Uniform Bounds for Maximal Flat Periods on $SL(n,\mathbb{R})$

Phillip Harris

Abstract.

Let $X$ be a compact locally symmetric space associated to $SL(n,\mathbb{R})$ and $Y\subset X$ a maximal flat submanifold, not necessarily closed. Using a Euclidean approximation, we give an upper bound in the spectral aspect for Maass forms integrated against a smooth cutoff function on $Y$ .

1. Introduction

Let $G$ be a noncompact semisimple Lie group with Iwasawa decomposition $G=NAK$ and Lie algebras $\mathfrak{g}=\mathfrak{n+a+k}$ . Let $G/K$ be the associated symmetric space and $X=\Gamma\backslash G/K$ a compact quotient. Let $(f_{i})_{i}\in L^{2}(X)$ be an orthonormal basis of Maass forms with spectral parameters $\nu_{i}\in\mathfrak{a}^{*}_{\mathbb{C}}$ . Let $Y\subset X$ be a maximal flat submanifold (not necessarily closed) and $b\in C_{c}^{\infty}(Y)$ a smooth cutoff function. We are interested in the growth of the flat periods

\mathcal{P}_{i}=\int_{Y}f_{i}(x)b(x)dx

with respect to $\nu_{i}$ .

There is a general bound for Laplace eigenfunctions on compact manifolds due to Zelditch [7] which in our case gives

\mathinner{\!\left\lvert\mathcal{P}_{i}\right\rvert}\ll\mathinner{\!\left\lVert\nu_{i}\right\rVert}^{(\dim X-\dim Y-1)/2}=\mathinner{\!\left\lVert\nu_{i}\right\rVert}^{(n^{2}-n-2)/4}.

Michels [5] was the first to study flat periods on higher rank locally symmetric spaces. His work implies¹¹1This result for Maass forms follows from applying the trace formula argument in Section 2.2 to Michels’ bound for spherical functions. an averaged estimate for flat periods, as follows. Let $X$ be a compact locally symmetric space of noncompact type, of rank $r$ . Let $\Sigma\subset\mathfrak{a}^{*}$ be the restricted roots of $X$ and $\Sigma^{+}$ be the positive roots. Define the set of “generic points” $(\mathfrak{a^{*}})^{\text{gen}}\subset\mathfrak{a^{*}}$ to be the set of points that are regular and that do not lie in any proper subspace spanned by roots. Fix a closed cone $D\subset(\mathfrak{a^{*}})^{\text{gen}}$ . Let $\beta\mathrel{\mathop{\ordinarycolon}}\mathfrak{a}^{*}\to\mathbb{R}$ be the Plancherel density. Then there exists $C>0$ such that uniformly for $\lambda\in D$ we have

\sum_{\mathinner{\!\left\lVert\operatorname{Re}\nu_{i}-\lambda\right\rVert}\leq C}\mathinner{\!\left\lvert\mathcal{P}_{i}\right\rvert}^{2}\ll\beta(\lambda)(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-r}.

In other words, consider all periods $\mathcal{P}_{i}$ such that $\operatorname{Re}\nu_{i}$ lies in a ball of fixed radius around $\lambda$ , then their average norm squared is $\ll(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-r}$ . For certain choices of $X$ and $Y$ enumerated in [5, Theorem 1.3] Michels proves a lower bound (replacing $\ll$ with $\asymp$ ).

For $X$ associated to $SL(n,\mathbb{R})$ , we give an estimate uniform in $\lambda$ , elucidating the behavior on the non-generic set.

Theorem 1.

Uniformly for $\lambda=\textup{diag}(\lambda_{1},\dots,\lambda_{n})\in\mathfrak{a}^{*}$ we have

\displaystyle\sum_{\mathinner{\!\left\lVert\operatorname{Re}\nu_{i}-\lambda\right\rVert}\leq 1}\mathinner{\!\left\lvert\mathcal{P}_{i}\right\rvert}^{2}\ll\tilde{\beta}(\lambda)(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(\lambda)

with the implied constant depending only on $n$ and the test function $b$ . Here $\tilde{\beta}(\lambda)=\prod_{\alpha\in\Sigma^{+}}1+\mathinner{\!\left\lvert\langle\lambda,\alpha\rangle\right\rvert}$ is essentially the maximum of $\beta$ taken over a ball around $\lambda$ . The function $L_{n}(\lambda)$ is Weyl invariant, and in the Weyl chamber ²²2This is opposite from the canonical positive chamber. with $\lambda_{1}\leq\dots\leq\lambda_{n}$ , we define it as follows. Define $\log^{\prime}x=\log(2+x)$ . For $n\neq 4$ ,

L_{n}(\lambda)\mathrel{\mathop{\ordinarycolon}}=\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\right)^{n-2},

and for $n=4$ ,

L_{4}(\lambda)\mathrel{\mathop{\ordinarycolon}}=\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{3}\right\rvert}}\right)^{2}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}}.

Since $\tilde{\beta}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{n(n-1)/2}$ we obtain $\mathinner{\!\left\lvert\mathcal{P}_{i}\right\rvert}\ll\mathinner{\!\left\lVert\nu_{i}\right\rVert}^{(n^{2}-3n+2)/4}L_{n}(\nu_{i})^{1/2}$ , an improvement on the bound for general manifolds. The factor $L_{n}(\lambda)$ can be given a geometric interpretation, namely, it grows when $\lambda$ is collinear with a root. In the $n=4$ case, it grows when $\lambda$ is collinear with a root or a sum of two orthogonal roots.

1.1. Outline of the proof

Using a standard pre-trace formula argument, we reduce the period estimate to an estimate for spherical functions: let $A\subset SL(n,\mathbb{R})$ be the diagonal subgroup and $b\in C^{\infty}_{c}(A)$ , then

(1)

\int_{A}\varphi_{\lambda}(g)b(g)dg\ll_{b}(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(\lambda).

We use a theorem of Duistermaat to linearize the problem, replacing the spherical function $\varphi_{\lambda}$ on the Lie group with a “Euclidean” approximation on the Lie algebra. The problem then reduces to the following random matrix lemma: Let $\lambda\in\mathfrak{a}^{*}$ be regular. For a real $n\times n$ matrix $A$ , let $d(A)=\sum_{1\leq i\leq n}A_{ii}^{2}$ be the size of the diagonal. Choose $k\in SO(n)$ at random according to the Haar measure, then

\textup{Prob}[d(k\lambda k^{-1})<1]\asymp(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(\lambda).

While we were unable to prove a lower bound in Theorem 1, this lemma and Michels’ lower bounds are at least suggestive that our upper bound is sharp.

