RATIO ASYMPTOTICS AND WEAK ASYMPTOTIC MEASURES FOR ORTHOGONAL POLYNOMIALS

ON THE REAL LINE

BARRY SIMON

Abstract. We study ratio asymptotics, that is, existence of the limit of P_{n+1}(z)/P_{n}(z) (P_{n} = monic orthogonal polynomial) and the existence of weak limits of p_{n}^{2} dμ (p_{n} = P_{n}/∥P_{n}∥) as n → ∞ for orthogonal polynomials on the real line. We show existence of ratio asymptotics at a single z_{0} with Im(z_{0}) ≠ 0 implies dμ is in a Nevai class (i.e., a_{n} → a and b_{n} → b where a_{n}, b_{n} are the offdiagonal and diagonal Jacobi parameters). For μ’s with bounded support, we prove p_{n}^{2} dμ has a weak limit if and only if lim b_{n}, lim a_{2n}, and lim a_{2n+1} all exist. In both cases, we write down the limits explicitly.

1. Introduction

In [7], Khrushchev asked two questions about orthogonal polynomials on the unit circle [4, 13, 14] and found the following remarkable theorems in terms of the monic orthogonal polynomials, Φ_{n}; the orthonormal polynomials, φ_{n} = Φ_{n}/∥Φ_{n}∥_{L^{2}}; and the Verblunsky coefficients, α_{n} = −\overline{Φ_{n+1}(0)}.

Theorem A. Φ_{n+1}^{∗}(z)/Φ_{n}^{∗}(z) has a limit uniformly in z over compact subsets of 𝔻 if and only if either

(i) For ℓ = 1, 2, …, lim_{n→∞} α_{n+ℓ}α_{n} = 0, or (ii) There is a ∈ (0, 1] and λ ∈ ∂𝔻 so that lim_{n→∞}|α_{n}| = a,

lim_{n→∞} ᾱ_{n+1}α_{n} = a^{2}λ.

Theorem B. |φ_{n}|^{2} dμ has a weak limit if and only if either

(i) For ℓ = 1, 2, …, lim_{n→∞} α_{n+ℓ}α_{n} = 0, or

Date: July 3, 2003. 1 Mathematics 253-37, California Institute of Technology, Pasadena, CA 91125.

E-mail: bsimon@caltech.edu. Supported in part by NSF grant DMS-0140592.

(ii) There exists a, a′ ∈ (0, 1], λ ∈ ∂𝔻, and integers k ≥ 1 and ℓ ∈ {0, 1, …, k − 1} so that

lim_{n→∞} |α_{2nk+ℓ+j}| = \begin{cases}a & if j = 0 \\ a′ & if j = k \\ 0 & if j = 1, …, k − 1, k + 1, …, 2k − 1 \end{cases}

lim_{n→∞} ᾱ_{2nk+ℓ}α_{2nk+k+ℓ} = aa′λ

Khrushchev [7] also describes explicitly the limits in both cases. Our goal in this paper is to find the analogs of these theorems for orthogonal polynomials on the real line. The answers and proofs are much simpler — the methods of Khrushchev which depend heavily on Schur functions do not seem to extend, nor does mapping bounded intervals on ℝ to ∂𝔻 (as in Szegő [14, Sect. 11.5]) seem to allow direct transfer.

Before stating our results, let us set up notation. Given a measure dμ on ℝ with ∫ x^{2n} dμ < ∞ for all n, we let P_{n}(x) be the monic orthogonal and p_{n}(x) the orthonormal polynomials. To define them, we suppose henceforth that dμ is nontrivial, that is, not supported on a finite set, and we will also assume throughout that μ(ℝ) = 1. Thus P_{n} is determined by P_{n}(x) = x^{n}+ lower order and ∫ x^{j}P_{n}(x) dμ(x) = 0 for j = 0, 1, …, n − 1. p_{n} = P_{n}/∥P_{n}∥ where ∥ · ∥ is the L^{2}(ℝ, dμ) norm.

It is well-known [14] that the P_{n}’s obey a three-term recursion relation. There are b_{j} ∈ ℝ and a_{j} ∈ (0, ∞) so that

xP_{n}(x) = P_{n+1}(x) + b_{n+1}P_{n}(x) + a_{n}^{2}P_{n−1}(x)  (1.1)

Our indexing of b_{1}, b_{2}, … and a_{1}, a_{2}, … is not common — often the labelling starts at b_{0} and a_{0}. We take this convention from [8] for reasons explained there. (1.1) implies inductively that

∥P_{n}∥ = a_{n} … a_{1}  (1.2)

and then that the p_{n} obey the recursion relation

xp_{n}(x) = a_{n+1}p_{n+1}(x) + b_{n+1}p_{n}(x) + a_{n}p_{n−1}(x)  (1.3)

In turn, (1.3) suggests we study the Jacobi matrix

J = (\begin{matrix}b_{1} & a_{1} & 0 & 0 & ⋯ \\ a_{1} & b_{2} & a_{2} & 0 & ⋯ \\ 0 & a_{2} & b_{3} & a_{3} & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix})  (1.4)

Since {p_{n}}_{n=0}^{∞} is an orthonormal set,

U : ∑_{j=0}^{N} v_{j}p_{j} → (v_{1} ; v_{2} ; ⋮)

is a unitary map of the closed span, S, of the p’s to ℓ^{2}(ℤ_{+}) (ℤ_{+} = {1, 2, … }) and for v ∈ S_{0} = span of p’s, we have U^{−1}JUv = (multiplication by x) v.

In case the moment problem is determinant [1, 12], J is selfadjoint, and dμ is just the spectral measure for J and vector δ = (1, 0, …).

We can now state our main results.

Theorem 1 (≡ Theorems 2.1 and 2.2). Suppose that at a single z_{0} ∈ ℂ\ℝ, we have

lim_{n→∞} \frac{P_{n+1}(z)}{P_{n}(z)} = f(z)  (1.5)

for z = z_{0}. Then, for some a ∈ [0, ∞) and b ∈ ℝ,

lim_{n→∞} a_{n} = a  lim_{n→∞} b_{n} = b  (1.6)

Conversely, if (1.6) holds and spectrum(J) is the spectrum of the operator J, then (1.5) holds for all z ∈ ℂ\spectrum(J) and

f(z) = \frac{(z − b) + \sqrt{(z − b)^{2} − 4a^{2}}}{2}  (1.7)

where the branch of the square root is taken with \sqrt{⋯} = z + O(1/z) near z = ∞.

