MONOTONE JACOBI PARAMETERS AND NON-SZEGŐ WEIGHTS

YURY KREIMER^{1}, YORAM LAST^{1,3}, AND BARRY SIMON^{2,3}

Abstract. We relate asymptotics of Jacobi parameters to asymptotics of the spectral weights near the edges. Typical of our results is that for a_{n} ≡ 1, b_{n} = −Cn^{−β} (0 < β < \frac{2}{3}), one has dμ(x) = w(x) dx on (−2, 2), and near x = 2, w(x) = e^{−2Q(x)} where

Q(x) = β^{−1}C^{\frac{1}{β}} \frac{Γ(\frac{3}{2})Γ(\frac{1}{β} − \frac{1}{2})(2 − x)^{\frac{1}{2} − \frac{1}{β}}}{Γ(\frac{1}{β} + 1)} (1 + O((2 − x)))

1. Introduction

Since the earliest days of the general theory of orthogonal polynomials on the real line (OPRL), it has been known that a key role is played by the Szegő condition [38] that if

dμ(x) = w(x) dx + dμ_{s}  (1.1)

where w is supported on [−2, 2] (we follow the spectral theorists’ convention related to a_{n} → 1, b_{n} → 0 rather than the [−1, 1] tradition in the OP literature), then

∫ log(w(x))(4 − x^{2})^{− \frac{1}{2}} dx > −∞  (1.2)

In this paper, we will examine asymptotics of log(w(x)) for typical cases where (1.2) fails. Recall [39, 5, 2, 31, 34] that, given μ, one

Date: November 7, 2007. 2000 Mathematics Subject Classification. 33C45,34L05,47B15. Key words and phrases. Orthogonal polynomials, Schrödinger operators, spec-

tral weights, Szegő condition.

1 Institute of Mathematics, The Hebrew University, 91904 Jerusalem, Israel. E-mail: yuryk@math.huji.ac.il; ylast@math.huji.ac.il. Supported in part by The Israel Science Foundation (grant no. 1169/06).

2 Mathematics 253-37, California Institute of Technology, Pasadena, CA 91125, USA. E-mail: bsimon@caltech.edu. Supported in part by NSF grant DMS-0140592.

3 Research supported in part by Grants No. 2002068 and No. 2006483 from the United States-Israel Binational Science Foundation (BSF), Jerusalem, Israel.

can define monic orthogonal and orthonormal polynomials P_{n}(x, dμ), p_{n}(x, dμ) and Jacobi parameters {a_{n}, b_{n}}_{n=1}^{∞} by (b_{n} real, a_{n} > 0)

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

and

∥P_{n}∥ = a_{1} ⋯ a_{n}  (1.4)

Favard’s theorem (see, e.g., [31, 34]) asserts a one-one correspondence between μ’s of compact but infinite support and bounded sets of a_{n}’s and b_{n}’s. Moreover, by Weyl’s theorem, if a_{n} → 1, b_{n} → 0, then the essential support of dμ is [−2, 2].

Roughly speaking, the boundary for (1.2) to hold is a_{n} − 1, b_{n} decaying faster than O(n^{−1}). Explicitly, Killip and Simon [11] proved a conjecture of Nevai [24] that ∑_{n=1}^{∞}(|a_{n} − 1| + |b_{n}|) < ∞ ⇒ (1.2), and there are examples of Pollaczek [25, 26, 27] where (1.2) fails because log(w(x)) ∼ (4 − x^{2})^{− \frac{1}{2}} near x = ±2 and b_{n} = 0, a_{n} = 1 − Cn^{−1} + O(n^{−2}).

Killip–Simon [11] discovered a relevant weaker condition than (1.2) they called the quasi-Szegő condition:

∫ log(w(x))(4 − x^{2})^{\frac{1}{2}} dx > −∞  (1.5)

and they proved that

(1.5) + ∑_{x∈supp(μ)\[−2,2]}(|x| − 2)^{\frac{3}{2}} < ∞ ⇔ ∑_{n=1}^{∞} |a_{n} − 1|^{2} + |b_{n}|^{2} < ∞ (1.6)

Our cases will include situations where (1.5) and (1.6) fail.

It is known (see [10, 20, 21, 22, 29, 40]) that when ∑_{n=1}^{∞}|a_{n} − 1|^{2} + |b_{n}|^{2} = ∞, dμ can stop having an a.c. component, so we will need an additional condition. What we will use is

Theorem 1.1. If a_{n} → 1, b_{n} → 0, and

∑_{n=1}^{∞} |a_{n+1} − a_{n}| + |b_{n+1} − b_{n}| < ∞  (1.7)

then (1.1) holds where w(x) is continuous on (−2, 2) and strictly positive there. Moreover, dμ_{s} is supported on ℝ \ (−2, 2).

The continuum Schrödinger analog of this is a theorem of Weidmann [41]; for OPRL, it is due to Dombrowski–Nevai [4] (see also [12, 8, 32]). Most references do not discuss continuity of w but it holds; for example, it follows immediately from Theorem 1 of [4], since w can be obtained as a uniform limit of continuous functions on any closed subinterval of (−2, 2).

In fact, we will focus on cases where {a_{n}} and {b_{n}} are monotone, so (1.7) is automatic. Typical is

a_{n} ≡ 1 b_{n} = −Cn^{−β}  (1.8)

where, roughly speaking, we will prove w(x) is singular at x = 2 (i.e., the integral in (1.5) diverges there) with

w(x) = e^{−2Q(x)}  (1.9)
Q(x) ∼ C_{1}(2 − x)^{\frac{1}{2} − \frac{1}{β}}  (1.10)

Indeed, in Section 5, we will obtain for (1.8) an asymptotic series for Q(x) near x = 2 up to terms of O(log(2 − x)); see (5.32).

Our interest in these problems was stimulated by a recent paper of Levin–Lubinsky [18] and their related earlier works on non-Szegő weights [16, 17]. They study the problem inverse to ours, namely, going from w (or Q) to a_{n}, b_{n} (which they call A_{n}, B_{n}). Unfortunately, they do not obtain even leading order asymptotics for a_{n}, b_{n} if Q(x) has the form (1.10) but instead require

Q(x) ∼ exp_{k}(1 − x^{2})^{−α}  (1.11)

with exp_{k}(x) = exp(exp_{k−1}(x)) and exp_{1}(x) = e^{x}. We will obtain inverse results to theirs in Section 5. We note that [16] does have asymptotics on the Rakhmanov–Mhaskar–Saff numbers when (1.10) holds and that their asymptotics should be connected to asymptotics of a_{n}, b_{n}.

It is hard to imagine strict if and only if results on Q(x) to a_{n}, b_{n} since there will typically be side conditions (a_{n}, b_{n} monotone and/or convex in n or Q(x) convex) that may not strictly carry over, but it is comforting (even with side conditions) to get results in both directions. It would be interesting to show that (1.9) and (1.10) (with extra conditions) lead to estimates on a_{n}, b_{n} with |a_{n} − 1| + |b_{n}| = O(n^{−β}). We suspect, with analyticity assumptions on Q, that this might be accessible with Riemann–Hilbert techniques.

Our key to going from (a_{n}, b_{n}) to (w, Q) is Carmona’s formula that relates dμ to the growth of p_{n}(x), namely,

Theorem 1.2. If p_{n} are the orthonormal polynomials for a measure

dμ, then dν^{(n)} ⟶^{w} dμ where

dν^{(n)}(x) = \frac{dx}{π(a_{n}^{2}p_{n}(x)^{2} + p_{n−1}^{2}(x))}  (1.12)

The continuum analog of this result is due to Carmona [1]. This theorem when a_{n} = 1 is stated without proof in Last–Simon [14] and later (with proof) in Krutikov–Remling [13] and Simon [33]. It implies:

Corollary 1.3. Suppose uniformly on some interval [α, β], we have for strictly positive continuous functions f_{±}(x) that

π^{−1}f_{−}(x) ≤ lim inf(a_{n}^{2}p_{n}(x)^{2} + p_{n−1}(x)^{2})

≤ lim sup(a_{n}^{2}p_{n}(x)^{2} + p_{n−1}(x)^{2}) ≤ π^{−1}f_{+}(x) (1.13)