2. Proof of theorem 1

2.1. Preliminaries and Notation

The notation $A\ll B$ means there exists $C>0$ such that $A\leq CB$ , and $A\asymp B$ means $A\ll B\ll A$ . The implied constant $C$ may always depend on the dimension $n$ and bump function $b$ ; any other dependency is indicated with a subscript. When $A$ is a matrix, $\mathinner{\!\left\lVert A\right\rVert}$ will denote the Frobenius norm $\mathinner{\!\left\lVert A\right\rVert}^{2}=\sum_{ij}\mathinner{\!\left\lvert A_{ij}\right\rvert}^{2}$ . Indicator functions are denoted $\mathbf{1}_{S}(x)$ where $S$ is a set or $\mathbf{1}[P(x)]$ where $P$ is a predicate.

Let $G=SL(n,\mathbb{R})$ . We write the Iwasawa decomposition $G=NAK$ , where $N$ is the upper triangular subgroup with 1’s on the diagonal, $A$ is the diagonal subgroup and $K$ is $SO(n)$ . We have the corresponding Lie algebra decomposition $\mathfrak{g=n+a+k}$ and the Cartan decomposition $\mathfrak{g=p+k}$ . Then $X=\Gamma\backslash G/K$ for some cocompact lattice $\Gamma\subset G$ , and $Y$ lies in the image of $gA$ for some $g\in G$ . Using a partition of unity we may assume the support of $b$ is small, say, having diameter $<R/100$ where $R$ is the injectivity radius of $X$ , and lift $b$ to $C^{\infty}_{c}(gA)$ .

2.2. Spherical Functions and the Trace Formula

Let $H\mathrel{\mathop{\ordinarycolon}}G\to\mathfrak{a}$ be the smooth map satisfying $g\in Ne^{H(g)}K$ for all $g\in G$ (sometimes called the Iwasawa projection [5]). Let $\rho$ be the half sum of the positive roots and $W$ be the Weyl group. Recall the spherical functions

\varphi_{\lambda}(g)=\int_{K}e^{(i\lambda+\rho)H(kg)}dk,

the Harish-Chandra transform

\widehat{f}(\lambda)=\int_{G}f(x)\varphi_{-\lambda}(x)dx\,\,\,\,\,\,\,\,\,f\in C^{\infty}_{c}(K\backslash G/K),

and its inverse

f(x)=\frac{1}{\mathinner{\!\left\lvert W\right\rvert}}\int_{\mathfrak{a^{*}}}\varphi_{\lambda}(x)\widehat{f}(\lambda)\beta(\lambda)d\lambda.

In order to prove Theorem 1 we use the pre-trace formula for $SL(n,\mathbb{R})$ . Let $k\in C^{\infty}_{c}(K\backslash G/K)$ be a bi- $K$ -invariant test function and

K(x,y)=\sum_{\gamma\in\Gamma}k(x^{-1}\gamma y)

be the corresponding automorphic kernel on $X$ . Then the pre-trace formula [6] states

\sum_{\gamma\in\Gamma}k(x^{-1}\gamma y)=\sum_{i}\widehat{k}(-\nu_{i})f_{i}(x)\overline{f_{i}(y)}

where $\widehat{k}$ is the Harish-Chandra transform of $k$ . Integrating against $b(x)b(y)\in C_{c}^{\infty}(gA\times gA)$ we obtain

(2)

\int_{gA}\int_{gA}\sum_{\gamma\in\Gamma}k(x^{-1}\gamma y)b(x)b(y)dxdy=\sum_{i}\widehat{k}(-\nu_{i})\mathinner{\!\left\lvert\mathcal{P}_{i}\right\rvert}^{2}.

Next we construct a $k$ such that $\widehat{k}$ concentrates around a given $\lambda_{0}$ . Recall that the spectral parameters of Maass forms satisfy $W\nu_{i}=W\overline{\nu_{i}}$ and $\mathinner{\!\left\lVert\operatorname{Im}\nu_{i}\right\rVert}\leq\mathinner{\!\left\lVert\rho\right\rVert}$ . Define $\Omega=\{\lambda\in\mathfrak{a}^{*}_{\mathbb{C}}\mathrel{\mathop{\ordinarycolon}}W\lambda=W\overline{\lambda},\mathinner{\!\left\lVert\operatorname{Im}\lambda\right\rVert}\leq\mathinner{\!\left\lVert\rho\right\rVert}\}$ [1, Lemma 4.1]. Using the Paley-Wiener theorem for the Harish-Chandra transform we construct $k$ with the following properties:

(1)

$k$ is supported in a ball of radius $R/100$ , where $R$ is the injectivity radius of $X$ .
(2)

$\widehat{k}(\lambda)\geq 0$ for $\lambda\in\Omega$ .
(3)

$\widehat{k}(\lambda)\geq 1$ for $\lambda\in\Omega$ with $\mathinner{\!\left\lVert\text{Re }\lambda-\lambda_{0}\right\rVert}\leq 1$ .
(4)

$\widehat{k}(\lambda)\ll_{N}(1+\mathinner{\!\left\lVert\lambda-\lambda_{0}\right\rVert})^{-N}$ for $\nu\in\Omega$ uniformly in $\lambda_{0}$ .

The details of this construction may be found in [1, Section 4.1].

By property (1), all terms except $\gamma=e$ on the geometric side drop out. By properties (2) and (3) we have

\int_{gA}\int_{gA}k(x^{-1}y)b(x)b(y)dxdy\geq\sum_{\mathinner{\!\left\lVert\operatorname{Re}\nu_{i}-\lambda_{0}\right\rVert}\leq 1}\mathinner{\!\left\lvert\mathcal{P}_{i}\right\rvert}^{2}.

Note that $x^{-1}y\in A$ . Defining $b_{1}(z)=\int_{gA}b(x)b(xz)dx\in C^{\infty}_{c}(A)$ and changing variables $z=x^{-1}y$ the left hand side becomes

	$\displaystyle\int_{gA}\int_{gA}k(x^{-1}y)b(x)b(y)dxdy$	$\displaystyle=\int_{A}k(z)b_{1}(z)\,dz$
		$\displaystyle=\int_{A}\left(\frac{1}{\mathinner{\!\left\lvert W\right\rvert}}\int_{\mathfrak{a^{*}}}\varphi_{\lambda}(z)\widehat{k}(\lambda)\beta(\lambda)\,d\mu\right)b_{1}(z)dz$
		$\displaystyle=\frac{1}{\mathinner{\!\left\lvert W\right\rvert}}\int_{\mathfrak{a^{*}}}\widehat{k}(\lambda)\beta(\lambda)\left(\int_{A}\varphi_{\lambda}(z)b_{1}(z)dz\right)\,d\lambda.$

The task is now to bound the integral of a spherical function $\varphi_{\lambda}$ against a smooth cutoff function on $A$ . In other words, we need the following bound: for $b\in C^{\infty}_{c}(A)$ ,

\int_{A}\varphi_{\lambda}(z)b(z)dz\ll_{b}(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(\lambda).