Theorem 2 (≡ Theorems 3.1 and 3.2). Let

dμ_{n} = p_{n}^{2}(x) dμ(x)  (1.8)

Suppose that for ℓ = 1, 2, and 4, lim_{n→∞} ∫ x^{ℓ} dμ_{n} exists. Then for a, c ∈ [0, ∞) and b ∈ ℝ, we have

lim_{n→∞} b_{n} = b  lim_{n→∞} a_{2n} = a  lim_{n→∞} a_{2n+1} = c  (1.9)

Conversely, if (1.9) holds, the dμ_{n} have supports lying in a fixed bounded interval and there is a measure dρ_{b;a,c} so that for any continuous f on ℝ (including f(x) = x^{ℓ}), we have

∫ f(x) dμ_{n}(x) → ∫ f(x) dρ(x)  (1.10)

dρ is a function of a, b, c only, and if dρ_{b;a,c} = dρ_{b′;a′c′}, we have b = b′ and either a = a′, c = c′ or a = c′, c = a′.

Theorem 1 is proven in Section 2 and Theorem 2 in Section 3. dρ is calculated in Section 5. Theorems 1 and 2 seem to be optimal in that the two pieces of real data (1.6) (i.e., a and b) correspond to one complex number f(z_{0}) while the three moments in Theorem 2 correspond to the real numbers, a, b, and c.

There is previous work of Nevai [9] on the subjects of Theorem 1 and Theorem 2. He proved that (1.6) implies (1.5) with f given by (1.7), and conversely proved that if (1.5) holds for all z ∈ ℂ\ℝ and f(z) given by (1.7), then (1.6) holds. He did not get a result depending on a single z_{0} nor, more importantly, did he show that (1.7) are the only possible limits in (1.5).

Nevai [9] also proved that if a_{n} → a and b_{n} → b, then dμ_{n} has a weak limit (he wrote down the explicit form of dρ_{b;a,a} for a = c, as we will in Section 5).

In [10], Nevai made a conjecture closely related to a special case of Theorem 2. Namely,

Nevai Conjecture 2.16 ([10]). If (1.10) holds for all bounded uniformly continuous functions on ℝ with dρ(x) = π^{−1}χ_{[−1,1]}(x)(1 − x^{2})^{−1/2} dx, then a_{n} → \frac{1}{2} and b_{n} → 0.

Corollary to Theorem 2. If one supposes supp(dμ) is bounded, then Nevai’s conjecture holds.

Proof. x^{ℓ} ↾ supp(dμ) is bounded, so ∫ x^{ℓ} dμ_{n} converges for ℓ = 1, 2, 4 to the same limit as for a_{n} ≡ \frac{1}{2} and b ≡ 0. Uniqueness of the limit (and the fact that a = c) completes the proof. □

Related to this is

Nevai Conjecture 2.17 ([10]). If for some A, we have ∫_{A}^{∞} dμ_{n} → 0, then for every ε > 0, [A + ε, α) ∩ supp(dμ) is finite.

We mention

Weaker Nevai Conjecture 2.17. If for some A > 0, μ_{n}({x | |x| > A}) → 0, then supp(dμ) is bounded.

Clearly, Nevai Conjecture 2.17 implies the weaker version. The point of this is that a positive solution of the Weaker Nevai Conjecture 2.17 plus the results of this paper would imply Nevai Conjecture 2.16.

It is a pleasure to thank Rowan Killip and Paul Nevai for cogent comments.

2. Ratio Asymptotics

The two main theorems on limits of P_{n+1}(x)/P_{n}(x) are as follows:

Theorem 2.1. Suppose a_{n} → a ∈ [0, ∞) and b_{n} → b ∈ ℝ. Then for all z ∈ ℂ\spectrum(J), we have that

lim_{n→∞} \frac{P_{n+1}(z)}{P_{n}(z)} = \frac{(z − b) + \sqrt{(z − b)^{2} − 4a^{2}}}{2}  (2.1)

Remarks. 1. In (2.1), we take the branch of the square root with \sqrt{⋯} ∼ z for |z| large, that is, as z → ∞.

2. For z ∉ ℝ, P_{n} is nonzero for all n. (2.1) for z_{0} ∈ ℝ\spectrum(J) includes the fact that for z_{0} fixed, P_{n}(z_{0}) ≠ 0 for all large n.

3. One can also show that for z ∈ spectrum(J)\[b − 2a, b + 2a] so that z is an eigenvalue of J,

lim_{n→∞} \frac{P_{n+1}(z)}{P_{n}(z)} = \frac{(z − b) − \sqrt{(z − b)^{2} − 4a^{2}}}{2}  (2.2)

Theorem 2.2. Suppose for one z_{0} with Im z_{0} ≠ 0, lim_{n→∞} P_{n+1}(z)/P_{n}(z) exists (and is finite). Then there exists a ∈ [0, ∞) with b ∈ ℝ so that a_{n} → a and b_{n} → b so that (2.1) holds. In particular, the only functions that can occur as ratio asymptotics are the ones in (2.1).

Theorem 2.1 is not new. In this generality, it is due to Nevai [9], who also proved a converse; namely, he showed that if (2.1) holds for all z ∈ ℂ\ℝ, then a_{n} → a and b_{n} → b. But we will sketch two proofs of Theorem 2.1 for the reader’s convenience. One uses transfer matrices and the other, operator theory.

