## 1 Introduction

In this note, we prove a conjecture of Hundertmark and Simon [2] concerning a sharp Lieb–Thirring inequality for Jacobi matrices. We denote the symmetric Jacobi matrix with diagonal entries $$\{ b_n\}_{n=-\infty }^\infty$$ and off-diagonal entries $$\{a_n\}_{n=-\infty }^\infty$$ by

\begin{aligned} J:= W(\{a_n\},\{b_n\})\ . \end{aligned}

It is assumed that the $$a_n$$ tend to 1 as $$n \rightarrow \pm \infty$$ which yields the interval $$[-2,2]$$ as the essential spectrum of this Jacobi matrix. We denote by $$E_j^+(J)$$ the eigenvalues of J that are larger than 2 and by $$E_j^-(J)$$ the eigenvalues of J that are less than $$-2$$. Hundertmark and Simon proved that

\begin{aligned} \sum _j (E_j^+(J)^2-4)^{1/2}+(E_j^-(J)^2-4)^{1/2} \le \sum _n |b_n| + 4\sum _n |a_n-1| \ , \end{aligned}
(1.1)

and observed that this inequality is sharp. Indeed, in the absence of the potential, they noted that the Jacobi matrix with all entries $$a_n=1$$ except for a single one that is chosen to be larger than one yields asymptotic equality in (1.1). They then conjectured the improved version of (1.1) in which $$|a_n-1|$$ is replaced by $$(a_n-1)_+$$, where we use the notation $$(a)_+$$ to mean a if $$a>0$$ and 0 if $$a \le 0$$. We have the following theorem.

### Theorem 1.1

Assume that $$\sum _n |b_n| < \infty$$, $$\sum _n (a_n-1)_+ < \infty$$ and $$\lim _{n \rightarrow \pm \infty } a_n = 1$$. Then we have the bound on the eigenvalue sum

\begin{aligned} \sum _j (E_j^+(J)^2-4)^{1/2}+(E_j^-(J)^2-4)^{1/2} \le \sum _n |b_n| + 4\sum _n (a_n-1)_+ \ . \end{aligned}
(1.2)

The following consequence provides another justification for this short note. Generally, the proof of Lieb–Thirring inequalities is patterned after the ones for the continuum in which case the kinetic energy is given by $$-\Delta$$. The discrete Laplacian requires that all $$a_n=1$$. It is, however, of some interest that in the case of Jacobi matrices this needs not be the case. This is a distinguishing feature of Lieb–Thirring inequalities for Jacobi matrices. Further, it was shown in [5], (see also [7]) that

\begin{aligned} \sum _j F(E_j^+(J)) + F(E_j^-(J)) \le \sum _n b_n^2 + 2 G(a_n)^2 , \end{aligned}
(1.3)

where $$G(a) = a^2-1-\log |a|^2$$ and $$F(E) = \beta ^2 -\beta ^{-2} - \log |\beta |^2$$ and $$E=\beta +\beta ^{-1}$$ with $$|\beta |>1$$.

In [6], one of us proved that if $$a_n\equiv 1$$ then the inequality (1.1) implies (1.3). Using this argument and Theorem 1.1, we obtain as a consequence

### Theorem 1.2

Let $$\gamma >1/2$$. Assume that $$\sum _n b_n^{\gamma +1/2} < \infty$$ and that $$a_n\ge 0$$ for all $$n \in {\mathord {{\mathbb {Z}}}}$$ with $$\lim _{n \rightarrow \pm \infty } a_n = 1$$ and $$\sum _n (a_n-1)_+^{\gamma +1/2}<\infty$$. Then with $$\mathrm {B}(x,y)$$ denoting the Beta function

\begin{aligned}&\sum _j \int _{2}^{|E_j^\pm (J)|}(t^2-4)^\frac{1}{2}(|E_j^\pm (J)|-t)^{\gamma -\frac{3}{2}} \,\mathrm{d}t\nonumber \\&\quad \le \mathrm {B}(\gamma -1/2,2) \sum _n (\pm [b_n]_\pm \pm [a_n-1]_+\pm [a_{n-1}-1]_+)_\pm ^{\gamma +\frac{1}{2}} \ . \end{aligned}
(1.4)

For $$\gamma =3/2$$, the left-hand side coincides with $$\frac{1}{2}\sum _{j}F(E_j^\pm (J))$$. The function $$G(a) \ge 0$$ equals zero if and only if $$a= \pm 1$$ and hence (1.4) is an improvement over (1.3) for the case where $$0 \le a_n \le 1$$.