With this bound, Theorem 1 follows from the rapid decay of $\widehat{k}$ away from $-\lambda_{0}$ (property 4) and the polynomial growth of $\beta$ .

2.3. Euclidean Approximation of Spherical Functions

Let $\mathfrak{g=k+p}$ be the Cartan decomposition. For $X\in\mathfrak{p},k\in K$ write $k.X=kXk^{-1}$ for the adjoint action of $K$ on $\mathfrak{p}$ . Let $\pi\mathrel{\mathop{\ordinarycolon}}\mathfrak{p}\to\mathfrak{a}$ be the orthogonal projection with respect to the Killing form. By a theorem of Duistermaat [3, Equation 1.11], there exists a nonnegative analytic function $a\in C^{\infty}(\mathfrak{p})$ such that

\varphi_{\lambda}(e^{X})=\int_{K}e^{\langle i\lambda,\pi(k.X)\rangle}a(k.X)dk

where $dk$ is the Haar measure. Changing variables $z=e^{X}$ and using Duistermaat’s formula gives

	$\displaystyle(*)\mathrel{\mathop{\ordinarycolon}}=\int_{A}\varphi_{\lambda}(z)b(z)dz$	$\displaystyle=\int_{\mathfrak{a}}\int_{K}e^{\langle i\lambda,\pi(k.X)\rangle}a(k.X)b(e^{X})dXdk$
		$\displaystyle=\int_{K}\left(\int_{\mathfrak{a}}e^{\langle ik^{-1}.\lambda,X\rangle}a(k.X)b(e^{X})dX\right)dk,$

where in the second line we extend $\lambda\in\mathfrak{a}^{*}$ to $\mathfrak{p}^{*}$ by pulling back along $\pi$ , and replace the adjoint action with the coadjoint action. Let $c(k,X)\mathrel{\mathop{\ordinarycolon}}=a(k.X)b(e^{X})$ and note that $c\in C^{\infty}_{c}(K\times\mathfrak{a})$ . Let $\widehat{c}(k,\xi)\in C^{\infty}(K\times\mathfrak{a}^{*})$ be the Fourier transform of $c$ in the second variable. Then the inner integral is just $\widehat{c}$ evaluated at the frequency $\xi=-\pi^{*}(k^{-1}.\lambda)$ , where $\pi^{*}\mathrel{\mathop{\ordinarycolon}}\mathfrak{p}^{*}\to\mathfrak{a}^{*}$ is restriction:

(*)=\int_{K}\widehat{c}(k,-\pi^{*}(k^{-1}.\lambda))dk.

Since $\widehat{c}(k,\xi)$ has rapid decay in $\xi$ for all $k$ and $K$ is compact, $\sup_{k}\widehat{c}(k,\xi)$ also has rapid decay in $\xi$ . Let $B$ be the indicator function of the unit ball on $\mathfrak{a^{*}}$ , pulled back along $\pi^{*}$ to $\mathfrak{p^{*}}$ . Taking the supremum in the first variable and upper bounding in the second variable by balls gives

\mathinner{\!\left\lvert(*)\right\rvert}\leq\sum_{r=1}^{\infty}d_{r}\int_{K}B(r^{-1}(k^{-1}.\lambda))dk

where the sequence $d_{r}$ has rapid decay. Essentially, we want to know how much the coadjoint orbit $K.\lambda$ can concentrate near $(\mathfrak{a^{*}})^{\perp}$ . It suffices to show the following lemma:

Lemma 1.

Let $\text{Sym}_{n}(\mathbb{R})$ be the set of real symmetric $n\times n$ matrices. For $X\in\text{Sym}_{n}(\mathbb{R})$ let $\pi(X)$ be the diagonal part of $X$ and $B(X)=\mathbf{1}[{\mathinner{\!\left\lVert\pi(X)\right\rVert}<1}]$ . Let $\lambda=\text{diag}(\lambda_{1},\dots,\lambda_{n})$ with $\lambda_{1}<\lambda_{2}<\dots<\lambda_{n}$ . Define

I_{n}(\lambda)\mathrel{\mathop{\ordinarycolon}}=\int_{SO(n)}B(k.\lambda)\,dk

and

A_{n}(\lambda)\mathrel{\mathop{\ordinarycolon}}=(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(\lambda).

Then if $\operatorname{\operatorname{Tr}}\lambda=0$ we have $I_{n}(\lambda)\asymp A_{n}(\lambda)$ .

Finally, since $(*)$ is continuous in $\lambda$ , we may perturb $\lambda$ to be regular (i.e. $\lambda_{i}<\lambda_{i+1}$ rather than $\leq)$ , then

	$\displaystyle\mathinner{\!\left\lvert(*)\right\rvert}$	$\displaystyle\leq\sum_{r=1}^{\infty}d_{r}(1+r^{-1}\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(r^{-1}\lambda)$
		$\displaystyle\ll(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(\lambda).$

3. The coadjoint orbit $K.\lambda$

The remainder of the paper will be proving Lemma 1.

3.1. Preliminaries

We note some properties of $I_{n}(\lambda)$ and $A_{n}(\lambda)$ .

•

$I_{n}(\lambda)$ is monotone under scaling: if $t\geq 1$ then $I_{n}(t\lambda)\leq I_{n}(\lambda)$ .
•

For a fixed $C\geq 0$ , we have $A_{n}(C\lambda)\asymp_{C}A_{n}(\lambda)$ , and $\mathinner{\!\left\lVert\lambda-\lambda^{\prime}\right\rVert}\leq C$ implies $A_{n}(\lambda)\asymp_{C}A_{n}(\lambda^{\prime})$ .

•

An estimate for $I_{n}(\lambda)$ when $\operatorname{\operatorname{Tr}}\lambda\neq 0$ easily follows from the tracefree case. Write $\lambda=\lambda_{0}+(\operatorname{\operatorname{Tr}}\lambda/n)I$ . Then

	$\displaystyle\mathinner{\!\left\lVert\pi(k.\lambda)\right\rVert}^{2}$	$\displaystyle=\mathinner{\!\left\lVert\pi(k.\lambda_{0})\right\rVert}^{2}+2\langle\pi(k.\lambda_{0}),(\operatorname{\operatorname{Tr}}\lambda/n)I\rangle+\mathinner{\!\left\lVert\pi((\operatorname{\operatorname{Tr}}\lambda/n)I)\right\rVert}^{2}$
		$\displaystyle=\mathinner{\!\left\lVert\pi(k.\lambda_{0})\right\rVert}^{2}+(\operatorname{\operatorname{Tr}}\lambda)^{2}/n,$

so $\mathinner{\!\left\lVert\pi(k.\lambda)\right\rVert}<1$ if and only if $\mathinner{\!\left\lVert\pi(k.\lambda_{0})\right\rVert}<\sqrt{1-(\operatorname{\operatorname{Tr}}\lambda)^{2}/n}$ . Thus

(3)

\displaystyle I_{n}(\lambda)=\begin{cases}I_{n}(\lambda_{0}/\sqrt{1-(\operatorname{\operatorname{Tr}}\lambda)^{2}/n})&\mathinner{\!\left\lvert\operatorname{\operatorname{Tr}}\lambda\right\rvert}\leq\sqrt{n}\\ 0&\mathinner{\!\left\lvert\operatorname{\operatorname{Tr}}\lambda\right\rvert}>\sqrt{n}\end{cases}.