As a preliminary, we need:

Proposition 2.3. Let {x_{j,n}}_{j=1}^{n} be the zeros of P_{n}(x) with

x_{1,n} < x_{2,n} < ⋯ < x_{n,n}  (2.3)

Then (i)

−\frac{P_{n−1}(z)}{P_{n}(z)} = ∑_{j=0}^{n} \frac{α_{j,n}}{x_{j,n} − z}  (2.4)

where α_{j,n} > 0 and

∑_{j=1}^{n} α_{j,n} = 1  (2.5)

(ii) If Im z > 0,

0 < − Im(\frac{P_{n−1}(z)}{P_{n}(z)}) ≤ \frac{1}{Im z}  (2.6)

and

|\frac{P_{n−1}(z)}{P_{n}(z)}| ≤ \frac{1}{Im z}  (2.7)

(iii) If Im z > 0,

Im(\frac{P_{n+1}}{P_{n}} ) ≥ Im z  (2.8)

Proof. (i) Since P_{n} is monic, P_{n}(z) = ∏_{j=1}^{n}(z − x_{j,n}). Since P_{n−1}/P_{n} has simple poles and goes to zero at infinity, (2.4) holds for some α_{j,n}. Multiplying by x_{i,n} − z and taking z to x_{i,n}, we find

α_{j,n} = \frac{∏_{ℓ=1}^{n−1}(x_{j,n} − x_{ℓ,n−1})}{∏_{ℓ=1; ℓ≠j}^{n}(z_{j,n} − x_{ℓ,n})}  (2.9)

For j = n, all factors in the right are positive. Since zeros of P_{n−1} and P_{n} interlace as we decrease j by one, both numerator and denominator each pick up a minus sign which cancel to prove α_{j,n} > 0.

The left side of (2.4) is −z^{−1} + O(z^{−2}) as z → ∞ since P_{n} is monic. The right side is −z^{−1}(∑_{j=0}^{n} α_{j,n}) + O(z^{−2}), so (2.5) holds.

(ii) This follows from (2.4) and (2.5) if one notes that for any x ∈ ℝ and z with Im z > 0,

0 < Im(\frac{1}{x − z} ) ≤ \frac{1}{|x − z|} ≤ \frac{1}{Im z}

(iii) This follows immediately from

\frac{P_{n+1}(z)}{P_{n}(z)} = z − b_{n+1} − a_{n}^{2} \frac{P_{n−1}(z)}{P_{n}(z)}  (2.10)

Since Im P_{n−1}/P_{n} < 0, (2.10) implies (2.8). □

Proof of Theorem 2.2. By replacing x by (x − Re z_{0})/ Im z_{0} (i.e., translating and scaling the measure), we can suppose for notational simplicity that z_{0} = i. Let

α = lim_{n→∞} \frac{P_{n+1}(i)}{P_{n}(i)}  (2.11)

By (2.8), Im α ≥ 1 so − Im(α^{−1}) = Im α/|α|^{2} > 0.

Taking imaginary parts of (2.10), we see that

a_{n}^{2} = \frac{[Im(P_{n+1}/P_{n}) − 1]}{Im(−P_{n−1}/P_{n})} → \frac{(Im α − 1)}{Im(−α^{−1})} ≡ a^{2} ≥ 0  (2.12)

proving a_{n} has a limit.

Taking real parts of (2.10) shows

b_{n+1} = −a_{n}^{2} Re(\frac{P_{n−1}}{P_{n}} ) − Re(\frac{P_{n+1}}{P_{n}} ) → −a^{2} Re(α^{−1}) − Re(α) ≡ b

(2.13) □

Our first proof of Theorem 2.1 is a simple consequence on the following theorem of Poincaré (see [3, 11, 13] for proofs):

Theorem of Poincaré. Let u_{j} ∈ ℂ solve the n-th order difference equation

u_{n+j} = a_{j,1}u_{n+j−1} + a_{j,2}u_{n+j−2} + ⋯ + a_{j,n}u_{j}  (2.14)

j = 1, 2, . . .. Suppose

(i) a_{j,n} ≠ 0 for all j. (ii) lim_{j→∞} a_{j,ℓ} = A_{ℓ} exists for ℓ = 1, …, n. Let λ_{1}, …, λ_{n} be the solutions of

A_{n} + A_{n−1}λ + ⋯ + A_{1}λ^{n−1} = λ^{n}  (2.15)

Suppose {λ_{j}}_{j=1}^{n} are distinct, and for j ≠ k, |λ_{j}| ≠ |λ_{k}|. Then, if u is not identically zero, we have for some k that

lim_{j→∞} \frac{u_{j+1}}{u_{j}} = λ_{k}  (2.16)

First Proof of Theorem 2.1. (1.1) is exactly of the form (2.14). The equation (2.15) becomes

λ^{2} = (z − b)λ − a^{2}  (2.17)

whose solutions are

λ_{±} = \frac{(z − b) ± \sqrt{(z − b)^{2} − 4a^{2}}}{2}  (2.18)

To prove (2.1), we must show that if z ∉ [b−2a, b+2a], then |λ_{+}| ≠ |λ_{−}| and then identify which root is taken by the ratio.

If α and β are complex numbers, |α+β| = |α−β| if and only if α and β are orthogonal as vectors in ℂ = ℝ^{2}, if and only if β = icα for c ∈ ℝ, if and only if β^{2} = −c^{2}α^{2}. Taking α = (z−b) and β = \sqrt{(z − b)^{2} − 4a^{2}}, we see |λ_{+}| = |λ_{−}| if and only if

−c^{2}(z − b)^{2} = (z − b)^{2} − 4a^{2}

or

z − b = ± \frac{2a}{\sqrt{1 + c^{2}}}  (2.19)

for c ∈ ℝ. (2.19) holds if and only if z ∈ [b − 2a, b + 2a].

Since |λ_{+}| ≠ |λ_{−}|, λ_{±}(z) are analytic in ℂ\[b−2a, b+2a] (as is obvious from (2.18)). Since λ_{+}λ_{−} = a^{2}, |λ_{+}| |λ_{−}| = a^{2}, and λ_{+} = z + O(1/z) as |z| → ∞, |λ_{+}| > a for |z| large, and so since |λ_{+}| |λ_{−}| = a^{2} and |λ_{+}| ≠ |λ_{−}| implies |λ_{±}| ≠ a, we see

|λ_{+}| > a |λ_{−}| < a all z ∈ ℂ\[b − 2a, b + 2a]  (2.20)

Thus for all z ∈ ℂ\[b − 2a, b + 2a], Poincaré’s theorem applies and P_{n+1}/P_{n} has a limit ℓ(z) where for each z, ℓ(z) = λ_{+} or ℓ(z) = λ_{−}.