### Remark

As proved in [6], the left-hand side in (1.4) is bounded from below by

\begin{aligned} \sum _j \int _2^{|E_j^\pm |}(t^2 - 4)^{\frac{1}{2}}(|E_j^\pm |-t)^{\gamma -\frac{3}{2}}\,\mathrm{d}t \ge 2\mathrm {B}(\gamma -1/2,3/2)\sum _{j}(|E_j^\pm |-2)^{\gamma } \end{aligned}

and by

\begin{aligned} \sum _j \int _2^{|E_j^\pm |}(t^2 - 4)^{\frac{1}{2}}(|E_j^\pm |-t)^{\gamma -\frac{3}{2}}\,\mathrm{d}t \ge \mathrm {B}(\gamma -1/2,2)\sum _{j}(|E_j^\pm |-2)^{\gamma +\frac{1}{2}}. \end{aligned}

Thus, (1.4) improves on the corresponding Lieb–Thirring inequalities in [2]. Note that in [2, p.121] an argument is given that allows to replace $$(a_n-1)$$ in their results by $$(a_n-1)_+$$ for $$\gamma \ge 1$$ but importantly not for $$1/2<\gamma <1$$ and not in the case of the main result (1.2).

In order to prove Theorem 1.1, we reduce the problem (as in [2]) to the discrete Schrödinger operator. When treating terms $$0 \le a_n \le 1$$, we use some additional convexity property.

## 2 The Proof of the Main Result

We follow Hundertmark and Simon except for one key step. Using norm resolvent convergence, we may assume that only finitely many of the $$a_n$$ are not equal to one and finitely many $$b_n$$ are not equal to zero. Likewise, we may assume that $$b_n \ge 0$$. We also write

\begin{aligned} W(\{a_n\},\{b_n\}) = A+B \end{aligned}

with the understanding that A contains only the off-diagonal terms and B the diagonal terms of the Jacobi matrix. If the off-diagonal terms are all equal to one, we denote the corresponding matrix by $$A_1$$. The essential spectrum is given by the interval $$[-2,2]$$ which follows from Weyl’s theorem. Let us denote the eigenvalues that are strictly greater than 2 by $$E^+_1(A+B) \ge E^+_2(A+B) \ge E^+_3(A+B) \ge \cdots$$. In what follows, the eigenvalues that are strictly less than $$-2$$ can be treated in a similar fashion.

Treating $$a_n>1$$: Consider the window of the matrix A that contains an off-diagonal term $$a>1$$ and use the elementary inequalities

\begin{aligned} \begin{pmatrix} -a+1&{}\quad 1\\ 1 &{}\quad -a+1 \end{pmatrix} \le \begin{pmatrix} 0 &{}\quad a\\ a &{}\quad 0 \end{pmatrix} \le \begin{pmatrix} a-1&{}\quad 1\\ 1 &{}\quad a-1 \end{pmatrix}. \end{aligned}

Applying it to all $$a_n>1$$, we obtain

\begin{aligned} {\widetilde{J}}^- = W(\{{\widetilde{a}}_n\}, \{{\widetilde{b}}^-_n\} \le J = W(\{a_n\}, \{b_n\}) \le W(\{{\widetilde{a}}_n\}, \{{\widetilde{b}}^+_n\} = {\widetilde{J}}^+, \end{aligned}

where

\begin{aligned} {\widetilde{a}}_n = {\left\{ \begin{array}{ll} a_n, &{}{} \mathrm {if} \,\,a_n\le 1\\ 1, &{}{} \mathrm {if} \,\,a_n >1 \end{array}\right. } \qquad {\widetilde{b}}_n^\pm = \pm [b_n]_\pm \pm [(a_{n-1}-1)_++ (a_n-1)_+] \ \mathrm {for \ all } \ n, \ \end{aligned}

where $$[x]_\pm = \max \{\pm x,0\}$$. This implies

\begin{aligned} \sum _j (E_j^+(J)^2-4)^{1/2} \le \sum _j (E_j^{+}(\widetilde{J}^+)^2-4)^{1/2}\ \end{aligned}
(2.1)

and similarly

\begin{aligned} \sum _j (E_j^{-}(J)^2-4)^{1/2} \le \sum _j (E_j^{-}(\widetilde{J}^-)^2-4)^{1/2}\ . \end{aligned}

This reduces the problem to the case $$a_n\le 1$$, $$n\in {\mathbb {Z}}$$.