We also note the following soft lower bound for $I_{n}(\lambda)$ .

Lemma 2.

$I_{n}(\lambda)\gg(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-\dim SO(n)}$ .

Proof.

An easy inductive argument shows that for all $\lambda$ with $\operatorname{\operatorname{Tr}}\lambda=0$ , there exists some $k_{0}\in SO(n)$ such that $\pi(k_{0}.\lambda)=0$ . Indeed, suppose $\lambda_{1}<0<\lambda_{2}$ , and let $R_{\theta}\in SO(2)$ act on the first two coordinates. Then acting by a quarter-turn swaps $\lambda_{1}$ and $\lambda_{2}$ :

\displaystyle R_{\pi/2}.\begin{bmatrix}\lambda_{1}&&\\ &\lambda_{2}&\\ &&\ddots\end{bmatrix}

\displaystyle=\begin{bmatrix}\lambda_{2}&&\\ &\lambda_{1}&\\ &&\ddots\end{bmatrix},

so by continuity there exists $\theta$ such that $R_{\theta}.\lambda$ has a 0 in the upper left corner, etc.

Choose $X\in\mathfrak{so}(n)$ with $\mathinner{\!\left\lVert X\right\rVert}<1/100$ . Taylor expansion gives

\displaystyle(\exp X)k_{0}.\lambda

\displaystyle=k_{0}.\lambda+X(k_{0}.\lambda)-(k_{0}.\lambda)X+\frac{1}{2}\left(X^{2}(k_{0}.\lambda)-2X(k_{0}.\lambda)X+(k_{0}.\lambda)X^{2}\right)+\dots

Thus $\mathinner{\!\left\lVert\pi((\exp X)k_{0}.\lambda)\right\rVert}\ll\mathinner{\!\left\lVert X\right\rVert}\mathinner{\!\left\lVert\lambda\right\rVert}$ , and $\mathinner{\!\left\lVert\pi(k.\lambda)\right\rVert}<1$ on a ball of radius $\gg\min(1/100,\mathinner{\!\left\lVert\lambda\right\rVert}^{-1})$ around $k_{0}$ . ∎

Since $(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-\dim SO(n)}\ll I_{n}(\lambda)\leq 1$ , in the rest of the proof we may assume that $\mathinner{\!\left\lVert\lambda\right\rVert}$ is greater than some constant depending only on $n$ and prove $I_{n}(\lambda)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}L_{n}(\lambda)$ instead of $I_{n}(\lambda)\asymp(1+\mathinner{\!\left\lVert\lambda\right\rVert})^{-n+1}L_{n}(\lambda)$ .

3.2. Inductive step

For $n=2$ , we write $\lambda=\begin{bmatrix}-a&0\\ 0&a\end{bmatrix},k=\begin{bmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{bmatrix}$ and easily evaluate the integral:

	$\displaystyle\int_{SO(2)}B(k.\lambda)dk$	$\displaystyle=\frac{1}{2\pi}\int_{0}^{2\pi}B\left(\left[\begin{matrix}-a\cos 2\theta&a\sin 2\theta\\ a\sin 2\theta&a\cos 2\theta\end{matrix}\right]\right)d\theta$
		$\displaystyle=\frac{1}{2\pi}\int_{0}^{2\pi}\mathbb{1}[2a^{2}\cos^{2}(2\theta)<1]d\theta$
		$\displaystyle\asymp(1+\mathinner{\!\left\lvert a\right\rvert})^{-1}.$

So in the remainder of the proof assume $n\geq 3$ .

Let $e_{1},\dots,e_{n}$ be basis vectors for $\mathbb{R}^{n}$ and let $SO(n-1)\subset SO(n)$ be the subgroup fixing $e_{n}$ . Let $\Pi\mathrel{\mathop{\ordinarycolon}}\text{Sym}_{n}(\mathbb{R})\to\text{Sym}_{n-1}(\mathbb{R})$ be the projection to the upper left $n-1\times n-1$ submatrix; we have $\Pi(k^{\prime}.X)=k^{\prime}.\Pi(X)$ for $k^{\prime}\in SO(n-1)$ . Let $\pi^{\prime}$ and $B^{\prime}$ be the $(n-1)$ -dimensional versions of $\pi$ and $B$ . Let $X=k.\lambda$ . We have $\mathinner{\!\left\lVert\pi(X)\right\rVert}^{2}=\mathinner{\!\left\lVert\pi^{\prime}(\Pi(X))\right\rVert}^{2}+X_{nn}^{2}$ , so $B(X)\leq B^{\prime}(\Pi(X))$ and

(4)

\displaystyle I_{n}(\lambda)\leq\int_{SO(n)}B^{\prime}(\Pi(k.\lambda))\,dk.

On the other hand, since $\operatorname{\operatorname{Tr}}X=0$ we have $X_{nn}=-\sum_{i=1}^{n-1}X_{ii}$ and by Cauchy-Schwarz $X_{nn}^{2}\leq(n-1)\sum_{i=1}^{n-1}X_{ii}^{2}=(n-1)\mathinner{\!\left\lVert\pi^{\prime}(\Pi(X))\right\rVert}^{2}$ , so $\mathinner{\!\left\lVert\pi(X)\right\rVert}\leq\sqrt{n}\mathinner{\!\left\lVert\pi^{\prime}(\Pi(X))\right\rVert}$ and

\displaystyle I_{n}(\lambda)\geq\int_{SO(n)}B^{\prime}(\sqrt{n}\Pi(k.\lambda))\,dk.

To do the induction we will convert the integral over $SO(n)$ to an integral over $SO(n-1)$ :

Lemma 3.

Define

\displaystyle\mathcal{M}\mathrel{\mathop{\ordinarycolon}}=\left\{\left[\begin{matrix}\mu_{1}&&\\ &\ddots&\\ &&\mu_{n-1}\end{matrix}\right]\mathrel{\mathop{\ordinarycolon}}\lambda_{i}<\mu_{i}<\lambda_{i+1}\right\}

and

\displaystyle J(\mu)\mathrel{\mathop{\ordinarycolon}}=\frac{\prod_{1\leq i<j\leq n-1}\mathinner{\!\left\lvert\mu_{i}-\mu_{j}\right\rvert}}{\prod_{\begin{subarray}{c}1\leq i\leq n\\ 1\leq j\leq n-1\end{subarray}}\mathinner{\!\left\lvert\lambda_{i}-\mu_{j}\right\rvert}^{1/2}}.