By (2.6), P_{n}/P_{n+1} is a normal family on ℝ\ℂ, so ℓ(z) is analytic on ℝ\ℂ. By (2.8), |ℓ(z)| > |Im z|, so for |z| large, ℓ(z) = λ_{+}(z) and thus, by analyticity, ℓ(z) = λ_{+}(z) for z ∈ ℂ\ℝ. This establishes (2.1) there.

For z ∈ ℝ\[b − 2a, b + 2a],

lim_{n→∞} \frac{p_{n+1}(z)}{p_{n}(z)} = lim_{n→∞} \frac{1}{a} \frac{P_{n+1}(z)}{P_{n}(z)} = \frac{λ_{±}(z)}{a}

If the limit is λ_{−}(z), lim|p_{n+1}/p_{n}| < 1, so p_{n} ∈ ℓ_{2} and z ∈ spectrum(J). Conversely, if it is λ_{+}, lim|p_{n+1}/p_{n}| > 1, so p_{n} ∉ ℓ_{2}. Thus (2.1) holds for z ∈ ℝ\spectrum(J) and (2.2) holds for z ∈ spectrum(J)\[b−2a, b+ 2a]. □

Second Proof of Theorem 2.1. Let J^{(n)} be the n × n matrix obtained from the first n rows and columns of J. As is well-known,

det(z − J^{(n)}) = P_{n}(z)

Thus, by Cramer’s rule,

\frac{P_{n−1}(z)}{P_{n}(z)} = (z − J^{(n)})_{nn}^{−1} = (z − J̃^{(n)})_{11}^{−1}  (2.21)

where J̃_{ij}^{(n)} = J̃_{n−i, n−j}^{(n)}, that is,

J̃^{(n)} = (\begin{matrix}b_{n} & a_{n−1} & 0 & … & … \\ a_{n−1} & b_{n−1} & a_{n−2} & … & … \\ 0 & a_{n−2} & b_{n−2} & … & … \\ … & … & … & … & … \\ … & … & … & a_{1} & b_{1} \end{matrix})  (2.22)

If J̃^{(∞)} is the constant Jacobi matrix a_{n} ≡ a, b_{n} ≡ b, it is clear as operators on ℓ^{2}(ℤ_{+}), J̃^{(n)} → J̃^{(∞)} strongly. It follows that as operators on ℓ^{2}(ℤ), (J̃^{(n)} − z)^{−1} → (J̃^{(∞)} − z)^{−1} strongly for Im z ≠ 0. Thus

\frac{P_{n−1}(z)}{P_{n}(z)} = (z − J̃^{(n)})_{11}^{−1} → (z − J̃^{(∞)})_{11}^{−1}  (2.23)

Let w solve

a(w + w^{−1}) + b = z  (2.24)

with |w| < 1. Let u_{n} = w^{n}. Thus u_{n} ∈ ℓ_{2} and

(z − J̃^{(∞)})u = (aw^{−1})δ_{1}

so

(z − J̃^{(∞)})_{11}^{−1} = aw^{−1}  (2.25)

Solving (2.24), we see that aw^{−1} = RHS of (2.1). This proves (2.1) on ℂ\ℝ.

On ℝ\spectrum(J), one shows the eigenvalues of J̃^{(n)} (equals zeros of P_{n}(z)) outside [b − 2a, b + 2a] converge to the eigenvalues of J so for z ∈ ℝ\spectrum(J), (J̃^{(n)} − z)^{−1} → (J̃^{(∞)} − z)^{−1} strongly also. □

3. Weak Asymptotic Limits

Let dμ_{n} = p_{n}^{2} dμ. In this section, the main theorems are

Theorem 3.1. Suppose

b_{n} → b  a_{2n} → a  a_{2n+1} → c  (3.1)

for b ∈ ℝ, a, c ∈ [0, ∞). Then as n → ∞, dμ_{n} has a weak limit dρ_{b;a,c}(x), and for every ℓ,

∫ x^{ℓ} dμ_{n} → ∫ x^{ℓ} dρ_{b;a,c}(x)  (3.2)

Remarks. 1. The hypotheses imply that dμ is supported in [inf b_{n} − 2 sup(a_{n}), sup b_{n} + 2 sup(a_{n})] which is bounded, so weak convergence is equivalent to convergence of the moments.

2. We will see below that dρ_{b′;a′,c′} = dρ_{b;a,c} implies b′ = b and either a′ = a, c′ = c, or a′ = c, c′ = a.

3. We will discuss the form of dρ_{b;a,c} in Section 5.

Theorem 3.2. Suppose for ℓ = 1, 2, and 4,

lim_{n→∞} ∫ x^{ℓ}dμ_{n} = A_{ℓ}  (3.3)

Then for some a, b, c, (3.1) holds. Moreover, A_{1}, A_{2}, A_{4} determine b, a + c, |a − c| (i.e., they determine b and the unordered pair (a, c)).

Remarks. 1. We will see A_{2} < ∞ implies sup(|b_{n}| + |a_{n}|) < ∞.

2. The final assertion proves the second remark after Theorem 3.1. Our proofs will depend on a graphical representation of ∫ x^{ℓ} dμ_{n}.

Consider the lattice ℤ_{+} = {0, 1, … }. We will consider a random walk on ℤ_{+} where at each step, one either stays at the site one is at or one jumps by a single site. Paths have unnormalized weights, products over the steps: b_{k+1} if one stays at site k, a_{k+1} is one move from k to k +1 or k + 1 to k. To be more precise, a path is a sequence ρ_{0}, ρ_{1}, …, ρ_{ℓ} ∈ ℤ_{+} so that |ρ_{m} − ρ_{m−1}| ≤ 1 and

W (ρ) = ∏_{j=0}^{ℓ−1} w(ρ_{j}, ρ_{j+1})  (3.4)

and

w(ρ_{j}, ρ_{j+1}) = \begin{cases}b_{k+1} & if ρ_{j+1} = ρ_{j} = k \\ a_{k+1} & if ρ_{j+1} = ρ_{j} + 1 = k + 1 \\ a_{k} & it ρ_{j+1} = ρ_{j} − 1 = k − 1 \end{cases}  (3.5)

Here is the key tool:

Proposition 3.3.