Treating $$a_n\le 1$$: Assuming $$a_n\le 1$$, we consider the Birman–Schwinger operator

\begin{aligned} K(A;\beta ) := B^{1/2} (\beta - A)^{-1} B^{1/2}, \end{aligned}

where $$\beta >2$$ and list the eigenvalues of the Birman–Schwinger operator, $$E_j(B^{1/2} (\beta - A)^{-1} B^{1/2})$$, in decreasing order. The Birman–Schwinger principle states that the j-th eigenvalue of $$B^{1/2} (E^+_j(A+B) - A)^{-1} B^{1/2}$$ is one.

Let us decompose the matrix A in a certain fashion. Consider the following window of the general matrix $$A_\kappa$$,

\begin{aligned} \begin{pmatrix} 0 &{}\quad a &{}\quad 0 &{}\quad 0 &{}\quad 0\\ a &{}\quad 0 &{}\quad \kappa &{}\quad 0 &{}\quad 0\\ 0 &{}\quad \kappa &{}\quad 0 &{}\quad c &{}\quad 0\\ 0 &{}\quad 0 &{}\quad c &{}\quad 0 &{}\quad d\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad d &{}\quad 0 \end{pmatrix}. \end{aligned}

The distinct notation $$\kappa$$ indicates that we concentrate on this particular position of the matrix. Denote by U the infinite diagonal matrix that consists of $$+1$$ on the diagonal above the position of $$\kappa$$ and of $$-1$$ below $$\kappa$$, i.e., its window is given by

\begin{aligned} \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad -1 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad -1 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad -1 \end{pmatrix}\ . \end{aligned}

It has the effect that the corresponding window of the matrix $$U \,A_\kappa \,U$$ is given by

\begin{aligned} \begin{pmatrix} 0 &{}\quad a &{}\quad 0 &{}\quad 0 &{}\quad 0\\ a &{}\quad 0 &{}\quad -\kappa &{}\quad 0 &{}\quad 0\\ 0 &{}\quad -\kappa &{}\quad 0 &{}\quad c &{}\quad 0\\ 0 &{}\quad 0 &{}\quad c &{}\quad 0 &{}\quad d\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad d &{}\quad 0 \end{pmatrix} \, , \end{aligned}

i.e., the entry $$\kappa$$ changes sign and all others are unchanged. Since U is unitary, the matrices A and $$UA\,U$$ are unitarily equivalent and have the same spectrum. With a slight abuse of notation, we now identify the matrices with their window. If we assume that $$0\le \kappa <1$$, we may write

\begin{aligned} A_\kappa = \frac{\kappa +1}{2} \, \begin{pmatrix} 0 &{}\quad a &{}\quad 0 &{}\quad 0 &{}\quad 0\\ a &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 1 &{}\quad 0 &{}\quad c &{}\quad 0\\ 0 &{}\quad 0 &{}\quad c &{}\quad 0 &{}\quad d\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad d &{}\quad 0 \end{pmatrix} + \frac{1-\kappa }{2} \begin{pmatrix} 0 &{}\quad a &{}\quad 0 &{}\quad 0 &{}\quad 0\\ a &{}\quad 0 &{}\quad -1 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad -1 &{}\quad 0 &{}\quad c &{}\quad 0\\ 0 &{}\quad 0 &{}\quad c &{}\quad 0 &{}\quad d\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad d &{}\quad 0 \end{pmatrix} \, , \end{aligned}

or if we denote by $$A'$$ the first matrix displayed above we can write

\begin{aligned} A_\kappa =\frac{1+\kappa }{2} A' + \frac{1-\kappa }{2} UA'U \ . \end{aligned}

Repeating this for all the off-diagonal elements that are strictly less than one we find

\begin{aligned} A = \sum _j \lambda _j U(j) \, A_1 \, U(j), \end{aligned}
(2.2)

where off-diagonal elements of $$A_1$$ are equal one and where $$\lambda _j \ge 0, \, \sum _j \lambda _j = 1$$. Since the matrices U(j) are diagonal and have the matrix elements $$\pm 1$$, the matrices $$A_1$$ and $$U(j) A_1 U(j)$$ have the same eigenvalues.