For any non-negative $f\mathrel{\mathop{\ordinarycolon}}\text{Sym}_{n-1}(\mathbb{R})\to\mathbb{R}$ , we have

(5)

\displaystyle\int_{SO(n)}f(\Pi(k.\lambda))\,dk=c_{n}\int_{\mathcal{M}}\int_{SO(n-1)}f(k^{\prime}.\mu)\,dk^{\prime}J(\mu)\,d\mu.

for some absolute constant $c_{n}>0$ .

Proof.

Let $\mathcal{A}\subset\text{Sym}_{n-1}(\mathbb{R})$ be the set of matrices conjugate to some $\mu\in\mathcal{M}$ , equipped with the measure inherited from $\mathbb{R}^{(n-1)^{2}}$ . We will show both sides are proportional to

(6)

\displaystyle\int_{\mathcal{A}}f(X)\prod_{\begin{subarray}{c}1\leq i\leq n\\ 1\leq j\leq n-1\end{subarray}}\mathinner{\!\left\lvert\lambda_{i}-\mu_{j}(X)\right\rvert}^{-1/2}\,dX

where $\mu_{j}(X)$ is the $j$ -th highest eigenvalue of $X$ , viewed as a function on $\mathcal{A}$ . Define a map

	$\displaystyle F$	$\displaystyle\mathrel{\mathop{\ordinarycolon}}SO(n)\to\text{Sym}_{n-1}(\mathbb{R})$
	$\displaystyle F(k)$	$\displaystyle=\Pi(k.\lambda).$

We begin by pushing the left hand side of (5) forward along $F$ to get (6). First we exclude the degenerate case where $\Pi(k.\lambda)$ shares an eigenvalue with $\lambda$ .

Lemma 4.

$\lambda_{i}$ is an eigenvalue of $\Pi(k.\lambda)$ if and only if $\langle e_{i},k^{-1}e_{n}\rangle=0$ .

Proof.

By replacing $\lambda$ with $\lambda-\lambda_{i}I$ , we may assume without loss of generality that $\lambda_{i}=0$ . Let $P=I-e_{n}e_{n}^{T}$ , the projection killing the $n$ -th coordinate. Then for any $A\in\text{Sym}_{n}(\mathbb{R})$ ,

\displaystyle PAP=\begin{bmatrix}\Pi(A)&0\\ 0&0\end{bmatrix},

so the eigenvalues of $PAP$ are the eigenvalues of $\Pi(A)$ , with an extra multiplicity at $0$ .

Suppose $\lambda_{i}$ is an eigenvalue of $\Pi(k.\lambda)$ , then $Pk\lambda k^{-1}Pv=0$ for some $v$ with $Pv\neq 0$ . If $\lambda k^{-1}Pv=0$ , then $k^{-1}Pv=ce_{i}$ for some $c\neq 0$ , and $\langle e_{i},k^{-1}e_{n}\rangle=c^{-1}\langle k^{-1}Pv,k^{-1}e_{n}\rangle=c^{-1}\langle Pv,e_{n}\rangle=c^{-1}\langle v,Pe_{n}\rangle=0$ . Otherwise, if $Pk\lambda k^{-1}Pv=0$ and $\lambda k^{-1}Pv\neq 0$ then $k\lambda k^{-1}Pv=ce_{n}$ for some $c\neq 0$ and $\langle e_{i},k^{-1}e_{n}\rangle=c^{-1}\langle e_{i},\lambda k^{-1}Pv\rangle=c^{-1}\langle\lambda e_{i},k^{-1}Pv\rangle=0$ .

Conversely, if $\langle e_{i},k^{-1}e_{n}\rangle=0$ then $Pke_{i}=ke_{i}$ and $Pk\lambda k^{-1}Pke_{i}=Pk\lambda e_{i}=0$ , so $ke_{i}$ is the desired eigenvector for $\Pi(k.\lambda)$ . ∎

Let $\mathcal{S}=\{k\in SO(n)\mathrel{\mathop{\ordinarycolon}}\langle e_{i},k^{-1}e_{n}\rangle=0\text{ for some $i$ }\}$ . Note that the quotient $SO(n-1)\backslash\mathcal{S}$ is the intersection of the sphere $S^{n}\simeq SO(n-1)\backslash SO(n)$ with the coordinate planes, so $\mathcal{S}$ is negligible. A calculation implicit in Fan and Pall [4, Theorem 1 and pp.300-301] shows that $F(SO(n)-\mathcal{S})=\mathcal{A}$ .

Lemma 5 (Fan-Pall).

Let $\mu_{1}\leq\dots\leq\mu_{n-1}$ and $z_{1},\dots,z_{n}\in\mathbb{R}$ . Suppose the matrix

(7)

\left[\begin{matrix}\mu_{1}&&&z_{1}\\ &\ddots&&\vdots\\ &&\mu_{n-1}&z_{n-1}\\ z_{1}&\dots&z_{n-1}&z_{n}\\ \end{matrix}\right]

has eigenvalues $\lambda_{1}\leq\dots\leq\lambda_{n}$ . Then $\lambda_{1}\leq\mu_{1}\leq\lambda_{2}\leq\dots\leq\mu_{n-1}\leq\lambda_{n}$ . Furthermore, if $\lambda_{i}\neq\mu_{j}$ for all $i,j$ then

z_{i}^{2}=\frac{\prod_{1\leq j\leq n}\mathinner{\!\left\lvert\lambda_{j}-\mu_{i}\right\rvert}}{\prod_{\begin{subarray}{c}1\leq j\leq n-1\\ i\neq j\end{subarray}}\mathinner{\!\left\lvert\mu_{i}-\mu_{j}\right\rvert}}

for $1\leq i\leq n-1$ , and taking traces gives $z_{n}=\sum_{i}\lambda_{i}-\sum_{j}\mu_{j}$ .

Conversely, for any $\mu$ and $\lambda$ satisfying $\lambda_{i}\neq\mu_{j}$ the above choice of $z_{i}$ gives a matrix with eigenvalues $\lambda$ .

We check that $F$ restricted to $SO(n)-\mathcal{S}$ is a smooth covering and compute its differential. Let $Y\in\mathcal{A}$ . Conjugating by $SO(n-1)$ we may assume $Y$ is diagonal. There are $2^{n-1}$ choices of $X\in\text{Sym}_{n}(\mathbb{R})$ conjugate to $\lambda$ such that $\Pi(X)=Y$ , corresponding to choices of sign for the $z_{i}$ in (7). Furthermore, since the centralizer of $\lambda$ in $SO(n)$ consists of reflections through an even number of coordinate axes, for each $X$ there are $2^{n-1}$ choices of $k\in SO(n)-\mathcal{S}$ such that $k.\lambda=X$ .