∫ x^{ℓ} dμ_{n} = ∑_{ρ∈Q_{n,ℓ}} W (ρ)  (3.6)

where Q_{n,ℓ} is the set of all paths of length ℓ with ρ_{0} = ρ_{ℓ} = n.

Proof. Since

xp_{n} = a_{n+1}p_{n+1} + b_{n+1}p_{n} + a_{n}p_{n−1}

we see immediately that, by induction in j,

x^{j}p_{n} = ∑ c_{j,m,n}p_{m}  (3.7)

where

c_{j,m,n} = ∑_{ρ∈Q_{n,m,j}} W (ρ)

and Q_{n,m,j} is all paths of length j with ρ_{0} = n and ρ_{j} = m. (3.6) follows since ∫ x^{ℓ} dμ_{n} = ⟨p_{n}, x^{ℓ}p_{n}⟩ = c_{ℓ,n,n}. □

Proof of Theorem 3.1. Under hypothesis (3.1), J is bounded, so dμ has a bounded support, so weak convergence is equivalent to (3.2). By Proposition 3.3, ∫ x^{ℓ} dμ_{n} is a finite sum over paths. This representation shows that if a_{n}, b_{n}, and ã_{n}, b̃_{n} are two sets of Jacobi parameters and lim_{n→∞}|a_{n} − ã_{n}| + |b_{n} − b̃_{n}| = 0, then |∫ x^{ℓ}dμ_{n} − ∫ x^{ℓ}dμ̃_{n}| → 0. Thus we need only prove (3.2) for b_{n} ≡ b, a_{2n} ≡ a, a_{2n+1} ≡ c.

Fix ℓ. So long as ℓ < 2n, there is a one-one correspondence between paths ρ ∈ Q_{n,ℓ} and ρ ∈ Q_{n,+1,ℓ} by U = T S,

T (ρ)_{j} = ρ_{j} + 1
S(ρ)_{j} = n − (ρ_{j} − n)

S reflects the path in n, T translates by 1. ℓ < 2n is needed to assure paths do not get mapped into ones that have ρ_{j} < 0, which is forbidden (and that the inverse does not do this), showing U is a bijection of Q_{n,ℓ} and Q_{n+1,ℓ}. The key point is that W (Uρ) = W (ρ), for if ρ_{j} = ρ_{j+1}, the weight is always b and both S and T interchange links with weight a and those with weight c. It follows that if b ≡ b, a_{2n} ≡ a, a_{2n+1} ≡ c, then ∫ x^{ℓ} dρ_{n} is independent of n once ℓ < 2n, so the limit exists. Once the moments exist, they provide a measure since the nonnegative Hankel matrices converge to nonnegative Hankel matrices. □

Proof of Theorem 3.2. ∫ xp_{n}^{2} dμ = b_{n+1}, so (3.4) for ℓ = 1 implies b_{n+1} → A_{1} ≡ b. Let dμ̃(x) = dμ(x + b). The Jacobi parameters of μ̃ are given by

ã_{n} = a_{n} b̃_{n} = b_{n} − b (3.8)

Moreover,

∫ x^{ℓ} dμ_{n} = ∑_{j=0}^{ℓ} \binom{ℓ}{j}(b)^{ℓ−j} ∫ x^{j} dμ̃_{n}  (3.9)

and

∫ x^{ℓ} dμ̃_{n} = ∑_{j=0}^{ℓ} \binom{ℓ}{j}(−b)^{ℓ−j} ∫ x^{j} dμ_{n}  (3.10)

Since b̃_{n} → 0 and every odd ℓ random walk has a b_{k} factor in it, ∫ x^{ℓ} dμ̃_{n} → 0 for all odd ℓ. Thus, (3.9) implies ∫ x^{3} dμ_{n} exists and then (3.10) that ∫ x^{4} dμ̃_{n} converges. Thus, without loss, we suppose A_{1} = 0 and b_{n} → 0.

If b_{n} → 0, any path with ρ_{j} = ρ_{j+1} contributes zero in the limit; so we can restrict to paths with |ρ_{j+1} − ρ_{j}| = 1. Thus, looking at the two such paths with ρ_{2} = ρ_{0} = n,

lim_{n→∞} a_{n+1}^{2} + a_{n}^{2} = A_{2}  (3.11)

In looking at paths with ρ_{0} = ρ_{4} = n, all those with ρ_{2} = n contribute (∫ x^{2}dμ_{n})^{2}, so

lim_{n→∞} a_{n+2}^{2}a_{n+1}^{2} + a_{n}^{2}a_{n−1}^{2} = A_{4} − A_{2}^{2}  (3.12)

Thus, using (x − y)^{2} = (x + y)^{2} − 4xy,

lim_{n→∞} (a_{n+2}^{2} − a_{n+1}^{2})^{2} + (a_{n}^{2} − a_{n−1}^{2})^{2} = 6A_{2}^{2} − 4A_{4}  (3.13)

Suppose a_{n} has a limit point, a, that is, a_{n(j)} → a as j → ∞ for a subsequence. Define c = \sqrt{A_{2} − a^{2}}. By (3.11), for any ℓ = 0, ±1, ±2, …,

a_{n(j)+ℓ} → \begin{cases}a & ℓ even \\ c & ℓ odd \end{cases}

In particular, by (3.13),

|a^{2} − c^{2}| = \sqrt{3A_{2}^{2} − 2A_{4}}  (3.14)

Since also

a^{2} + c^{2} = A_{2}  (3.15)

there are at most two solutions of (3.14), (3.15):

a^{2} = \frac{1}{2} [A_{2} + \sqrt{3A_{2}^{2} − 2A_{4}} ]  (3.16)
c^{2} = \frac{1}{2} [A_{2} − \sqrt{3A_{2}^{2} − 2A_{4}} ]  (3.17)

and the one with a, c, reversed. Thus the right sides of (3.16) and (3.17) are the only limit points of a_{n}^{2}.