The key observation is the following lemma.

### Lemma 2.1

Let $$\beta I > X$$. Then the function

\begin{aligned} X \rightarrow (\beta I - X)^{-1} \end{aligned}

is operator convex, i.e., if $$0\le \lambda \le 1$$ then

\begin{aligned} (\beta -\lambda X_1 - (1-\lambda ) X_2)^{-1} \le \lambda (\beta - X_1)^{-1}+ (1 -\lambda ) (\beta - X_2)^{-1} \ . \end{aligned}

### Proof

We follow [1]. Let $$Y_j = \beta - X_j$$, $$j=1,2$$. It amounts to showing that for two positive and invertible self-adjoint operators $$Y_1$$ and $$Y_2$$, we have

\begin{aligned} (\lambda Y_1 +(1-\lambda )Y_2)^{-1} \le \lambda Y_1^{-1} + (1-\lambda ) Y_2^{-1} \ . \end{aligned}

This is equivalent to

\begin{aligned} \left[ Y_2^{1/2} \left( \lambda Y_2^{-1/2} Y_1 Y_2^{-1/2} +(1-\lambda )I\right) Y_2^{1/2}\right] ^{-1} \\ \qquad \le Y_2^{-1/2} \left[ \lambda Y_2^{1/2} Y_1^{-1} Y_2^{1/2} +(1-\lambda )I\right] Y_2^{-1/2} \end{aligned}

or

\begin{aligned} Y_2^{-1/2} \left( \lambda Y_2^{-1/2} Y_1 Y_2^{-1/2} +(1-\lambda )I\right) ^{-1} Y_2^{-1/2}\\ \qquad \le Y_2^{-1/2} \left[ \lambda Y_2^{1/2} Y_1^{-1} Y_2^{1/2} +(1-\lambda )I\right] Y_2^{-1/2} \end{aligned}

which is equivalent to

\begin{aligned} \left( \lambda Y_2^{-1/2} Y_1Y_2^{-1/2} +(1-\lambda )I\right) ^{-1}\le \lambda Y_2^{1/2} Y_1^{-1} Y_2^{1/2} +(1-\lambda ) I \ . \end{aligned}

This is an inequality in terms of the positive, invertible and self-adjoint operator $$Y= Y_2^{-1/2} Y_1Y_2^{-1/2}$$, i.e.,

\begin{aligned} (\lambda Y+(1-\lambda )I)^{-1} \le \lambda Y^{-1} +(1-\lambda )I, \end{aligned}

which reduces the whole problem to positive numbers on account of the spectral theorem. For positive numbers, the inequality is obvious. $$\square$$

Applying now Lemma 2.1 to (2.2), we find

\begin{aligned} K(A;\beta ) \le \sum _j \lambda _j U(j) K(A_1; \beta ) U(j) \ . \end{aligned}

If we set $$\beta = \mu + \frac{1}{\mu }$$ and introduce the operator,

\begin{aligned} L_\mu (A) := (\beta ^2-4)^{1/2} K(A;\beta ) \end{aligned}

we find

\begin{aligned} L_\mu (A) \le \sum _j \lambda _j U(j) L_\mu (A_1) U(j) \ . \end{aligned}

The operator $$L_\mu (A_1)$$ has the matrix representation

\begin{aligned}{}[L_\mu (A_1)]_{m,n} = b_m^{1/2} \mu ^{|n-m|} b_n^{1/2} \end{aligned}

where once more

\begin{aligned} \beta = \mu + \frac{1}{\mu } \ , \mu < 1 \ . \end{aligned}

Denote by $$S_n(\mu )$$ the sum of the n largest eigenvalues of $$\sum _j \lambda _j U(j) L_\mu (A_1) U(j)$$.

### Lemma 2.2

\begin{aligned} S_n(\mu ) \le S_n(\nu ) \end{aligned}

for $$\nu \ge \mu$$.

We will present two proofs of this Lemma.