Let $\mathbb{E_{ij}}$ be the $n\times n$ matrix with a $1$ at position $(i,j)$ , a $-1$ at $(j,i)$ and $0$ elsewhere. Then $\{\mathbb{E_{ij}}k\mathrel{\mathop{\ordinarycolon}}1\leq i<j\leq n\}$ is an orthonormal basis for $T_{k}SO(n)$ (up to some constant). Let $\mathbb{F_{ij}}$ be the $n-1\times n-1$ matrix with a $1$ at $(i,j)$ and $(j,i)$ (or a single $1$ if $i=j$ ). Then $\{\mathbb{F_{ij}}\mathrel{\mathop{\ordinarycolon}}1\leq i\leq j\leq n-1\}$ is a orthonormal basis for $T_{F(k)}\mathcal{A}$ .

We calculate

	$\displaystyle D_{k}F(\mathbb{E_{ij}}k)$	$\displaystyle=\lim_{t\to 0}\frac{1}{t}\left[\Pi((\exp t\mathbb{E_{ij}})k.\lambda)-\Pi(k.\lambda)\right]$
		$\displaystyle=\Pi([\mathbb{E_{ij}},k.\lambda])$
		$\displaystyle=\begin{cases}(\mu_{j}-\mu_{i})\mathbb{F}_{ij}&j<n\\ \sum_{h=1}^{n-1}z_{h}(1+\delta_{ih})\mathbb{F}_{ih}&j=n\end{cases}.$

The Jacobian determinant has one nonzero term (the one that associates $\mathbb{E_{ij}}$ to $\mathbb{F_{ij}}$ for $j<n$ and $\mathbb{F_{ii}}$ for $j=n$ ). Hence

\displaystyle\det D_{k}F=\prod_{1\leq i<j\leq n-1}(\mu_{j}-\mu_{i})\prod_{1\leq i\leq n-1}2z_{i}

\displaystyle=\pm 2^{n-1}\prod_{\begin{subarray}{c}1\leq i\leq n\\ 1\leq j\leq n-1\end{subarray}}\mathinner{\!\left\lvert\lambda_{i}-\mu_{j}\right\rvert}^{1/2},

which is nonvanishing everywhere. Thus $F$ defines a smooth $2^{2n-2}$ -fold covering, and pushing forward along $F$ we see that the left hand side of (5) is proportional to (6).

On the other hand, define

	$\displaystyle G\mathrel{\mathop{\ordinarycolon}}SO(n-1)\times\mathcal{M}\to\mathcal{A}$
	$\displaystyle G(k^{\prime},\mu)=k^{\prime}.\mu.$

It suffices to compute the differential at $k^{\prime}=1$ . Take $\{\mathbb{E_{ij}^{\prime}}\mathrel{\mathop{\ordinarycolon}}1\leq i<j\leq n-1\}$ as a basis for $T_{1}SO(n-1)$ and let $(e_{i}^{\prime})_{1\leq i\leq n-1}$ be the obvious basis for $T_{\mu}\mathcal{M}$ . Then

	$\displaystyle D_{(1,\mu)}G(\mathbb{E_{ij}^{\prime}},0)$	$\displaystyle=[\mathbb{E_{ij}^{\prime}},\mu]=(\mu_{j}-\mu_{i})\mathbb{F_{ij}^{\prime}}$
	$\displaystyle D_{(1,\mu)}G(0,e_{i}^{\prime})$	$\displaystyle=\mathbb{F_{ii}^{\prime}}$

and the Jacobian determinant is $\pm\prod_{1\leq i<j\leq n-1}\mathinner{\!\left\lvert\mu_{i}-\mu_{j}\right\rvert}$ , so the right hand side of (5) pushes forward to (6).

∎

Define

J_{n}(\lambda)\mathrel{\mathop{\ordinarycolon}}=\int_{\mathcal{M}}A_{n-1}(\mu)J(\mu)\,d\mu.

The next step is to show $I_{n}(\sqrt{n-1}\lambda)\ll J_{n}(\lambda)\ll I_{n}(\lambda/\sqrt{n})$ . Then it will suffice to prove $J_{n}(\lambda)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}L_{n}(\lambda)$ . Applying Lemma 3 to Equation (4) we get

	$\displaystyle I_{n}(\lambda)$	$\displaystyle\ll\int_{\mathcal{M}}\int_{SO(n-1)}B^{\prime}(k^{\prime}.\mu)\,dk^{\prime}J(\mu)\,d\mu$
		$\displaystyle=\int_{\mathcal{M}}I_{n-1}(\mu)J(\mu)\,d\mu.$

By Equation (3) we may restrict to the region where $\mathinner{\!\left\lvert\operatorname{\operatorname{Tr}}\mu\right\rvert}<\sqrt{n-1}$ , in which we have $I_{n-1}(\mu)=I_{n-1}(\mu_{0}/\sqrt{1-(\operatorname{\operatorname{Tr}}\mu)^{2}/(n-1)})$ , where $\mu_{0}$ is the tracefree part of $\mu$ . By monotonicity, this is $\leq I_{n-1}(\mu_{0})$ , by induction $I_{n-1}(\mu_{0})\asymp A_{n-1}(\mu_{0})$ and since $\mathinner{\!\left\lVert\mu_{0}-\mu\right\rVert}$ is bounded, $A_{n-1}(\mu_{0})\asymp A_{n-1}(\mu)$ . The upper bound now reads

I_{n}(\lambda)\ll\int_{\begin{subarray}{c}\mu\in\mathcal{M}\\ \mathinner{\!\left\lvert\operatorname{\operatorname{Tr}}\mu\right\rvert}<\sqrt{n-1}\end{subarray}}A_{n-1}(\mu)J(\mu)d\mu.

Scaling $\lambda$ and $\mu$ by $\sqrt{n-1}$ this is equivalent to

\displaystyle I_{n}(\sqrt{n-1}\lambda)

\displaystyle\ll\int_{\begin{subarray}{c}\mu\in\mathcal{M}\\ \mathinner{\!\left\lvert\operatorname{\operatorname{Tr}}\mu\right\rvert}<1\end{subarray}}A_{n-1}(\sqrt{n-1}\mu)J(\mu)d\mu,

and since $A_{n-1}(\sqrt{n-1}\mu)\asymp A_{n-1}(\mu)$ the RHS is $\asymp J_{n}(\lambda)$ , so $I_{n}(\sqrt{n-1}\lambda)\ll J_{n}(\lambda)$ .