The lemma below completes the proof. □

Lemma 3.4. Let x_{n} be a sequence so that for some α, β ∈ ℝ,

lim_{n→∞} x_{n} + x_{n+1} = α + β  (3.18)
lim_{n→∞} |x_{n} − x_{n+1}| = |α − β|  (3.19)

Then either

lim_{n} x_{2n} = α  lim_{n} x_{2n+1} = β

or

lim_{n} x_{2n} = β  lim_{n} x_{2n+1} = α

Proof. By replacing x_{n} by x_{n} − \frac{1}{2}(α + β), we can suppose α = −β ≥ 0. If α = β = 0, the result is trivial, so suppose α = −β > 0. Pick N so that for n > N,

|x_{n} + x_{n+1}| < α
|x_{n} − x_{n+1}| > α

Thus, since |x_{n} − x_{n+1}| > |x_{n} + x_{n+2}|, x_{n} and x_{n+1} have opposite signs for all n > N, that is, for n > N, either (−1)^{n}x_{n} > 0 or (−1)^{n+1}x_{n} > 0. Since ±α are the only allowed limit points if (−1)^{n}x_{n} > 0, x_{2n} → α, x_{2n+1} → β = −α, and if (−1)^{n+1}x_{n} > 0, x_{2n} → β, x_{2n+1} → α. □

4. Ratio Asymptotics for p_{n+1}/p_{n}

In this section and the next, we discuss two further issues related to our results: what about lim p_{n+1}/p_{n} (p_{n} rather than P_{n}) and we calculate the measures dρ_{b;a,c} of (3.2) (already well known if a = c).

Let

R_{n}(z) = \frac{P_{n+1}(z)}{P_{n}(z)}  r_{n}(z) = \frac{p_{n+1}(z)}{p_{n}(z)}  (4.1)

Since p_{n} = (a_{1} … a_{n})^{−1}P_{n},

r_{n}(z) = a_{n+1}^{−1}R_{n}(z)  (4.2)

so we immediately see with Theorem 2.1 that if a_{n} → a ≠ 0 and b_{n} → b,

lim_{n→∞} r_{n}(z) = \frac{(z − b) + \sqrt{(z − b)^{2} − 4a^{2}}}{2a}  (4.3)

We want to address a converse. One problem we find is that while |R_{n}(z)| > Im|z|, without an a priori upper bound on a_{n}, we do not have a bound for |r_{n}(z)|, so it is not obvious that existence of the limit implies a_{n} is bounded.

Example 4.1. Let a_{n} = e^{n!}, b_{n} = arbitrary bounded sequence, especially one without a limit. Then, by (1.3) and

r_{n}(z) = \frac{z − b_{n+1}}{a_{n+1}} − \frac{a_{n}}{a_{n+1}} [r_{n−1}(z)]^{−1}
= \frac{z − b_{n+1}}{a_{n+1}} − \frac{a_{n}^{2}}{a_{n+1}} [R_{n−1}(z)]^{−1}  (4.4)

by (4.2). By (2.7),

|r_{n}(z)| ≤ \frac{|z − b_{n+1}|}{a_{n+1}} + \frac{a_{n}^{2}}{a_{n+1}} |Im z|^{−1} → 0  (4.5)

since a_{n} is chosen so a_{n}^{2}/a_{n+1} → 0. Thus for this example, r_{n}(z) → 0. □

Because of this example, we will need to suppose that if lim r_{n}(z) exists, it has nonzero imaginary part. Here is a result that requires two points rather than one, with some extra conditions:

Theorem 4.2. Suppose sup_{n} a_{n} < ∞. Suppose z_{1}, z_{2} are in {z | Im z > 0} and let

lim_{n→∞} r_{n}(z_{j}) = λ_{j}  (4.6)

with

(a)  Im λ_{j} > 0

and either

(b1)  \frac{Im λ_{1}}{Im z_{1}} ≠ \frac{Im λ_{2}}{Im z_{2}}

or

(b2)  \frac{Im(λ_{1}^{−1})}{Im z_{1}} ≠ \frac{Im(λ_{2}^{−1})}{Im z_{2}}

Then a_{n} → a ≠ 0 and b_{n} → b.

Proof. By (1.3),

a_{n+1}r_{n}(z) = (z − b_{n+1}) − a_{n}[r_{n−1}(z)]^{−1}  (4.7)

Since Im(−r_{n−1}(z))^{−1} > 0, (4.7) implies

a_{n+1} Im r_{n}(z_{j}) ≥ Im z_{j}

which implies

lim inf a_{n} ≥ \frac{Im z_{j}}{Im λ_{j}} > 0

so the a’s are bounded above and below.

Let (a, c) be a limit point of (a_{n+1}, a_{n}). By (4.7),

a Im λ_{j} = Im z_{j} + c(Im(−λ_{j})^{−1})  (4.8)

If Im λ_{1}/ Im z_{1} ≠ Im λ_{2}/ Im z_{2}, (4.8) implies

a[\frac{Im λ_{1}}{Im z_{1}} − \frac{Im λ_{2}}{Im z_{2}} ] = c[\frac{Im(−λ_{1})^{−1}}{Im z_{1}} − \frac{Im(−λ_{2})^{−1}}{Im z_{2}} ]

so we can solve for a as a multiple of c, and then for c in (4.8) for j = 1. If Im(λ_{1})^{−1}/ Im z_{1} ≠ Im(λ_{2})^{−1}/ Im z_{2}, we solve for c as a multiple of a. Either way, we see (4.8) has a unique solution for (a, c) so (a_{n+1}, a_{n}) → (a, c). But then (a_{n+2}, a_{n+1}) → (a, c) so a = c and a_{n} → a ≠ 0. This implies lim R_{n}(z_{1}) = lim a_{n}c_{n}(z) exists. So, by Theorem 4.2, b_{n} → b. □

We have a second remark about the existence of lim_{n→∞} r_{n}(z). In the OPUC case, existence of φ_{n+1}^{∗}(z)/φ_{n}^{∗}(z) for all z ∈ 𝔻 implies the same of Φ_{n+1}^{∗}(z)/Φ_{n}^{∗}(z) (and, by taking α_{j} = 0 if j ≠ n^{2}, α_{n^{2}} = \frac{1}{2}, not conversely) for φ_{n+1}^{∗}(0)/φ_{n}^{∗}(0) = ρ_{n}^{−1}, so existence of the φ ratio limit at z = 0 implies ρ_{n} → ρ_{∞} and then, since Φ_{n+1}^{∗}(z)/Φ_{n}^{∗}(z) = ρ_{n}φ_{n+1}^{∗}(0)/φ_{n}^{∗}(0), we get the Φ_{n}^{∗} ratio limits.