Then dμ is purely absolutely continuous on (α, β) and

\frac{1}{f_{+}(x)} ≤ w(x) ≤ \frac{1}{f_{−}(x)}  (1.14)

there. In particular, if (1.13) holds for each compact interval [α, β] in (x_{0}, 2),

f_{±}(x) = exp(2(g(x) ± h(x)))  (1.15)

then (1.9) holds with

|Q(x) − g(x)| ≤ h(x)  (1.16)

Proof. By Theorem 1.1, for any positive continuous function, η(x), on [α, β] supported on (α, β), we have

∫ \frac{η(x)}{πf_{+}(x)} dx ≤ ∫ η(x) dμ(x) ≤ ∫ \frac{η(x)}{πf_{−}(x)} dx  (1.17)

from which absolute continuity of μ ↾ (α, β) and (1.14) are immediate. This in turn implies (1.15) and (1.16). □

Thus, we need to show a_{n}^{2}p_{n}^{2} + p_{n−1}^{2} is bounded as n → ∞, but with bounds that diverge as x ↑ 2. The difference equation is

\binom{p_{n+1}}{a_{n+1}p_{n}} = \frac{1}{a_{n+1}} (\begin{matrix}x − b_{n+1} & −1 \\ a_{n+1}^{2} & 0 \end{matrix}) \binom{p_{n}}{a_{n}p_{n−1}}
≡ A_{n+1}(x)\binom{p_{n}}{a_{n}p_{n−1}}  (1.18)

Here

det(A_{n}) = 1 tr(A_{n}) = x − b_{n}  (1.19)

In a case like (1.8) where b_{n} is negative and monotone increasing, a fundamental object is the turning point, the integer, N(x), with

x − b_{n} ≥ 2 if n ≤ N(x)  (1.20)
x − b_{n} < 2 if n > N(x)  (1.21)

If γ_{n}(x) is defined by γ_{n} ≥ 0 and

x − b_{n} = 2 cosh(γ_{n}(x)) (n ≤ N(x))  (1.22)

then one expects some kind of exponential growth as exp(∑_{j=1}^{n} γ_{j}(x)), and we will prove that

exp( ∑_{j=1}^{N} γ_{j}(x)) ≤ p_{N}(x) ≤ (N + 1) exp( ∑_{j=1}^{N} γ_{j}(x))  (1.23)

As one expects, there is an intermediate region N(x) ≤ n ≤ N_{1}(x) and an oscillatory region n ≥ N_{1}(x). We will see that so long as one is willing to accept O((b_{N+2} − b_{N+1})^{−1}) errors (and they will typically be very small compared to exp(∑_{j=1}^{N} γ_{j}(x))), one can actually take N_{1} = N +2 (!) and use the method of proof for Theorem 1.1 to control the region n ≥ N_{1}. Thus, the key will be (1.23) and we will get (1.16) where

g(x) = ∑_{j=1}^{N} γ_{j}(x)  (1.24)

and

h(x) = O(max(log(N), log((b_{N+2} − b_{N+1})^{−1})))  (1.25)

The discussion of turning points sounds like WKB—and the reader might wonder if one can’t obtain our result via standard WKB techniques. There is some literature on discrete WKB [6, 35, 36, 37], but we have not seen how to apply them to this situation (for a different application to OPRL, see [7]) or, because of a double n → ∞, x → 2 limit, how to use the continuum WKB theory (on which there is much more extensive literature) to the continuum analog of our problem here. That said, the current paper should be regarded as a WKB-like analysis.

In Section 2, we discuss the case a_{n} ≡ 1, b_{n} < b_{n+1} < 0. In Section 3, we discuss b_{n} ≡ 0, a_{n} < a_{n+1} < 1. It is likely one could handle mixed a_{n}, b_{n} cases with more effort. In Section 4, we discuss some Schrödinger operators. Finally, in Section 5, we discuss examples including (1.8) and (1.11).

It is a pleasure to thank Fritz Gesztesy, Uri Kaluzhny, and Doron Lubinsky for useful discussions. B. S. would like to thank Ehud de Shalit for the hospitality of the Einstein Institute of Mathematics at the Hebrew University where some of this work was done. Y. L. would like to thank Matthias Flach for the hospitality of the Department of Mathematics at Caltech where some of this work was done.

2. Monotone b_{n}

In this section, we will prove:

Theorem 2.1. Let dμ be the spectral measure associated with a Jacobi matrix having a_{n} ≡ 1 and

b_{n} ≤ b_{n+1} < 0 b_{n} → 0 as n → ∞

Define N(x) for x in (0, 2) and near 2 by (1.20)/ (1.21) and γ_{n}(x) by (1.22). Then dμ is purely absolutely continuous on (−2, 2), where w = \frac{dμ}{dx} is continuous and nonvanishing on (−2, 2),

C_{1}(x + 2) ≤ w(x) ≤ C_{2}(x + 2)^{−1} for x ∈ (−2, 0]  (2.1)

and on (0, 2),

w(x) = e^{−2Q(x)}  (2.2)

where

|Q(x) − g(x)| ≤ h(x)  (2.3)

where

g(x) = ∑_{j=1}^{N(x)} γ_{j}(x)  (2.4)

and h(x) is given by

e^{h(x)} = CN(x)(b_{N(x)+2} − b_{N(x)+1})^{−1}(2 − x)^{\frac{1}{2}}  (2.5)

for an explicit constant C (dependent on sup|b_{n}| but not on x).

Remark. Typically, h is much smaller than g. For example, if b_{n} is given by (1.8), g(x) = O((2−x)^{\frac{1}{2} − \frac{1}{β}}) and e^{h(x)} = O(N(x)^{2+β}(2−x)^{\frac{1}{2}} ) = O((2 − x)^{−( \frac{1}{2} + \frac{2}{β})}), so h(x) = O(log(2 − x)^{−1}).

As we explained in the introduction, we need to study the asymptotics of p_{n}(x) as x ↑ 2 with some uniformity in n. Given that a_{n} ≡ 1,

p_{n+1}(x) = (e^{γ_{n+1}} + e^{−γ_{n+1}})p_{n}(x) − p_{n−1}(x)  (2.6)
p_{−1}(x) = 0 p_{0}(x) = 1  (2.7)

which suggests we define for n ≤ N(x),

ψ_{n}(x) = e^{− ∑_{j=1}^{n} γ_{j}}p_{n}(x)  (2.8)

so ψ_{n} obeys

ψ_{n+1}(x) = (1 + e^{−2γ_{n+1}})ψ_{n} − e^{−(γ_{n}+γ_{n+1})}ψ_{n−1}  (2.9)
ψ_{−1}(x) = 0 ψ_{0}(x) = 1  (2.10)

Lemma 2.2. For 0 ≤ n < N(x),

ψ_{n+1} ≥ ψ_{n}  (2.11)

In particular,

ψ_{n}(x) ≥ 1  (2.12)

Proof. As a preliminary, we note that b_{n} ≤ b_{n+1} implies x − b_{n} ≥ x − b_{n+1}, so

0 ≤ γ_{n+1} ≤ γ_{n}  (2.13)

By (2.9),

(ψ_{n+1} − ψ_{n}) = e^{−2γ_{n+1}}ψ_{n} − e^{−(γ_{n}+γ_{n+1})}ψ_{n−1}
= e^{−2γ_{n+1}}(ψ_{n} − ψ_{n−1}) + e^{−γ_{n+1}}(e^{−γ_{n+1}} − e^{−γ_{n}})ψ_{n−1} (2.14)

For n = 0, ψ_{n} − ψ_{n−1} = 1 ≥ 0 and ψ_{n−1} = 0 ≥ 0. By (2.14) and (2.13) (which implies e^{−γ_{n+1}}−e^{−γ_{n}} ≥ 0), we see inductively that ψ_{n+1}−ψ_{n} ≥ 0, and so, ψ_{n+1} ≥ ψ_{n} ≥ 0, proving (2.11). □

Lemma 2.3. Define for n = 0, 1, 2, …, N(x) − 1,

W_{n} = e^{γ_{n+1}}ψ_{n} − e^{−γ_{n}}ψ_{n−1}  (2.15)

Then

W_{n} ≤ e^{γ_{n+1}}  (2.16)

Proof. W_{0} = e^{γ_{1}} ≤ e^{γ_{1}}, starting an inductive proof of (2.16). By (2.9),

ψ_{n+1} = e^{−γ_{n+1}}W_{n} + e^{−2γ_{n+1}}ψ_{n}

so

W_{n+1} = e^{(γ_{n+2}−γ_{n+1})}(W_{n} + e^{−γ_{n+1}}ψ_{n}) − e^{−γ_{n+1}}ψ_{n}
= e^{(γ_{n+2}−γ_{n+1})}W_{n} + e^{−γ_{n+1}}(e^{(γ_{n+2}−γ_{n+1})} − 1)ψ_{n}  (2.17)
≤ e^{(γ_{n+2}−γ_{n+1})}W_{n}  (2.18)

since (2.13) implies e^{γ_{n+2}} ≤ e^{γ_{n+1}} and ψ_{n} ≥ 0, (e^{γ_{n+2}−γ_{n−1}} − 1)ψ_{n} ≤ 0. Thus, W_{n} ≤ e^{γ_{n+1}} implies W_{n+1} ≤ e^{γ_{n+2}} and (2.16) holds inductively.