On the other hand, for the lower bound we have

	$\displaystyle I_{n}(\lambda)$	$\displaystyle\gg\int_{\mathcal{M}}\int_{SO(n-1)}B^{\prime}(\sqrt{n}k^{\prime}.\mu)\,dk^{\prime}J(\mu)\,d\mu$
		$\displaystyle=\int_{\mathcal{M}}I_{n-1}(\sqrt{n}\mu)J(\mu)\,d\mu,$

and we may restrict the integral to the region where $\operatorname{\operatorname{Tr}}\mu<1/\sqrt{n}$ , in which

	$\displaystyle I_{n-1}(\sqrt{n}\mu)$	$\displaystyle=I_{n-1}(\sqrt{n}\mu_{0}/\sqrt{1-(\operatorname{\operatorname{Tr}}\sqrt{n}\mu)^{2}/(n-1)})$
		$\displaystyle\geq I_{n-1}(\sqrt{n}\mu_{0}/\sqrt{(n-2)/(n-1)})$
		$\displaystyle\asymp A_{n-1}(\mu_{0})$
		$\displaystyle\asymp A_{n-1}(\mu).$

\displaystyle I_{n}(\lambda)\gg\int_{\begin{subarray}{c}\mu\in\mathcal{M}\\ \mathinner{\!\left\lvert\operatorname{\operatorname{Tr}}\mu\right\rvert}<1/\sqrt{n}\end{subarray}}A_{n-1}(\mu)J(\mu)\,d\mu,

and scaling $\lambda$ and $\mu$ by $1/\sqrt{n}$ we get $I_{n}(\lambda/\sqrt{n})\gg J_{n}(\lambda)$ .

Our strategy for showing $J_{n}(\lambda)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}L_{n}(\lambda)$ will be to simplify $J_{n}(\lambda)$ to a convolution. Define

	$\displaystyle\mathcal{H}$	$\displaystyle\mathrel{\mathop{\ordinarycolon}}={\textstyle\{\mu\in\mathcal{M}\mathrel{\mathop{\ordinarycolon}}\mathinner{\!\left\lvert\sum_{i}\mu_{i}\right\rvert}<1\}}$
	$\displaystyle F_{i}(x)$	$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\mathbf{1}_{[\lambda_{i},\lambda_{i+1}]}\mathinner{\!\left\lvert\lambda_{i}-x\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{i+1}-x\right\rvert}^{-1/2}$
	$\displaystyle\kappa_{i,j}$	$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\mathinner{\!\left\lvert\mu_{i}-\mu_{j}\right\rvert}\mathinner{\!\left\lvert\lambda_{i}-\mu_{j}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{j+1}-\mu_{i}\right\rvert}^{-1/2}.$

Then

(8)

\displaystyle J_{n}(\lambda)=\int_{\mathcal{H}}(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}L_{n-1}(\mu)\prod_{1\leq i<j\leq n-1}\kappa_{i,j}\prod_{1\leq i\leq n-1}F_{i}(\mu_{i})\,d\mu.

Note that

F_{i}(x)\asymp\mathbb{1}_{[\lambda_{i},\lambda_{i+1}]}\mathinner{\!\left\lvert\lambda_{i}-\lambda_{i+1}\right\rvert}^{-1/2}\left(\mathinner{\!\left\lvert\lambda_{i}-x\right\rvert}^{-1/2}+\mathinner{\!\left\lvert\lambda_{i+1}-x\right\rvert}^{-1/2}\right),

implying $\int F_{i}(x)dx\asymp 1$ . Also, $\kappa_{i,j}\leq 1$ on $\mathcal{H}$ . If we can eliminate the factors $(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}$ , $L_{n-1}(\mu)$ , and $\kappa_{i,j}$ then the integral becomes simply a convolution evaluated at 0:

	$\displaystyle\int_{\mathcal{H}}\prod_{1\leq i\leq n-1}F_{i}(\mu_{i})\,d\mu$	$\displaystyle=\int\dots\int{\textstyle\mathbf{1}_{[-1,1]}\left(-\sum_{i}\mu_{i}\right)}\prod_{1\leq i\leq n-1}F_{i}(\mu_{i})\,d\mu_{n-1}\dots d\mu_{1}$
		$\displaystyle=\left(\mathbf{1}_{[-1,1]}\left(\operatorname{\text{\scalebox{1.5}{$\ast$}}}_{1\leq i\leq n-1}F_{i}\right)\right)(0).$

Let $d=\lambda_{n}-\lambda_{1}$ and let $d_{i}=\lambda_{i+1}-\lambda_{i}$ be the “spectral gaps.” Then Lemma 1 follows from the following cases:

First, by Lemma 6, we have unconditionally that $J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ .

•
Let $I$ be the index of the largest gap. Suppose $d_{I}>(1-1/100n)d$ . Then:
- –
  
  If $(n,I)\neq(4,2)$ , then $L_{n}(\lambda)\asymp 1$ and it remains to show $J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ . See Lemma 9.
- –
  
  If $(n,I)=(4,2)$ , then the exceptional term term in $L_{n}(\lambda)$ is large. See Lemma 10.
•
Otherwise, let $I<J$ be the two largest gaps, breaking ties arbitrarily. Then $(1-1/100n)d\geq d_{I},d_{J}\geq d/100n^{2}$ .
- –
  
  If $(I,J)\neq(1,n-1)$ then $L_{n}(\lambda)\asymp 1$ and it remains to show $J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ . See Lemma 11.
- –
  
  If $(I,J)=(1,n-1)$ then the main term in $L_{n}(\lambda)$ is large. See Lemma 13.

4. Convolution Bounds

4.1. The general lower bound

Lemma 6.

For all $\lambda$ with $\lambda_{1}<\lambda_{2}<\dots<\lambda_{n}$ , we have $J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ .

Proof.

Applying the inequalities $L_{n}(\mu)\gg 1$ and $(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}$ to the definition of $J_{n}(\lambda)$ gives

J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}\int_{\mathcal{H}}\prod_{1\leq i<j\leq n-1}\kappa_{i,j}\prod_{1\leq i\leq n-1}F_{i}(\mu_{i})\,d\mu.

We define a subset of $\mathcal{H}$ on which the $\kappa_{i,j}$ are bounded away from 0. Let $\theta=-\lambda_{1}/d$ . From $(n-1)\lambda_{1}+\lambda_{n}\leq\lambda_{1}+\lambda_{2}+\dots+\lambda_{n}=0$ we deduce $\lambda_{1}\leq-d/n$ and similarly $\lambda_{1}+(n-1)\lambda_{n}\geq\lambda_{1}+\lambda_{2}+\dots+\lambda_{n}=0\implies(n-1)d\geq-n\lambda_{1}$ . Thus $1/n\leq\theta\leq 1-1/n$ . Let $m_{i}=\theta\lambda_{i}+(1-\theta)\lambda_{i+1}$ . Then $\sum_{1\leq i\leq n-1}m_{i}=0$ . Define intervals $M_{i}\subset[\lambda_{i},\lambda_{i+1}]$ centered on the $m_{i}$ , given by

M_{i}=[m_{i}-d_{i}/2n,m_{i}+d_{i}/2n]

and restrict the integral to the set

\mathcal{H}^{\prime}=\{\mu\mathrel{\mathop{\ordinarycolon}}\mu_{i}\in M_{i},{\textstyle\mathinner{\!\left\lvert\sum_{i}\mu_{i}\right\rvert}}<1\}\subset\mathcal{H}.