The same is true here, but alas, the analog of z = 0 for OPUC is z = ∞ here. The following captures the idea, without the need for hypotheses (b) of Theorem 4.2.

Theorem 4.3. Suppose a_{n} and |b_{n}| are bounded and r_{n}(z) converges to a nonzero limit as n → ∞ for all z in a small neighborhood of z_{0} ∈ ℂ\ℝ. Then a_{n} → a and b_{n} → b for some a ≠ 0, b ∈ ℝ.

Proof. By (4.4),

r_{n}(z) = \frac{z}{a_{n+1}} − \frac{b_{n+1}}{a_{n+1}} + O(\frac{1}{z^{−1}} )  (4.9)

Thus

r_{n}(z)^{−1} = \frac{a_{n+1}}{z} + \frac{b_{n+1}}{z^{2}} + O(\frac{1}{z^{3}} )  (4.10)

If supp(dμ) ⊂ [−c, c] (take c = sup_{n}|b_{n}| + 2 sup|a_{n}|), then for ρ > c, by the Cauchy formula for every ℓ ∈ ℤ and the fact that r_{n} has its zeros in [−c, c], \frac{1}{2πi} ∮_{|z|=ρ} r_{n}(z)^{−1}z^{ℓ} dz is ρ independent for ρ > c. Taking ρ to infinity, we get, by (4.9),

\frac{1}{a_{n+1}} = \frac{1}{2πi} ∮_{|z|=c+1} r_{n}(z)^{−1} dz  (4.11)
−\frac{b_{n+1}}{a_{n+1}} = \frac{1}{2πi} ∮_{|z|=c+1} r_{n}(z)^{−1}z dz  (4.12)

Thus uniform convergence of r_{n}(z) to a limit on |z| = c + 1 implies convergence of a_{n} and b_{n}. Therefore, we are reduced to showing convergence of r_{n}^{−1} on a single compact subset of ℂ\ℝ implies convergence on all compact subsets of ℂ\[−c, c].

By (2.4) and (2.5) and x_{i,n} ∈ [−c, c], we have

|R_{n}(z)^{−1}| ≤ sup_{x∈[−c,c]} |z − x|^{−1}

and thus, by (4.2),

|r_{n}(z)^{−1}| ≤ [sup_{n} |a_{n}|] sup_{x∈[−c,c]} |z − x|^{−1}  (4.13)

Thus convergence of r_{n}(z) on a compact implies, by Vitali’s theorem and (4.13), uniform convergence of r_{n}(z)^{−1} on compact sets in ℂ\[−c, c]. □

5. Calculation of dρ_{b;a,c}(x)

In this section, we compute the weak limit ρ_{b;a,c} of p_{n}^{2} dμ for μ, the measure corresponding to the Jacobi matrix a_{2n} = c, a_{2n+1} = a, b_{n} = b. We will also find dμ_{b;a,c}, the measure associated to the Jacobi matrix

J = (\begin{matrix}b & a \\ a & b & c \\ & c & b & a \\ & & a & b & c \\ & & & ... & ... & ... \end{matrix})  (5.1)

We begin by computing

f(z; b; a, c) = ∫ \frac{dμ_{b;a,c}(x)}{z − x}  (5.2)

and

G(z; b; a, c) = ∫ \frac{dρ_{b;a,c}(x)}{z − x}  (5.3)

This calculation is not unrelated to calculations in Khrushchev [7] for the measure associated to a period 2 Verblunsky coefficient. As he does, we could ask for the Jacobi coefficients for the measure dρ and show they converge exponentially fast to those for dμ (with a need to interchange a and c depending on the sign of a − c).

We begin with a result about a finite Jacobi matrix

J^{[1,n]} = (\begin{matrix}b_{1} & a_{1} \\ a_{1} & b_{1} & a_{2} \\ & ... & ... & ... \\ & & ... & ... & ... \\ & & & ... & ... & a_{n−1} \\ & & & & a_{n−1} & b_{n} \end{matrix})  (5.4)

We let J^{[j,k]} for 1 ≤ j ≤ k ≤ n denote the (k − j + 1) × (k − j + 1) matrix we get by keeping rows and columns between j and k (inclusive). We refer to the row number of J^{[j,k]} as j, j + 1, … so, for example, (J^{[j,k]})_{jj} = b_{j}.

Here is a key lemma that appears in Gesztesy-Simon [5] although closely related formulae have appeared elsewhere; in particular, the k = 1, n results go back to Jacobi:

Proposition 5.1. For 2 ≤ k ≤ n − 1,

\begin{matrix}[ & (z − J^{[1,n]})^{−1}]_{kk} \\ = 1/\{z − b_{k} − a_{k−1}^{2}[(z − J^{[1,k−1]})^{−1}]_{k−1,k−1} − a_{k}^{2}[(z − J^{[k+1,n]})^{−1}]_{k+1,k+1}\} \end{matrix} ; (5.5)

For k = 1,

[(z − J^{[1,n]})^{−1}]_{11} = 1/{z − b_{1} − a_{1}^{2}[(z − J^{[2,n]})^{−1}]_{22}}  (5.6)

Proof. Let 2 ≤ k ≤ n − 1. Here is a proof that is more direct than that in [5], although the essence is the same. If row and column k are removed, the resulting matrix is J^{[1,k−1]} ⊕ J^{[k+1,n]} so, by Cramer’s rule,

[(z − J^{[1,n]})^{−1}]_{kk} = \frac{det(z − J^{[1,k−1]}) det(z − J^{[k+1,n]})}{det(z − J^{[1,n]})}  (5.7)

Expanding det(z − J^{[1,n]}) in minors in row k,

det(z − J^{[1,n]}) = (z − b_{k})d_{1} − a_{k−1}^{2}d_{2} − a_{k}^{2}d_{3}  (5.8)

where

d_{1} = det(z − J^{[1,k−1]}) det(z − J^{[k+1,n]})  (5.9)
d_{2} = det(z − J^{[1,k−2]}) det(z − J^{[k+1,n]})  (5.10)
d_{3} = det(z − J^{[1,k−1]}) det(z − J^{[k+2,n]})  (5.11)

(where det(z − J^{[1,0]}) and det(z − J^{[n+1,n]}), which occur if k = 2 or k = n − 1, are interpreted as 1).