□

Lemma 2.4. For n = 0, 1, 2, …, N(x) − 2,

ψ_{n+1} ≤ 1 + ψ_{n}  (2.19)

So, in particular, for 0 ≤ n < N(x),

ψ_{n} ≤ n + 1  (2.20)

Proof. By (2.15),

ψ_{n+1} = e^{−γ_{n+2}}W_{n+1} + e^{−(γ_{n+1}+γ_{n+2})}ψ_{n}
≤ 1 + ψ_{n}

since e^{−γ_{n+2}}W_{n+1} ≤ 1 by (2.16) and γ_{j} ≥ 0 implies e^{−(γ_{n+1}+γ_{n+2})} ≤ 1. This proves (2.19), which inductively implies (2.20). □

We summarize with:

Proposition 2.5. For any n with 1 ≤ n < N(x),

e^{∑_{j=1}^{n} γ_{j}(x)} ≤ p_{n}(x) ≤ (n + 1)e^{∑_{j=1}^{n} γ_{j}(x)}  (2.21)

In particular, if

η_{n}(x) = p_{n−1}(x)^{2} + p_{n}(x)^{2}  (2.22)

then

e^{2 ∑_{j=1}^{n} γ_{j}(x)} ≤ η_{n}(x) ≤ 2(n + 1)^{2}e^{2 ∑_{j=1}^{n} γ_{j}(x)}  (2.23)

Proof. (2.21) is an immediate consequence of (2.8), (2.12) and (2.20). □

Suppose x ∈ (0, 2). For n > N(x), define κ_{n}(x) by 0 ≤ κ_{n} < \frac{π}{2} and

x − b_{n} = 2 cos κ_{n}(x)  (2.24)

so 0 > b_{n+1} ≥ b_{n} implies

0 ≤ κ_{n} ≤ κ_{n+1}

and b_{n} → 0 implies

κ_{n} → κ_{∞} = cos^{−1}(\frac{x}{2})  (2.25)

For later reference, we note

sin(κ_{∞}) = (1 − (\frac{x}{2})^{2})^{\frac{1}{2}} = \frac{1}{2} (4 − x^{2})^{\frac{1}{2}}  (2.26)

So as x ↑ 2,

κ_{∞} = (2 − x)^{\frac{1}{2}} + O((2 − x)^{\frac{3}{2}} )  (2.27)

We first present a matrix method following Kooman [12] to control the region [N(x) + 2, ∞). At the end, we will discuss an alternate method using scalar Prüfer-like variables.

By (1.18), for n > N, A_{n} has eigenvalues e^{±iκ_{n}}. In fact,

(\begin{matrix}2 cos κ & −1 \\ 1 & 0 \end{matrix}) (\begin{matrix}1 \\ e^{∓iκ} \end{matrix}) = e^{±iκ} (\begin{matrix}1 \\ e^{∓iκ} \end{matrix})  (2.28)

so if

Y (κ) = (\begin{matrix}1 & 1 \\ e^{−iκ} & e^{iκ} \end{matrix})  (2.29)

and

V (κ) = (\begin{matrix}e^{iκ} & 0 \\ 0 & e^{−iκ} \end{matrix})  (2.30)

then

A_{n}(x) = Y (κ_{n})V (κ_{n})Y (κ_{n})^{−1}  (2.31)

Next, notice that

Y (κ)^{−1} = \frac{1}{2i sin κ} (\begin{matrix}e^{iκ} & −1 \\ −e^{−iκ} & −1 \end{matrix})  (2.32)

Following Kooman [12], we write for n > ℓ > N(x),

T_{n}(x) ≡ A_{n} ⋯ A_{ℓ+1}  (2.33)
= Y (κ_{n})V_{n}Y (κ_{n})^{−1}Y (κ_{n−1})V_{n−1} ⋯ Y (κ_{ℓ+1})^{−1}

and since ∥V_{n}(κ)∥ = 1,

∥T_{n}∥ ≤ ∥Y (κ_{n})∥ ∥Y (κ_{ℓ+1})^{−1}∥ ∏_{j=ℓ+1}^{n−1} ∥Y (κ_{j+1})^{−1}Y (κ_{j})∥  (2.34)

This prepares us for two critical estimates:

Lemma 2.6. We have

∥Y (κ_{j+1})^{−1}Y (κ_{j})∥ ≤ 1 + \frac{|e^{iκ_{j+1}} − e^{iκ_{j}}|}{sin(κ_{j+1})}  (2.35)

so, in particular,

∥Y (κ_{j+1})^{−1}Y (κ_{j})∥ ≤ 1 + \frac{|κ_{j+1} − κ_{j}|}{sin(κ_{j})}  (2.36)

Proof. By (2.29) and (2.32),

Y (κ_{j+1})^{−1}Y (κ_{j}) − 1 = \frac{1}{2 sin(κ_{j+1})} (\begin{matrix}e^{−iκ_{j+1}} − e^{−κ_{j}} & e^{iκ_{j+1}} − e^{iκ_{j}} \\ e^{−iκ_{j}} − e^{−iκ_{j+1}} & e^{iκ_{j}} − e^{iκ_{j+1}} \end{matrix})

(2.37)

If A = (a_{ij}) is a 2 × 2 matrix,

|⟨φ, Aψ⟩| ≤ max(|a_{ij}|)(|φ_{1}| + |φ_{2}|)(|ψ_{1}| + |ψ_{2}|)
≤ 2 max(|a_{ij}|)(|φ_{1}|^{2}) + |φ_{2}|^{2})^{\frac{1}{2}} (|ψ_{1}|^{2} + |ψ_{2}|^{2})^{\frac{1}{2}}

since (|x| + |y|) ≤ \sqrt{2}(|x|^{2} + |y|^{2})^{\frac{1}{2}} , so

∥Y (κ_{j+1})^{−1}Y (κ_{j}) − 1∥ ≤ \frac{1}{sin(κ_{j+1})} |e^{iκ_{j+1}} − e^{iκ_{j}}|

which implies (2.35).

(2.35) implies (2.36) since \frac{π}{2} > κ_{j+1} ≥ κ_{j} implies sin(κ_{j+1}) ≥ sin(κ_{j}). □

Remark. That (2.36) holds with a 1 in front of |κ_{j+1} − κ_{j}|/ sin(κ_{j}) is critical. Lest it seem a miracle of Kooman’s method, we give an alternate calculation at the end of this section.

Lemma 2.7. We have that

∏_{j=ℓ+1}^{∞}(1 + \frac{|κ_{j+1} − κ_{j}|}{sin(κ_{j})}) ≤ \frac{κ_{∞}}{κ_{ℓ+1}} exp(κ_{∞}e(κ_{∞}))  (2.38)

where

e(y) = sup_{0<x≤y} (\frac{1}{sin(x)} − \frac{1}{x})  (2.39)

Remark. Since sin(x) = x − \frac{x^{3}}{6} + O(x^{5}), \frac{1}{sin(x)} = \frac{1}{x} + \frac{x}{6} + O(x^{3}) and since sin(x) < x, we see e(y) is finite and

e(y) = O(\frac{y}{6}) as y ↓ 0  (2.40)

Proof. We have

\frac{1}{sin(κ_{j})} ≤ \frac{1}{κ_{j}} + e(κ_{∞})  (2.41)

so, since κ_{j+1} ≥ κ_{j},

1 + \frac{|κ_{j+1} − κ_{j}|}{sin(κ_{j})} ≤ \frac{κ_{j+1}}{κ_{j}} + (κ_{j+1} − κ_{j})e(κ_{∞})  (2.42)
≤ \frac{κ_{j+1}}{κ_{j}} (1 + (κ_{j+1} − κ_{j})e(κ_{∞}))  (2.43)
≤ \frac{κ_{j+1}}{κ_{j}} exp((κ_{j+1} − κ_{j})e(κ_{∞}))  (2.44)

from which (2.38) is immediate if we note that κ_{∞} − κ_{ℓ} ≤ κ_{∞}. □

Proof of Theorem 2.1. By (2.34) and Lemmas 2.6 and 2.7, if T_{k}(x) is the transfer matrix from N(x) + 2 to k > N(x) + 2, then uniformly in k,

∥T_{n}∥ ≤ 2(sin(κ_{N(x)+2}))^{−1} \frac{κ_{∞}}{κ_{N(x)+2}} exp(κ_{∞}e(κ_{∞}))  (2.45)

where we also used ∥Y (κ_{k})∥ ≤ 2 and ∥Y (κ_{N(x)+2})^{−1}∥ ≤ 2/2 sin(κ_{N(x)+2}).

