine Erweiterung des Stieltjesschen Momentenproblems, Math. Ann. 81 (1920), 235–319; 82 (1921), 120–164, 168–187.

9. M.E.H. Ismail and D. Masson, q-hermite polynomials, biorthogonal rational functions, and q-beta integrals, Trans. Amer. Math. Soc. 346 (1994), 63–116.

10. M.E.H. Ismail and M. Rahman, The q-Laguerre polynomials and related moment problems, to appear in J. Math. Anal. Appl.

11. S. Jitomirskaya and Y. Last, Power law subordinacy and singular spectra, I. Half-line operators, in preparation.

12. T. Kato, Perturbation Theory for Linear Operators, 2nd ed., Springer, New York, 1980.

13. Y. Katznelson, An Introduction to Harmonic Analysis, 2nd corrected ed., Dover Publications, New York, 1976.

14. U. Keich, private communication.

15. M.G. Krein, Chebyshev’s and Markov’s ideas in the theory of limiting values of integrals and their further development, Uspekhi matem. Nauk 44 (1951), in Russian.

16. M.G. Krein, On a generalization of an investigation by Stieltjes, Dokl. Akad. Nauk SSSR 87 (1952), in Russian.

17. M.G. Krein, Some new problems in the theory of oscillations of Sturm systems, Priklad. matem. i mekh. 16 (1952), 555–568, in Russian.

18. M.G. Krein, On an extrapolation problem due to Kolmogorov, Dokl. Akad. Nauk SSSR 46 (1945), 306–309.

19. H. Landau, The classical moment problem: Hilbertian proofs, J. Funct. Anal. 38 (1980), 255-272.

20. R.W. Langer, More determinacy theory for the Livšic moments problem, J. Math. Anal. Appl. 56 (1976), 586–616.

21. M.S. Livšic, On an application of the theory of hermitian operators to the generalized problem of moments, Dokl. Akad. Nauk SSSR 26 (1940), 17–22.

22. J.J. Loeffel, A. Martin, B. Simon, and A.S. Wightman, Padé approximants and the anharmonic oscillator, Phys. Lett. 30B (1969), 656–658.

23. D. Masson and W. McClary, Classes of C^{∞}-vectors and essential self-adjointness, J. Funct. Anal. 10 (1972), 19–32.

24. M.A. Naimark, Self-adjoint extensions of the second kind of a symmetric operator, Izv. Akad. Nauk SSSR 4 (1940), 90–104.

25. M.A. Naimark, Spectral functions of a symmetric operator, Izv. Akad. Nauk SSSR 4 (1940), 309–318.

THE CLASSICAL MOMENT PROBLEM 107

26. M.A. Naimark, On spectral functions of a symmetric operator, Izv. Akad. Nauk SSSR 7 (1943), 285–296.

27. M.A. Naimark, On extremal spectral functions of a symmetric operator, Dokl. Akad. Nauk SSSR 54 (1946), 7–9.

28. E. Nelson, Analytic vectors, Ann. of Math. 70 (1959), 572–615.

29. R. Nevanlinna, Asymptotische Entwickelungen beschränkter Funktionen und das Stieltjessche Momentenproblem, Ann. Acad. Sci. Fenn. A 18 (1922).

30. A. Nussbaum, Quasi-analytic vectors, Ark. Mat. 6 (1965), 179–191.

31. A. Nussbaum, A note on quasi-analytic vectors, Studia Math. 33 (1969), 305–310.

32. D.B. Pearson, Quantum Scattering and Spectral Theory, Academic Press, London, 1988.

33. M. Reed and B. Simon, Methods of Modern Mathematical Physics, I. Functional Analysis, Academic Press, New York, 1972.

34. M. Reed and B. Simon, Methods of Modern Mathematical Physics, II. Fourier Analysis, Self-Adjointness, Academic Press, New York, 1975.

35. J.A. Shohat and J.D. Tamarkin, The problem of moments, Amer. Math. Soc. Surveys, No. 1 (1943).

36. B. Simon, Coupling constant analyticity for the anharmonic oscillator (with an appendix by A. Dicke), Ann. Phys. 58 (1970), 76–136.

37. B. Simon, A canonical decomposition for quadratic forms with applications to monotone convergence theorems, J. Funct. Anal. 28 (1978), 377–385.

38. T. Stieltjes, Recherches sur les fractions continues, Anns. Fac. Sci. Univ. Toulouse (1894–95), 8, J1–J122; 9, A5–A47.

39. M.H. Stone, Linear Transformations in Hilbert Space, Amer. Math. Soc. Colloq. Publ. XV, New York, 1932.