### Proof of Lemma 2.2 following Hundertmark and Simon

[2]. Pick any bounded sequence $$\{ \mu _n\}_{n=-\infty }^\infty$$ and consider the matrix

\begin{aligned} M_{\{\mu _n\}} := \sum _j \lambda _j U(j) L_{\{\mu _n\}} (A_1) U(j)\ , \end{aligned}

where

\begin{aligned} (L_{\{\mu _n\}})_{k,\ell } = {\left\{ \begin{array}{ll} b_k^{1/2} \mu _k \cdots \mu _{\ell -1} b_\ell ^{1/2} &{}\mathrm{if } \ k \le \ell \\ (L_{\{\mu _n\}})_{\ell , k} &{} \mathrm{if} \ k > \ell \ . \end{array}\right. } \end{aligned}

Now we fix some integer n and set $$\mu _n = \mu$$. All the other ones are fixed. The matrix $$\sum _j \lambda _j U(j) L_{\{\mu _n\}}U(j)$$ is an affine function of $$\mu$$ with a diagonal that is independent of $$\mu$$ and hence the matrix is of the form

\begin{aligned} \left[ \begin{array}{cc} A &{}\quad 0 \\ 0 &{}\quad B \end{array} \right] +\mu \left[ \begin{array}{cc} 0 &{}\quad C \\ C^* &{}\quad 0 \end{array} \right] \ . \end{aligned}

Now we consider the sum of the top n eigenvalues of this matrix and denote this function by $$f(\mu )$$. This function is convex. Moreover, if we consider the diagonal matrix $$V: \ell ^2 \rightarrow \ell ^2$$ given by

\begin{aligned}{}[V\phi ]_n = (-1)^n \phi _n \end{aligned}

we find that

\begin{aligned} V\sum _j \lambda _j U(j) L_{\{\mu _n\}}(A_1) U(j) V = \sum _j \lambda _j V U(j) L_{\{\mu _n\}}(A_1)U(j) V \\ = \sum _j \lambda _j U(j)V L_{\{\mu _n\}}(A_1)VU(j) \end{aligned}

since the matrix V is also diagonal and commutes with the U(j). This matrix has the same spectrum but is of the form

\begin{aligned} \left[ \begin{array}{cc} A &{}\quad 0 \\ 0 &{}\quad B \end{array} \right] -\mu \left[ \begin{array}{cc} 0 &{}\quad C \\ C^* &{}\quad 0 \end{array} \right] \end{aligned}

and hence $$f(\mu )$$ is even. Thus, f, being convex, is monotone for $$\mu >0$$. $$\square$$

### Remark

The above proof is closely related to the proof of a similar result from [4] except for using some symmetry property rather than perturbation at the spectral point zero.

### Proof of Lemma 2.2 following Hundertmark, Laptev and Weidl

[3]. We aim to use the following abstract result of [3] concerning the sum $$\Vert T\Vert _n$$ of the largest n singular values of a compact operator T,

\begin{aligned} \Vert T\Vert _n=\sum _{j=1}^n\sqrt{E_j(T^*T)}. \end{aligned}

The result is an immediate consequence of $$\Vert T\Vert _n$$ defining a norm by Ky-Fan’s inequality.

### Lemma 2.3

Let T be a non-negative compact operator on a Hilbert space $$\mathcal {G}$$, let g be a probability measure on $$\Omega$$, and let $$\{V(k)\}_{k\in \Omega }$$ be a family of unitary operators on $$\mathcal {G}$$. Then, for any $$n\in {\mathord {{\mathbb {N}}}}$$,

\begin{aligned} \left\| \int _{\Omega }V(k)^*TV(k)\,\mathrm{d}g(k)\right\| _n\le \Vert T\Vert _n. \end{aligned}

To apply the above result to $$L_\mu (A_1)$$, we recall the unitary map $$\mathcal {F}:L^2([-\pi ,\pi ])\rightarrow \ell ^2({\mathord {{\mathbb {Z}}}})$$ onto the Fourier coefficients

\begin{aligned} (\mathcal {F}u)_n=\widehat{u}_n=\frac{1}{\sqrt{2\pi }}\int _{-\pi }^{\pi }\mathrm{e}^{\mathrm{i}n k}u(k)\mathrm{d} k,\qquad (\mathcal {F}^*u)(k) =\frac{1}{\sqrt{2\pi }}\sum _{n} u_n\mathrm{e}^{-\mathrm{i}n k}. \end{aligned}