As x ↑ 2, κ_{∞} → 0. Indeed, by (2.27), κ_{∞} = (2 − x)^{\frac{1}{2}} + O((2 − x)^{\frac{3}{2}} ). Moreover, by the definition of N(x),

x − b_{N+1} < 2  (2.46)

while

x − b_{N+2} = 2 cos(κ_{N+2})  (2.47)

so

2(1 − cos(κ_{N+2})) > b_{N+2} − b_{N+1}  (2.48)

Since N(x) → ∞, b_{N(x)+2} → 0 so κ_{N+2}(x) → 0 and (2.48) implies

κ_{N+2}(x)^{2} > (1 + o(1))(b_{N+2} − b_{N+1})  (2.49)

Thus, in (2.45), [κ_{N(x)+2} sin(κ_{N+2})]^{−1} ≤ (1 + o(1))(b_{N+2} − b_{N+1}) and (2.45) becomes

sup_{n≥N(x)+2} ∥T̃_{n}∥ ≤ C(2 − x)^{\frac{1}{2}} (b_{N+2} − b_{N+1})^{−1} ≡ A(x)  (2.50)

where now T̃_{n} transfers from N − 1 to n and we use the boundedness from N − 1 to N + 2. Using

∥T̃_{n}∥^{−2}(|p_{n+1}|^{2} + |p_{n}|^{2}) ≤ |p_{N}|^{2} + |p_{N−1}|^{2} ≤ ∥T̃_{n}^{−1}∥^{2}(|p_{n+1}|^{2} + |p_{n}|^{2})

(2.51)

and (2.23), we obtain for all n > N,

C_{1}A(x)^{−2}e^{2 ∑_{1}^{N} γ_{j}(x)} ≤ (|p_{n}|^{2} + |p_{n+1}|^{2}) ≤ CA(x)^{2}N(x)^{2}e^{2 ∑_{1}^{N} γ_{j}(x)}

(2.52)

which, given Corollary 1.3, implies (2.2)–(2.4).

In going from (2.51) to (2.52), we used

det(T̃_{n}) = 1 ⇒ ∥T̃_{n}^{−1}∥ = ∥T̃_{n}∥

We also need to control the region x > −2 with 2 − x small. By replacing x by −x (and p_{n}(x) by (−1)^{n}p_{n}(−x)), this is the same as looking at x + b_{n} with still b_{n} < b_{n+1} < 0. We define θ_{n}(x) by

2 cos(θ_{n}(x)) = x + b_{n}  (2.53)

so

θ_{1} ≥ θ_{2} ≥ ⋯ ≥ θ_{∞} = κ_{∞} = (2 − x)^{\frac{1}{2}} + O((2 − x)^{\frac{3}{2}} )  (2.54)

As above, we have (2.35), so

∥Y (θ_{j+1})^{−1}Y (θ_{j})∥ ≤ 1 + \frac{|θ_{j+1} − θ_{j}|}{sin(θ_{j+1})}  (2.55)

but since θ_{j+1} < θ_{j}, we have

1 + \frac{|θ_{j+1} − θ_{j}|}{|θ_{j+1}|} = \frac{θ_{j+1} + (θ_{j} − θ_{j+1})}{θ_{j+1}} = \frac{θ_{j}}{θ_{j+1}}  (2.56)

and we find that, with T_{n} being the transfer matrix from 1 to n,

∥T_{n}∥ ≤ \frac{θ_{1}}{θ_{∞}} 2 \frac{2}{2 sin(θ_{1})} ≤ \frac{C}{θ_{∞}} ≤ C(2 − x)^{\frac{1}{2}} (1 + o(1))  (2.57)

This bound on the transfer matrix and Corollary 1.3 yield (2.1). □

Remark. It might be surprising that (2.1) has (x + 2), (x + 2)^{−1} rather than (x + 2)^{\frac{1}{2}} , (x + 2)^{− \frac{1}{2}} (because Carmona’s bound relates w(x) to ∥T_{n}∥^{2} and sup ∥T_{n}∥ goes like (2 − x)^{\frac{1}{2}} ). Even in the free case, bounds from Carmona’s formula give the wrong behavior: sin(nθ) + sin^{2}((n + 1)θ) have oscillations that cause the actual square root behavior in the free case, and bounds based only on ∥T_{n}∥ lose that.

That completes the proof of Theorem 2.1, the main result of this paper. Here is an alternate approach to controlling p_{n} for n > N, using the complex quantities:

Φ_{n} = p_{n} − e^{−iκ_{n}}p_{n−1}  (2.58)

so, since p_{j} is real,

sin(κ_{n})|p_{n−1}| = |Im(−Φ_{n})|
≤ |Φ_{n}|  (2.59)

By (2.24), we have

p_{n+1} = (e^{iκ_{n+1}} + e^{−iκ_{n+1}})p_{n} − p_{n−1}  (2.60)

so

Φ_{n+1} = e^{iκ_{n+1}}[p_{n} − e^{−iκ_{n+1}}p_{n−1}]
= e^{iκ_{n+1}}Φ_{n} + e^{iκ_{n+1}}(e^{−iκ_{n}} − e^{−iκ_{n+1}})p_{n−1}  (2.61)

Using (2.59),

|Φ_{n+1}| ≤ |Φ_{n}| + \frac{|κ_{n} − κ_{n+1}|}{sin(κ_{n})} |Φ_{n}|  (2.62)

and similarly,

|Φ_{n+1}| ≥ |Φ_{n}| − \frac{|κ_{n} − κ_{n+1}|}{sin(κ_{n})} |Φ_{n}|  (2.63)

These replace (2.36) and imply, via Lemma 2.7 and the analysis in (2.46), that

C_{1}(2 − x)^{− \frac{1}{2}} (b_{N+2} − b_{N+1}) ≤ \frac{|Φ_{n}|}{|Φ_{N+2}|} ≤ C(2 − x)^{\frac{1}{2}} (b_{N+2} − b_{N+1})^{−1}

Since

|Φ_{n}|^{2} ≤ |p_{n}|^{2} + |p_{n−1}|^{2}

and

2|Φ_{n}|^{2} ≥ sin^{2}(κ_{n+1})(|p_{n}|^{2} + |p_{n−1}|^{2})

we can go from this to Theorem 2.1.

3. Monotone a_{n}

In this section, we will consider

b_{n} ≡ 0 a_{n+1} ≤ a_{n} ≤ 1 a_{n} → 1  (3.1)

The weight will be symmetric, the measure purely absolutely continuous (i.e., no eigenvalues outside [−2, 2]), and so for non-Szegő weights, the integral will diverge at both ends. Here is the main result:

Theorem 3.1. Let dμ(x) = w(x) dx be the measure associated with Jacobi parameters obeying (3.1). For any x ∈ (−2, 2), define N(x) by

2a_{n} ≤ |x| for n ≤ N(x)  2a_{n} > |x| for n > N(x)  (3.2)

and γ_{n}(x) for n ≤ N(x) by

\frac{|x|}{a_{n}} = 2 cosh(γ_{n}(x))  (3.3)

Then

w(x) = e^{−2Q(x)}  (3.4)

where

|Q(x) − g(x)| ≤ h(x)

g(x) = ∑_{j=1}^{N(x)} γ_{j}(x)

and h(x) is given by

e^{h(x)} = CN(x)(a_{N(x)+2} − a_{N(x)+1})^{−1}  (3.5)

The proof will closely mimic the proof of Theorem 2.1, so we will only indicate the changes. By symmetry, without loss, we can suppose x > 0. The recursion relation becomes

p_{n+1}(x) = (e^{γ_{n+1}(x)} + e^{−γ_{n+1}(x)})p_{n}(x) − \frac{a_{n}}{a_{n+1}} p_{n−1}(x)  (3.6)

where we note, by (3.3), that

\frac{a_{n}}{a_{n+1}} = \frac{cosh(γ_{n+1}(x))}{cosh(γ_{n}(x))}  (3.7)

Define ψ_{n}(x) by (2.8), so (2.9) becomes

ψ_{n+1}(x) = (1 + e^{−2γ_{n+1}(x)})ψ_{n}(x) − \frac{a_{n}}{a_{n+1}} e^{−(γ_{n}(x)+γ_{n+1}(x))}ψ_{n−1}(x) (3.8)

(2.10) still holds.