Finally, note that, by Cramer’s rule again,

[(z − J^{[1,k−1]})^{−1}]_{k−1,k−1} = \frac{d_{2}}{d_{1}}  (5.12)
[(z − J^{[k+1,n]})^{−1}]_{k+1,k+1} = \frac{d_{3}}{d_{1}}  (5.13)

(5.7)–(5.13) imply (5.5). (5.6) is proven in a similar way (but, e.g., the analog of (5.8) has only two terms). □

Corollary 5.2. For the functions f and G of (5.2)/ (5.3), we have that for z ∈ ℂ\ℝ,

[G(z; b; a, c)]^{−1} = z − b − c^{2}f(z; b; a, c) − a^{2}f(z; b; c, a)  (5.14)

Remark. This formula makes it evident once again that G is symmetric in a and c.

Proof. Let J be given by (5.1). On ℓ^{2}(ℤ^{+}), J^{[1,n]} ⊕0 converges strongly to J. Thus

f(z; b; a, c) = [(z − J)^{−1}]_{11} = lim_{n→∞} [(z − J^{[1,n]})^{−1}]_{11}  (5.15)

Moreover,

∫ \frac{p_{k}(x)^{2} dμ_{b;a,c}(x)}{z − x} = [(z − J)^{−1}]_{kk} = lim_{n→∞} [(z − J^{[1,n]})^{−1}]_{kk} (5.16)

so

G(z; b; a, c) = lim_{k→∞} [ lim_{n→∞}[(z − J^{[1,n]})^{−1}]_{kk}]  (5.17)

(5.14) follows by using (5.5), (5.15), and (5.17) together with the structure of J (e.g., J^{[1+2ℓ,n+2ℓ]} = J^{[1,n]} and J^{[1+2ℓ+1,n+2ℓ+1]} is J̃^{[1,n]} for J̃, the matrix with a and c reversed). □

Remark. The limit of (5.5) as n → ∞ is a precise analog of Khrushchev’s formula [6] that for the unit circle case, the Schur function of |φ_{n}|^{2} dμ is b_{n}f_{n}. It would be interesting to see if one could translate our proof here to a proof of Khrushchev’s formula using the CMV matrix [2, 13] in place of the Jacobi matrix.

Corollary 5.3.

[f(z; b; a, c)]^{−1} = z − b − a^{2}f(z; b; c, a)  (5.18)

Proof. This is identical to the last proof using (5.6) in place of (5.5). Of course, this is a special case of the well-known Stieltjes continued fraction expansion for the Stieltjes transform of the measure associated to a Jacobi matrix. □

Henceforth, for simplicity, we take b = 0. Since G(z; b; a, c) = G(z − b; 0; a, c), and similarly for f, it is easy to go from this case to the case of general b.

As a warmup, consider the case a = c. Then (5.18) becomes

1 = f(z − a^{2}f)  (5.19)

which is solved by

f(z; 0; a, a) = \frac{z − \sqrt{z^{2} − 4a^{2}}}{2a^{2}}  (5.20)

where the branch of the square root is taken (with \sqrt{⋯} = z + O(1/z) as |z| → ∞ consistent with f(z) ∼ 1/z + O(z^{−2})). By (5.14), we get

G(z; 0; a, a) = \frac{1}{\sqrt{z^{2} − 4a^{2}}}  (5.21)

lim_{ε↓0} Im G(x + iε; 0; a, a) is only nonzero for z ∈ [−2a, 2a] so

dρ_{b;a,a}(x) = \frac{1}{π} lim_{ε↓0} Im G(x + iε; b; a, c) dx  (5.22)

implies the well-known

dρ_{b;a,a} = \frac{1}{π\sqrt{4a^{2} − (x − b)^{2}}} χ_{[b−2a,b+2a]}(x) dx  (5.23)

consistent with Nevai’s conjecture.

Here is the main result of this section:

Theorem 5.4. Define

I(z; a, c) = (c^{2} − a^{2} + z^{2})^{2} − 4z^{2}c^{2}  (5.24)

Then

(i)  f(z; 0; a, c) = \frac{c^{2} − a^{2} + z^{2} − \sqrt{I(z; a, c)}}{2c^{2}z}  (5.25)
(ii)  G(z; 0; a, c) = \frac{1}{\sqrt{I(z; a, c)} + \sqrt{I(z; c, a)}}  (5.26)
(iii)  dρ_{b;a,c,}(x) = [χ_{J_{2}}(x) + χ_{J_{2}}(x)]w(x) dx  (5.27)

where

J_{1} = [b + |c − a|, b + a + c]  J_{2} = −J_{1}  (5.28)

and for x ∈ J_{1} ∪ J_{2},

w(x) = π^{−1}[\sqrt{−I(x − b; a, c)} + \sqrt{−I(x − b; c, a)} ]^{−1}  (5.29)

Proof. (i) Iterating (5.15) once, we get a quadratic equation for f(z; 0; a, c) whose solution (the one with f = 1/z +O(1/z^{2}) at infinity) is (5.25).

(ii) Follows from (5.14) and (5.25).

(iii) I(z; a, c) = 0 if and only if c^{2} − a^{2} + z^{2} = ±2zc if and only if (z ± c)^{2} = a^{2} if and only if z = ±a ± c (independent ±) which say \sqrt{I} has a branch cut on J_{1} ∪ J_{2} and G is purely imaginary there. (5.22) completes the proof. □

We note there is another proof of the theorem using Floquet theory and the theory of periodic whole-line Jacobi matrices.

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