By means of the transform $$\mathcal {F}$$, the free operator W with $$a_n\equiv 1, b_n\equiv 0$$ is unitarily equivalent to the operator $$2\cos k$$ on $$L^2([-\pi ,\pi ])$$. Defining $$g_\mu$$ to be the non-negative function

\begin{aligned} g_\mu (k)=\frac{1}{\sqrt{2\pi }}\frac{\frac{1}{\mu }-\mu }{-2\cos k+\frac{1}{\mu }+\mu } \end{aligned}

and denoting by T the rank one operator onto $$b_n^{1/2}$$, and by V(k) the unitary operator $$(V(k)u)_n=\mathrm{e}^{-\mathrm{i}n k} u_n$$, we can thus write

\begin{aligned} L_{\mu }(A_1)=\int _{-\pi }^\pi V(k)^*T V(k) \frac{{g}_{\mu }(k)}{\sqrt{2\pi }}\mathrm{d}k. \end{aligned}

To obtain some more properties of $$g_\mu$$, we note that its Fourier transform is given by $$(\widehat{g}_\mu )_n=\mu ^{|n|}$$. This can in particular be used to establish the aforementioned matrix representation of $$L_\mu (A_1)$$. For our purposes, we note that for $$0<\mu<\nu < 1$$ clearly

\begin{aligned} (\widehat{g}_{\mu })_0=1,\qquad \widehat{g}_{\nu }\widehat{g}_{\mu /\nu }=\widehat{g}_{\mu }. \end{aligned}
(2.3)

Since $${g}_{\mu }$$ is smooth and periodic in k, the pointwise identity

\begin{aligned} g_{\mu }(k)=\frac{1}{\sqrt{2\pi }}\sum _{n} (\widehat{g}_{\mu })_n\mathrm{e}^{-\mathrm{i}n k} \end{aligned}

holds and the properties (2.3) imply

\begin{aligned} \int _{-\pi }^\pi \frac{g_{\mu }(k)}{\sqrt{2\pi }}\, \mathrm{d}k=1, \qquad \frac{g_{\nu }}{\sqrt{2\pi }}* \frac{g_{\mu /\nu }}{\sqrt{2\pi }} =\frac{g_{\mu }}{\sqrt{2\pi }} \end{aligned}

for all $$0<\mu<\nu < 1$$. The convolution identity is understood in the sense that

\begin{aligned} \int _{-\pi }^\pi \frac{g_{\nu }(k-k')}{\sqrt{2\pi }}\frac{g_{\mu /\nu }(k')}{\sqrt{2\pi }}\,\mathrm{d}k'=\int _{-\pi }^\pi \frac{g_{\nu }(k')}{\sqrt{2\pi }}\frac{g_{\mu /\nu }(k-k')}{\sqrt{2\pi }}\,\mathrm{d}k'=\frac{g_{\mu }(k)}{\sqrt{2\pi }} \end{aligned}

which is well-defined since all three functions are periodic. Using this identity together with the fact that $$V(k'+k'')=V(k')V(k'')$$ and that $$V(k')$$ and U(j) commute as both are multiplication operators, we obtain

\begin{aligned}&\sum _{j}\lambda _j U(j)L_{\mu }(A_1)U(j)\\ {}&\quad =\sum _{j}\lambda _j\int _{-\pi }^\pi \int _{-\pi }^\pi U(j)V(k)^*T V(k)U(j) \frac{g_{\nu }(k-k')}{\sqrt{2\pi }}\frac{g_{\mu /\nu }(k')}{\sqrt{2\pi }}\,d k'\,d k\\ {}&\quad =\int _{-\pi }^\pi V(k')^*\left( \sum _{j}\lambda _j U(j)\int _{-\pi -k'}^{\pi -k'} V(k'')^*T V(k'') \frac{g_{\nu }(k'')}{\sqrt{2\pi }}\,d k''U(j)\right) \\ {}&\qquad \times V(k')\frac{g_{\mu /\nu }(k')}{\sqrt{2\pi }}\,d k'. \end{aligned}

By periodicity, the operator

\begin{aligned} \int _{-\pi -k'}^{\pi -k'} V(k'')^*T V(k'') \frac{g_{\nu }(k'')}{\sqrt{2\pi }}\,d k'' =\int _{-\pi }^{\pi } V(k'')^*T V(k'') \frac{g_{\nu }(k'')}{\sqrt{2\pi }}\,\mathrm{d}k''=L_{\nu }(A_1) \end{aligned}

is independent of $$k'$$ and thus we can apply Lemma 2.3 to obtain the desired monotonicity. The special case $$\nu =1$$ is an immediate consequence of Lemma 2.3. $$\square$$

### Remark

The above proof is closely related to the continuous case [3] with the established convolution identity for $$g_\mu$$ replacing the fact that the Cauchy distribution is a convolution semigroup. Using the parametrization $$\mu =\mathrm{e}^{-\sigma }, \nu =\mathrm{e}^{-\tau }$$ with $$\sigma>\tau >0$$, the above identity may also be written in the more similar form $$\frac{g_{\tau }}{\sqrt{2\pi }}*\frac{g_{\sigma -\tau }}{\sqrt{2\pi }}=\frac{g_{\sigma }}{\sqrt{2\pi }}$$.