Lemma 3.2. ψ_{n+1} ≥ ψ_{n}, so ψ_{n}(x) ≥ 1 for n ≥ 0.

Proof. We still have (2.13), and (2.14) becomes

ψ_{n+1} − ψ_{n} = e^{−2γ_{n+1}}(ψ_{n} − ψ_{n−1}) + e^{−γ_{n+1}}(e^{−γ_{n+1}} − \frac{a_{n}}{a_{n+1}} e^{−γ_{n}})ψ_{n−1}

(3.9)

Since a_{n} ≤ a_{n+1}, \frac{a_{n}}{a_{n+1}} < 1, and so

\frac{a_{n}}{a_{n+1}} e^{−γ_{n}} ≤ e^{−γ_{n}} ≤ e^{−γ_{n+1}}

14 Y. KREIMER, Y. LAST, AND B. SIMON

Thus, by (2.9), ψ_{n+1} − ψ_{n} ≥ 0 and ψ_{n+1} ≥ 0 inductively. □ Lemma 3.3.

e^{γ_{n+2}} ≤ e^{γ_{n+1}} \frac{cosh(γ_{n+2})}{cosh(γ_{n+1})}  (3.10)

Proof. This is equivalent to

e^{γ_{n+2}+γ_{n+1}} + e^{γ_{n+2}−γ_{n+1}} ≤ e^{γ_{n+2}+γ_{n+1}} + e^{γ_{n+1}−γ_{n+2}}  (3.11)

so to γ_{n+2} − γ_{n+1} ≤ 0, so to (2.13). □ Lemma 3.4. Define

W_{n} = e^{γ_{n+1}}ψ_{n} − \frac{a_{n}}{a_{n+1}} e^{−γ_{n}}ψ_{n−1}  (3.12)

Then

W_{n} ≤ e^{γ_{n+1}}  (3.13)

Proof. (3.13) holds for n = 0 by (3.12) for n = 0, so we can try an inductive proof. The analog of (2.17) is

W_{n+1} = e^{(γ_{n+2}−γ_{n+1})}W_{n} + e^{−γ_{n+1}}(e^{(γ_{n+2}−γ_{n+1})} − \frac{a_{n+1}}{a_{n+2}} )ψ_{n} (3.14)

By (3.7) and (3.10),

e^{(γ_{n+2}−γ_{n+1})} − \frac{a_{n+1}}{a_{n+2}} ≤ 0

so (3.14) says

W_{n+1} ≤ e^{(γ_{n+2}−γ_{n+1})}W_{n} ≤ e^{γ_{n+2}}

by induction. □ Lemma 3.5. ψ_{n+1} ≤ 1 + ψ_{n} so inductively, ψ_{n} ≤ n + 1. Proof. By (3.12) and (3.13),

ψ_{n+1} = e^{−γ_{n+2}}W_{n+1} + \frac{a_{n+1}}{a_{n+2}} e^{−γ_{n+2}−γ_{n+1}}ψ_{n}
≤ 1 + ψ_{n}

since \frac{a_{n+1}}{a_{n+2}} ≤ 1. □

If now

η_{n}(x) = p_{n−1}(x)^{2} + a_{n}^{2}p_{n}(x)^{2}  (3.15)

then we have proven (2.23) for large n.

To control the region n ≥ N(x) + 2, we use the scalar variable technique from the end of Section 2. Define κ_{n} for n ≥ N(x) + 1 by (recall x > 0)

\frac{x}{a_{n}} = 2 cos(κ_{n}(x))  (3.16)

so a_{n+1} ≥ a_{n} implies

κ_{n}(x) ≤ κ_{n+1}(x)  (3.17)

Define

Φ_{n} = p_{n} − e^{−iκ_{n}}p_{n−1}  (3.18)

Then

Lemma 3.6. (i)

|p_{n−1}| ≤ \frac{|Φ_{n}|}{sin(κ_{n})}  (3.19)

(ii)

\frac{|Φ_{n+1}|}{|Φ_{n}|} ≤ 1 + \frac{|e^{iκ_{n}} cos(κ_{n}) − e^{iκ_{n+1}} cos(κ_{n+1})|}{cos(κ_{n}) sin(κ_{n})}  (3.20)
≤ 1 + \frac{|κ_{n+1} − κ_{n}|}{\frac{1}{2} sin(2κ_{n})}  (3.21)

Proof. (i) This comes from |Im Φ_{n}| = sin(κ_{n})(p_{n−1}).

(ii) From

p_{n+1} = (e^{iκ_{n+1}} + e^{−iκ_{n+1}})p_{n} − \frac{a_{n}}{a_{n+1}} p_{n−1}

we obtain

|Φ_{n+1} − e^{iκ_{n+1}}Φ_{n}| = |e^{iκ_{n}} − \frac{a_{n}}{a_{n+1}} e^{iκ_{n+1}} | p_{n−1}  (3.22)

By (3.16),

\frac{a_{n}}{a_{n+1}} = \frac{cos(κ_{n+1})}{cos(κ_{n})}  (3.23)

so (3.22) and (3.19) imply (3.20). This in turn implies (3.21) since

e^{iκ_{n}} cos(κ_{n}) − e^{iκ_{n+1}} cos(κ_{n+1}) = \frac{1}{2} (e^{2iκ_{n}} − e^{2iκ_{n+1}})  (3.24)

□

With this formula, we can mimic the proof of Theorem 2.1 to complete the proof of Theorem 3.1.

4. Schrödinger Operators

In this section, we consider Schrödinger operators H = −\frac{d^{2}}{dx^{2}} + V (x) on L^{2}([0, ∞)) where one places u(0) = 0 boundary conditions. H is unitarily equivalent to multiplication by E on L^{2}(ℝ, dμ(E)), where dμ is the conventional spectral measure (see [3, 19, 23]). If u(x, E) obeys

−u′′ + V u = Eu u(0, E) = 0, u′(0, E) = 1  (4.1)

then Carmona’s formula [1] takes the form

\frac{π^{−1}dE}{(|u(x, E)|^{2} + |u′(x, E)|^{2})} ⟶^{w} dμ(E)  (4.2)

In particular, if uniformly in compact subsets of E ∈ (0, ∞),

exp(2(g(E) − h(E))) ≤ lim inf_{x→∞} (|u(x, E)|^{2} + |u′(x, E)|^{2})

≤ lim sup_{x→∞} (|u(x, E)|^{2} + |u′(x, E)|^{2})

≤ exp(2(g(E) + h(E)))  (4.3)

then dμ is purely absolutely continuous on (0, ∞), dμ(E) = e^{−2Q(E)} dE, and

|Q(E) − g(E)| ≤ h(E)  (4.4)

We want to assume the following conditions on V:

(a) V is C^{1} on [0, ∞).