Because of the Birman–Schwinger principle, $$E_j(K(A;E_j^+))$$, the j-the eigenvalue of $$K(A; E_j^+)$$, equals 1 and hence

\begin{aligned}&\sum _{j=1}^n (E_j^{+2}(J)-4)^{1/2} = \sum _{j=1}^n (E_j^{+2}(J)-4)^{1/2} E_j( K(A; E^+_j) ) = \sum _{j=1}^n E_j(L_{\mu _j}(A))\\&\quad \le \sum _{j=1}^n E_j( \sum _k \lambda _kU(k) L_{\mu _j}(A_1) U(k)) \le S_n(\mu _n) \end{aligned}

where $$\mu _j+\frac{1}{\mu _j} = E^+_j$$. Now, again, we proceed as in Hundertmark–Simon [2] and get the estimate

\begin{aligned} S_{N^+}(\mu _n) \le \mathrm{Tr} \sum _j \lambda _j U(j) L_{\mu =1}(A_1) U(j) = \mathrm{Tr} B \ , \end{aligned}

where $$S_{N^+}(\mu _n)$$ includes all the eigenvalues of $$\widetilde{J}$$ that are greater than 2. In other words

\begin{aligned}&\sum _j (E_j^{+2}({\widetilde{J}}^+)-4)^{1/2} \le \sum _n [b_n]_+ + (a_{n-1}-1)_++ (a_n-1)_+\\&\quad = \sum _n [b_n]_+ + 2 (a_n-1)_+ \ . \end{aligned}

As shown in [2], the Jacobi matrices

\begin{aligned} W(\{a_n\},\{b_n\}) \ \mathrm{and } -W(\{a_n\}, \{-b_n\}) \end{aligned}

are unitarily equivalent and hence it follows that

\begin{aligned}&\sum _j (E_j^{-2}({\widetilde{J}}^-)-4)^{1/2} \le \sum _n [b_n]_- + (a_{n-1}-1)_++ (a_n-1)_+\\&\quad = \sum _n [b_n]_-+ 2 (a_n-1)_+ \ , \end{aligned}

which together with the previous estimate proves Theorem 1.1.

As a corollary, we obtain Theorem 1.2.

### Proof of Theorem 1.2

Let $$\chi _{\{c,d\}}$$ be the characteristic function of the interval (cd). Then, recalling $$J\le \widetilde{J}^+$$, we have

\begin{aligned}&\sum _j \int _2^{E_j^+}(t^2 - 4)^{\frac{1}{2}}(E_j^+-t)^{\gamma -\frac{3}{2}} \, \mathrm{d}t \\&\quad = \sum _j \int _0^\infty \left( (E_j^+ (J-s) )^2 - 4\right) ^{\frac{1}{2}} s^{\gamma -\frac{3}{2}}\chi _{_{\{E_j^+(J -s) \ge 2\}}} (s) \, \mathrm{d}s\\&\quad \le \sum _j \int _0^\infty \left( (E_j^+ (\widetilde{J}^+-s) )^2 - 4\right) ^{\frac{1}{2}} s^{\gamma -\frac{3}{2}}\chi _{_{\{E_j^+(\widetilde{J}^+ -s) \ge 2\}}} (s) \, \mathrm{d}s. \end{aligned}

Applying first the variational principle and then the main result, we obtain

\begin{aligned}&\sum _j \int _2^{E_j^+}(t^2 - 4)^{\frac{1}{2}}(E_j^+-t)^{\gamma -\frac{3}{2}} \, \mathrm{d}t \\&\quad \le \sum _{n}\int _0^\infty (\widetilde{b}^+_n-s)_+s^{\gamma -\frac{3}{2}}\,\mathrm{d}s = \mathrm {B}(\gamma -1/2,2)\, \sum _n (\widetilde{b}^+_n)_+^{\gamma +\frac{1}{2}} \end{aligned}

and the proof is complete. $$\square$$