(b) V is positive and strictly monotone decreasing on [0, ∞). Indeed,

V′(x) < 0  (4.5)

(c)

lim_{x→∞} V (x) = 0  (4.6)

Of course, the canonical example is

V (x) = (x + x_{0})^{−β}  (4.7)

Our main result in this section is:

Theorem 4.1. Let V obey (a), (b), (c) so dμ(E) = e^{−2Q(E)} dE. Define for E < V (0),

N(E) = V^{−1}(E)

so

\begin{aligned}V (x) & > E & if x < N(E) \\ V (x) & < E & if x > N(E) \end{aligned}  (4.8)

For x < N(E), define

γ(x, E) = (V (x) − E)^{\frac{1}{2}}  (4.9)

Then (4.4) holds where for E < V (0),

g(E) = ∫_{0}^{N(E)} γ(x, E) dx  (4.10)

and for E < V (0),

e^{h(E)} = CN(E) (V (N(E)) − V (N(E) + 1))^{−1}E^{\frac{1}{2}}  (4.11)

This proof will illuminate the proofs of the previous two sections. We begin with an analysis of the region x < N(E). We define

ψ(x) = u(x, E) exp(− ∫_{0}^{x} γ(y, E) dy)  (4.12)

and are heading towards

0 ≤ ψ′(x) ≤ 1  (4.13)

Lemma 4.2. For 0 < E < V (0) and x < N(E), we have

(a)  u′(x) ≥ 1  (4.14)
(b)  u(x) ≥ x  (4.15)

Proof. u′′ = γ^{2}u, so u′′ > 0. This implies u′(x) ≥ u′(0) = 1, and then u(x) = ∫_{0}^{x} u′(y) dy ≥ x. □

Lemma 4.3. For E < V (0) and x < N(E),

ψ′(x) ≥ 0  (4.16)

Proof. Let

f(x) = u′(x) − γ(x)u(x)  (4.17)

so

ψ′(x) = f(x) exp(− ∫_{0}^{x} γ(y, E) dy)  (4.18)

and (4.16) is equivalent to f ≥ 0. Note that

f′ + γf = u′′ − γu′ − γ′u + γu′ − γ^{2}u
= −γ′u  (4.19)

since (4.1) says

u′′ = γ^{2}u  (4.20)

(4.5) implies

γ′(y) ≤ 0  (4.21)

so (4.19) says

(f exp(∫_{0}^{x} γ(y) dy))′ ≥ 0  (4.22)

which, given f(0) = 1, implies f ≥ 0 and so ψ′ ≥ 0. □

Lemma 4.4. Let

W (x) = ψ′(x) + 2γ(x)ψ(x)  (4.23)

Then W′(x) ≤ 0 and so

ψ′(x) ≤ 1  (4.24)

Proof. By (4.18),

ψ′ + 2γ(x)ψ = (u′ + γ(x)u)e^{− ∫_{0}^{x} γ(y) dy}  (4.25)

so

W′(x) = (u′′ + γu′ + γ′u − γu′ − γ^{2}u)e^{− ∫_{0}^{x} γ(y) dy}
= γ′ue^{− ∫_{0}^{x} γ(y) dy}  (4.26)
≤ 0

by (4.21). But W (x = 0) = ψ′(0) = 1, so

W (x) ≤ 1  (4.27)

and thus

ψ′ = W − 2γψ ≤ 1  (4.28)
□

Proposition 4.5. If E is such that N(E) > 1, then

e^{−2V (0)}e^{2 ∫_{0}^{N(E)} γ(y) dy} ≤ u(N(E))^{2}+u′(N(E))^{2} ≤ (N(E)^{2}+1)e^{2 ∫_{0}^{N(E)} γ(y) dy} (4.29)

Proof. Since γ(N(E)) = 0,

ψ′(N(E)) = u′(N(E))e^{− ∫_{0}^{N(E)} γ(y) dy}

so 0 ≤ ψ′ ≤ 1 and ψ(0) = 0 yield the upper bound in (4.29).

For the lower bound, (4.15) implies u(1) ≥ 1. So, since γ(y) ≤ γ(0) ≤ V (0),

ψ(1) ≥ e^{−V (0)}  (4.30)

which, given that ψ′ > 0 and N(E) > 1, implies

u(N(E)) ≥ e^{−V (0)}e^{∫_{0}^{N(E)} γ(y) dy}  □

In the region [N(E), N(E) + 1], we note that since

∥(\begin{matrix}1 & V (x) − E \\ 1 & 0 \end{matrix})∥ ≤ 1 + |E| + |V (0)|

the matrix form of the Schrödinger equation implies that if C(x) = |u(x)|^{2} + |u′(x)|^{2}, then

e^{−2(1+|E|+V (0))|x−y|}C(y) ≤ C(x) ≤ e^{2(1+|E|+V (0))|x−y|}C(y)

giving a constant term in e^{h(E)} in (4.11).

Finally, in the region [N(E)+1, ∞), we use the method of Appendix 2 of Simon [30] (see also Hinton–Shaw [9]). Define for x > N(E),

κ(x, E) = \sqrt{E − V (x)}  (4.31)

MONOTONE JACOBI PARAMETERS AND NON-SZEGŐ WEIGHTS 19

and define

u_{±}(x) = exp(±i ∫_{N(E)}^{x} κ(y) dy)  (4.32)

If

F (x) = \frac{i}{2} V′(x)(E − V (x))^{− \frac{1}{2}}  (4.33)

and if a(x), b(x) are defined by

u(x) = a(x)u_{+}(x) + b(x)u_{−}(x)  (4.34)
u′(x) = a(x)u′_{+}(x) + b(x)u′_{−}(x)  (4.35)

then u′′ = −κ^{2}u is equivalent to (see Problem 98 on p. 395 of [28])

\binom{a(x)}{b(x)}′ = M(x)\binom{a(x)}{b(x)}  (4.36)

where

M(x) = w(x)^{−1} (\begin{matrix}−F (x) & u_{−}^{2}(x)F (x) \\ u_{+}^{2}(x)F (x) & −F (x) \end{matrix})  (4.37)

with

w(x) = u′_{+}(x)u_{−}(x) − u′_{−}(x)u_{+}(x)
= 2iκ(x)  (4.38)

Proposition 4.6. Let M(x) be given by (4.37). Then

∫_{N(E)+1}^{∞} ∥M(x)∥ dx ≤ log(\frac{κ(∞, E)}{κ(N(E) + 1, E)})  (4.39)

Proof. Since |u_{±}| = 1,

∥M(x)∥ ≤ |w(x)|^{−1}∥(\begin{matrix}|F (x)| & |F (x)| \\ |F (x)| & |F (x)| \end{matrix})∥
= 2|w(x)|^{−1}|F (x)|
= −\frac{1}{2} V′(x)(E − V (x))^{−1}
= \frac{d}{dx} log((E − V (x))^{\frac{1}{2}} )  (4.40)

from which (4.39) follows. □ Proof of Theorem 4.1. Let

Y (x) = (\begin{matrix}u_{+}(x) & u_{−}(x) \\ u′_{+}(x) & u′_{−}(x) \end{matrix})  (4.41)

Let T (x, y) be the \binom{u}{u′} transfer matrix from x to y and T̃(x, y) be the \binom{a}{b} transfer matrix. For y > N(E) + 1, we have just seen

∥T̃(N(E) + 1, y)∥ ≤ exp(∫_{N(E)+1}^{∞} ∥M(x)∥ dx)
= \frac{κ(∞, E)}{κ(N(E) + 1, E)}  (4.42)

On the other hand,

∥Y (y)∥ ≤ 1 + κ ≤ 2  (4.43)

for κ small while

∥Y (y)^{−1}∥ = |det(Y)^{−1}| ∥Y ∥ ≤ κ(y)^{−1}  (4.44)

and

T (x, y) = Y (y)T̃(x, y)Y (x)^{−1}

so

∥T (N(E) + 1, y)∥ ≤ \frac{2κ(∞, E)}{κ(N(E) + 1, E)^{2}}  (4.45)

Since E = V (N(E)),

κ(N(E) + 1, E)^{2} = V (N(E)) − V (N(E) + 1)  (4.46)

and we have the bound (4.4) with the error built from e^{−V (0)}, N(E), (4.39), and (4.45). □

It is interesting that the differential equation methods of this section lead to terms that are identical to what we found in the discrete case.

5. Examples

We start with the continuum case.

Example 5.1.

V (x) = C_{0}x^{−β} β < 2 (5.1)

Technically this does not fit into Theorem 4.1 since V (0) = ∞, but when β < 2, it is easy to extend the analysis. The spectral measure is e^{−2Q(E)} dE where (4.4) holds.

N(E) = (\frac{E}{C_{0}} )^{− \frac{1}{β}}  (5.2)

V (N(E)) − V (N(E) + 1) ∼ V′(N(E))
∼ N(E)^{−1}V (N(E))
= EN(E)^{−1}  (5.3)

so h(E) = O(log(N(E)^{2}E^{− \frac{1}{2}} )) = O(log(E)). On the other hand, letting y = x/N(E),

g(E) = ∫_{0}^{N(E)}(V (x) − E)^{\frac{1}{2}} dx  (5.4)
= N(E)E^{\frac{1}{2}} ∫_{0}^{1}(y^{−β} − 1)^{\frac{1}{2}} dy  (5.5)
= E^{\frac{1}{2}} N(E)β^{−1} ∫_{0}^{1}(1 − u)^{\frac{1}{2}} u^{\frac{1}{β} − \frac{3}{2}} du

= E^{\frac{1}{2}} N(E)β^{−1} \frac{Γ(\frac{3}{2})Γ(\frac{1}{β} − \frac{1}{2})}{Γ(\frac{1}{β} + 1)}  (5.6)

using a u = y^{β} change of variables. Thus,

g(E) = c_{1}C0^{\frac{1}{β}} E^{\frac{1}{2} − \frac{1}{β}}  c_{1} = β^{−1} \frac{Γ(\frac{3}{2})Γ(\frac{1}{β} − \frac{1}{2})}{Γ(\frac{1}{β} + 1)}  (5.7)

Since β < 2, g(E) → ∞ and is much larger than the log(E) error. β = 1, the Coulomb case, has g(E) = C_{0}c_{1}E^{− \frac{1}{2}} and β = \frac{1}{2}, the quasi-Szegő borderline, has g(E) = C_{0}^{2}c_{1}E^{− \frac{3}{2}} . We emphasize that g occurs in an exponential, so w is very small near E = 0. □

Example 5.2.

V (x) = C_{0}(x + x_{0})^{−β}  β < 2  (5.8)

We claim that the changes from Example 5.1 are small compared to log(E) errors in h; explicitly,

g(E) = c_{1}C0^{\frac{1}{β}} E^{\frac{1}{2} − \frac{1}{β}} + O(1) + O(E^{\frac{1}{2}} )  (5.9)

For in this case,

N(E) = (\frac{E}{C_{0}} )^{− \frac{1}{β}} − x_{0}  (5.10)

and one changes variables to y = (x+x_{0})/(N(E)+x_{0}), so (5.5) becomes

g(E) = N(E)E^{\frac{1}{2}} ∫_{s(E)}^{1}(y^{−β} − 1)^{\frac{1}{2}} dy  (5.11)

where

s(E) = y(x = 0) = \frac{x_{0}}{N(E) + x_{0}}  (5.12)

Then

N(E)E^{\frac{1}{2}} ∫_{0}^{s(E)}(y^{−β} − 1)^{\frac{1}{2}} = N(E)E^{\frac{1}{2}} O(s(E)^{1− \frac{β}{2}} )

22 Y. KREIMER, Y. LAST, AND B. SIMON

= O(1)  (5.13)

by (5.10) and (5.12), so

g(E) = c_{1}N(E)E^{\frac{1}{2}} + O(1)
= c_{1}C0^{\frac{1}{β}} E^{\frac{1}{2} − \frac{1}{β}} + O(1) + O(E^{\frac{1}{2}} )  (5.14)

as claimed. □

Now we turn to the discrete case. Example 5.3 (= (1.8)).

a_{n} ≡ 1 b_{n} = −Cn^{−β}  (5.15)

Define

δ = 2 − x δ_{n} = Cn^{−β} − δ  (5.16)

so

x − b_{n} = 2 + δ_{n}  (5.17)

We have (with [y] = maximal integer ≤ y)

N(x) = [(C^{−1}δ)^{− \frac{1}{β}}]  (5.18)

We have b_{N+2} − b_{N+1} = O(N^{−β−1}), so the RHS of (2.5) is of order CN(x)^{β+2}δ^{\frac{1}{2}} = O(δ^{− \frac{1}{2} − \frac{2}{β}}) and thus, h(x) = O(log(2 − x)) and we need to compute g(x) = ∑_{j=1}^{N(x)} γ_{j}(x) up to O(log δ) terms.

We will suppose below that C ≤ 1 and explain at the end what to change if C > 1.

Define c_{ℓ} to be the Taylor coefficients in

cosh^{−1}(1 + \frac{z}{2}) = \sqrt{z} ∑_{ℓ=0}^{∞} c_{ℓ}z^{ℓ}  (5.19)

so, courtesy of Mathematica,

c_{0} = 1 c_{1} = −\frac{1}{24} c_{2} = \frac{3}{640} c_{3} = −\frac{5}{7168}

and, for example,

c_{20} = 34, 461, 632, 205/12, 391, 489, 651, 049, 749, 040, 738, 304 (assuming that we managed to copy it without a typo). Thus,

g(x) = ∑_{ℓ=0}^{∞} c_{ℓ} ∑_{j=1}^{N(x)} δ_{j}^{ℓ+ \frac{1}{2}}  (5.20)

Notice that since δ > 0,

δ_{j} ≤ Cj^{−β}  (5.21)

so, if β(ℓ + \frac{1}{2}) > 1, a crude δ-independent bound of ∑_{j=1}^{N(x)} δ_{j}^{ℓ+ \frac{1}{2}} can be summed independently of N(x). Moreover, if F is the function in (5.19), then

2\sqrt{z} \frac{dF}{dz} = \frac{1}{\sqrt{1 + \frac{z}{4}}}  (5.22)

so the c_{ℓ} power series has radius of convergence 4 and so ∑|c_{ℓ}| < ∞. Thus, if

ℓ_{0} = [\frac{1}{β} − \frac{1}{2}] + 1  (5.23)

then

∑_{ℓ=ℓ_{0}}^{∞} |c_{ℓ}| ∑_{j=1}^{N(x)} δ_{j}^{ℓ+ \frac{1}{2}} ≤ ( ∑_{0}^{∞} |c_{ℓ}|) ∑_{j=1}^{∞} j^{−β(ℓ_{0}+1)}  (5.24)

(since C ≤ 1) so

∑_{j=1}^{N} γ_{j} = ∑_{0≤ℓ≤ \frac{1}{β} − \frac{1}{2}} c_{ℓ} ∑_{j=1}^{N} δ_{j}^{ℓ+ \frac{1}{2}} + O(1)  (5.25)

If ℓ = \frac{1}{β} − \frac{1}{2} occurs, then

∑_{j=1}^{N} δ_{j}^{\frac{1}{β} − \frac{1}{2} + \frac{1}{2}} = ∑_{j=1}^{N} δj^{\frac{1}{β}}

= ∑_{j=1}^{N}(C \frac{1}{j^{β}} − δ)^{\frac{1}{β}}

≤ C^{\frac{1}{β}} ∑_{j=1}^{N} j^{−1}

= O(log N)  (5.26)

On the other hand, if ℓ < \frac{1}{β} − \frac{1}{2}, then

∑_{j=1}^{N} δ_{j}^{ℓ+ \frac{1}{2}} = ∑_{j=1}^{N}(Cj^{−β} − δ)^{ℓ+ \frac{1}{2}}

= C^{ℓ+ \frac{1}{2}} ∑_{j=1}^{N}(\frac{1}{j^{β}} − \frac{1}{N^{β}})^{ℓ+ \frac{1}{2}} + O(1)

= C^{ℓ+ \frac{1}{2}} ∑_{j=1}^{N} j^{−β(ℓ+ \frac{1}{2} )}(1 − (\frac{j}{N})^{β})^{ℓ+ \frac{1}{2}} + O(1)

= C^{ℓ+ \frac{1}{2}} ∫_{1}^{N} x^{−β(ℓ+ \frac{1}{2} )}(1 − (\frac{x}{N})^{β})^{ℓ+ \frac{1}{2}} + O(1)  (5.27)

= C^{ℓ+ \frac{1}{2}} β^{−1}N^{1−(ℓ+ \frac{1}{2} )β} ∫_{N^{−β}}^{1} u^{( \frac{1}{β} −ℓ− \frac{3}{2} )}(1 − u)^{ℓ+ \frac{1}{2}} du + O(1) (5.28)

= C^{ℓ+ \frac{1}{2}} β^{−1}N^{1−(ℓ+ \frac{1}{2} )β} ∫_{0}^{1} u^{( \frac{1}{β} −ℓ− \frac{3}{2} )}(1 − u)^{ℓ+ \frac{1}{2}} du + O(1) (5.29)

= C^{ℓ+ \frac{1}{2}} β^{−1} \frac{Γ(ℓ + \frac{3}{2})Γ(\frac{1}{β} − \frac{1}{2} − ℓ)}{Γ(\frac{1}{β} + 1)} N^{1−(ℓ+ \frac{1}{2} )β} + O(1)

In the above, (5.27) comes from the fact that the function in the integrand is monotone decreasing, and if f(x) is monotone, then

f(j) ≥ ∫_{j}^{j+1} f(y) dy ≥ f(j + 1)

so

∑_{j=1}^{N−1} f(j) ≥ ∫_{1}^{N} f(y) dy ≥ ∑_{j=2}^{N} f(j)

and

|∫_{1}^{N} f(y) dy − ∑_{j=1}^{N} f(j)| ≤ f(1)  (5.30)

(5.28) is the change of variables u = (\frac{x}{N})^{β}. Finally, (5.29) comes from the same cancellation that occurred in (5.13).

Since |N − C^{\frac{1}{β}} δ^{− \frac{1}{β}} | ≤ 1 and 0 < 1 − (ℓ + \frac{1}{2})β < 1,

N^{1−(ℓ+ \frac{1}{2} )β} = (C^{\frac{1}{β}} δ^{− \frac{1}{β}})^{1−(ℓ+ \frac{1}{2} )β} + o(1)  (5.31)

Thus, we find

Q(x) = β^{−1}C^{\frac{1}{β}} ∑_{0≤ℓ<( \frac{1}{β} − \frac{1}{2} )} c_{ℓ} \frac{Γ(ℓ + \frac{3}{2})Γ(\frac{1}{β} − \frac{1}{2} − ℓ)}{Γ(\frac{1}{β} + 1)} δ^{− \frac{1}{β} +ℓ+ \frac{1}{2}} + O(log δ)

(5.32)

If C > 1, we should not expand the power series of cosh^{−1} for small j (actually, as noted, the power series has radius of convergence 4 so we need only worry if C ≥ 4). Instead, we do not expand for those j with Cj^{−β} > 1. That is only finitely many terms, so it adds O(1) errors to ∑_{1}^{N} γ_{j}(x). We add back these small j terms to (5.25), again making O(1) errors. The final result does not change. □

Finally, we will explore examples that lead to Q’s roughly of the type (1.11) to link to work of Levin–Lubinsky [18]. We suppose

a_{n} = 1 − f(log(n + 1))  (5.33)

where the f’s we have in mind are typically

f(x) = (1 + x)^{−α}  (5.34)

or

f(x) = log_{k}(x + c_{k})  (5.35)

an iterated log (where c_{k} is chosen to keep all log’s that enter positive). We will need

Proposition 5.4. Let f be defined and C^{2} on [log 2, ∞) and obey

(i)  f(x) > 0, f′(x) < 0, f′′(x) > 0  (5.36)
(ii)  lim_{n→∞} f(n) = 0  (5.37)
(iii)  lim_{N→∞} N^{ε}(−f′(log N))^{\frac{1}{2}} = ∞  (5.38)
(iv)  lim_{ε↓0} (lim sup_{k→∞} \frac{−f′((1 − ε)k)}{−f′(k)}) = 1  (5.39)

Let

S_{N} = ∑_{j=2}^{N} \sqrt{f(log j) − f(log N)}  (5.40)

Then

lim_{N→∞} \frac{S_{N}}{N(−f′(log N))^{\frac{1}{2}}} = \frac{\sqrt{π}}{2}  (5.41)

Remark. It is easy to see that if f(x) = e^{−kx} (i.e., f(log(n + 1)) ∼ (n + 1)^{−k}), then (5.41) fails. In this case, both (5.38) and (5.39) fail, but they hold for the f’s of (5.34) and (5.35).

Proof. Since (−f′)′ < 0 and if x < y,

f(x) − f(y) = ∫_{x}^{y}(−f′(s)) ds  (5.42)

we have,

(y − x)(−f′(y)) ≤ f(x) − f(y) ≤ (y − x)(−f′(x))  (5.43)

We thus get a lower bound

f(log j) − f(log N) ≥ (−f′(log N))(− log(\frac{j}{N}))  (5.44)

so

S_{N} ≥ N(−f′(log N))^{\frac{1}{2}} ∑_{j=2}^{N} \frac{1}{N} (− log(\frac{j}{N}))^{\frac{1}{2}}  (5.45)

As N → ∞, the sum converges to ∫_{0}^{1}(− log(x))^{\frac{1}{2}} dx = \frac{\sqrt{π}}{2} (courtesy of Mathematica). Thus,

lim inf(LHS of (5.41)) ≥ \frac{\sqrt{π}}{2}  (5.46)

For the upper bound, fix ε > 0 and break S_{N} = S_{N}^{(1)} + S_{N}^{(2)} where S_{N}^{(1)} has j ≤ N^{1−ε} and S_{N}^{(2)} has j > N^{1−ε}. Clearly,

S_{N}^{(1)} ≤ f(log 2)N^{1−ε}  (5.47)

so, by hypothesis (5.38), it contributes 0 to the ratio in (5.41) as N → ∞.

For S_{N}^{(2)}, we use the upper bound when j > N^{1−ε}

f(log j) − f(log N) ≤ −f′((1 − ε) log N)(− log(\frac{j}{N}))

which yields (since the Riemann sum still converges to the integral)

lim sup(LHS of (5.41)) ≤ \frac{\sqrt{π}}{2} lim sup_{k→∞} (\frac{−f′((1 − ε)k)}{−f′(k)})^{\frac{1}{2}}

Since ε is arbitrary, we can use (5.39) to complete the proof of (5.41). □

Example 5.5. Let a_{n} have the form (5.31) where f obeys all the hypotheses of Proposition 5.4. By (3.2) and (3.3), N(x) roughly solves

\frac{x}{1 − f(log(N + 1))} = 2  (5.48)

namely,

N(x) = [exp(f^{−1}(1 − \frac{x}{2}))] − 1  (5.49)

For example, if f is (5.34), then

N(x) = [exp((1 − \frac{x}{2})^{−α} − 1)] − 1  (5.50)

Next, define z by \frac{x}{2a} = 1 + \frac{z}{2}, namely,

z = \frac{x}{a} − 2  (5.51)

where \frac{x}{a} > 2. Writing x = 2 − δ and a = 1 − f, we see

z = −δ + 2f + O(f^{2}) + O(fδ)  (5.52)

Taking into account that N(x) is such that

2f(log(N + 2)) ≤ δ ≤ 2f(log(N + 1))

and that (5.19) says

cosh^{−1}(\frac{x}{2a}) = \sqrt{z} + O(z^{\frac{3}{2}} )

we see that

γ_{j}(x) = \sqrt{2f(log(j + 1)) − δ} + O(f^{\frac{3}{2}} ) + O(f^{\frac{1}{2}} δ)

and thus

g(x) = ∑_{j=1}^{N(x)} γ_{j}(x)

is asymptotically the same as \sqrt{2} S_{N}. Thus,

|Q(x) − g(x)| ≤ h(x)  (5.53)

where

g(x) = \sqrt{\frac{π}{2}} N(x)(−f′(log N(x)))^{\frac{1}{2}} (1 + o(1))  (5.54)

and

h(x) = O(log N(x)) + O(log(1 − \frac{2}{x}))

N(x) is huge, so while log N(x) ∼ (1 − \frac{x}{2})^{−α} in case (5.34), it is still small relative to g(x).

The reader may be puzzled in comparing our results with those of Levin–Lubinsky [18]. They have no \sqrt{\frac{π}{2}} and their relations (after making the modifications from [−1, 1] to [−2, 2]) suggest

1 − a_{n} = (log n)^{− \frac{1}{2}} (1 + o(1))  (5.55)

should correspond to

Q(x) = exp((1 − \frac{x}{2})^{−α})  (5.56)

so there is no sign of (−f′(log N(x)))^{\frac{1}{2}} either.

The mystery is solved by the fact that multiple Q’s lead to the same leading asymptotics for a_{n}. In their scheme, after corrections to move to [−2, 2], leading asymptotics for f are given by

n = Q(1 − 2(f(n)(1 + o(1))))  (5.57)

If

Q(x) = e^{1/(1− \frac{x}{2} )}  (5.58)

then

n = exp((f(n))^{−1})  (5.59)

solved by

f(n) = \frac{1}{log n} (1 + o(1))  (5.60)

Changing (5.58) to

Q(x) = \frac{π}{2}(1 − \frac{x}{2}) exp((1 − \frac{x}{2})^{−1})

is solved by

f(n) = 1/(log(\frac{2n}{π} log n) + O(log log n))

Since

log \frac{2n}{π} log n = log n + log_{2} n + log(\frac{2}{π})

(5.60) still holds! □

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