Then for $\mu\in\mathcal{H}^{\prime}$ we have

\frac{\mathinner{\!\left\lvert\mu_{i}-\mu_{j}\right\rvert}}{\mathinner{\!\left\lvert\mu_{i}-\lambda_{j+1}\right\rvert}}=\frac{\mathinner{\!\left\lvert\mu_{i}-\lambda_{j}\right\rvert}+\mathinner{\!\left\lvert\lambda_{j}-\mu_{j}\right\rvert}}{\mathinner{\!\left\lvert\mu_{i}-\lambda_{j}\right\rvert}+\mathinner{\!\left\lvert\lambda_{j}-\lambda_{j+1}\right\rvert}}\geq\frac{\mathinner{\!\left\lvert\lambda_{j}-\mu_{j}\right\rvert}}{\mathinner{\!\left\lvert\lambda_{j}-\lambda_{j+1}\right\rvert}}\geq\frac{1}{2n},

and similarly $\mathinner{\!\left\lvert\mu_{i}-\mu_{j}\right\rvert}/\mathinner{\!\left\lvert\lambda_{i}-\mu_{j}\right\rvert}\geq 1/2n$ , so $\kappa_{ij}\geq 1/2n$ . Also, for $\mu_{i}\in M_{i}$ we have

\displaystyle F_{i}(\mu_{i})\asymp\mathinner{\!\left\lvert\lambda_{i}-\lambda_{i+1}\right\rvert}^{-1/2}\left(\mathinner{\!\left\lvert\lambda_{i}-\mu_{i}\right\rvert}^{-1/2}+\mathinner{\!\left\lvert\lambda_{i+1}-\mu_{i}\right\rvert}^{-1/2}\right)\gg d_{i}^{-1}.

Applying the lower bounds for $\kappa_{i,j}$ and $F_{i}$ gives

J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}\int_{\mathcal{H}^{\prime}}\prod_{1\leq i\leq n-1}d_{i}^{-1}\mathbf{1}_{M_{i}}(\mu_{i})d\mu_{i},

and this is equivalent to a convolution

	$\displaystyle J_{n}(\lambda)$	$\displaystyle\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}G(0)$
	$\displaystyle G$	$\displaystyle=\mathbf{1}_{[-1,1]}\left(\operatorname{\text{\scalebox{1.5}{$\ast$}}}_{1\leq i\leq n-1}d_{i}^{-1}\mathbf{1}_{M_{i}}\right).$

Since $\sum_{i}m_{i}=0$ we can translate each factor to be centered at 0,

\displaystyle G

\displaystyle=\mathbf{1}_{[-1,1]}*\left(\operatorname*{\text{\scalebox{1.5}{$\ast$}}}_{1\leq i\leq n-1}d_{i}^{-1}\mathbf{1}_{[-d_{i}/2n,d_{i}/2n]}\right).

Then by the following Lemma, $G(x)$ is maximized at $x=0$ .

Lemma 7.

For a sequence of reals $a_{i}\geq 0$ let

f_{n}(x)=\operatorname*{\text{\scalebox{1.5}{$\ast$}}}_{1\leq i\leq n}\mathbf{1}_{[-a_{i},a_{i}]}(x)

then $f_{n}$ is maximized at 0 for all $n$ .

Proof.

The $f_{n}$ are obviously even. We prove a stronger statement, that $f_{n}$ is nonincreasing on $[0,\infty)$ . The case $n=1$ is trivial. For $n>1$ and $x>0$ we have

	$\displaystyle f_{n}(x)$	$\displaystyle=\int\mathbf{1}_{[-a_{n},a_{n}]}(x-y)f_{n-1}(y)dy$
		$\displaystyle=\int_{x-a_{n}}^{x+a_{n}}f_{n-1}(y)dy$

implying, for $0<x<x^{\prime}$

	$\displaystyle f_{n}(x)-f_{n}(x^{\prime})$	$\displaystyle=\int_{x-a_{n}}^{x+a_{n}}f_{n-1}(y)dy-\int_{x^{\prime}-a_{n}}^{x^{\prime}+a_{n}}f_{n-1}(y)dy$
		$\displaystyle=\int_{x-a_{n}}^{x^{\prime}-a_{n}}f_{n-1}(y)dy-\int_{x+a_{n}}^{x^{\prime}+a_{n}}f_{n-1}(y)dy$
		$\displaystyle=\int_{x}^{x^{\prime}}f_{n-1}(y-a_{n})-f_{n-1}(y+a_{n})dy,$

and by induction the integrand is $\geq 0$ . ∎

Finally, note that $\int G(x)dx=2\prod_{i}\mathinner{\!\left\lVert d_{i}^{-1}\mathbf{1}_{M_{i}}\right\rVert}_{1}\gg 1$ and $G(x)$ is supported in an interval of length $\ll\mathinner{\!\left\lVert\lambda\right\rVert}$ . Thus, $G(0)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}$ and we are done. ∎

4.2. Monotone rearrangement

To upper bound various convolutions, we introduce the following convenient tool [2]. For $f\mathrel{\mathop{\ordinarycolon}}\mathbb{R}\to[0,\infty)$ define the level set

\{f>t\}=\{x\in\mathbb{R}\mathrel{\mathop{\ordinarycolon}}f(x)>t\},

and the “layer cake representation” of $f$ ,

f(x)=\int\mathbf{1}_{\{f>t\}}(x)\,dt.

For a set $S\subset\mathbb{R}$ define the rearranged set $S^{*}=[0,\mu(S))$ . Finally define the monotone rearrangement

f^{*}(x)=\int\mathbf{1}_{\{f>t\}^{*}}(x)\,dt.

For example, the monotone rearrangement of $\mathbf{1}_{[a,b]}$ is $\mathbf{1}_{[0,b-a]}$ , and the monotone rearrangement of

\displaystyle f(x)

\displaystyle=\mathbf{1}_{[a,b]}\mathinner{\!\left\lvert a-x\right\rvert}^{-1/2}\mathinner{\!\left\lvert b-x\right\rvert}^{-1/2}

	$\displaystyle f^{*}(x)$	$\displaystyle=\mathbf{1}_{[0,b-a]}(x/2)^{-1/2}((b-a)-x/2)^{-1/2}$
		$\displaystyle\ll(b-a)^{-1/2}x^{-1/2}.$

We note basic properties such as $f\leq g\implies f^{*}\leq g^{*}$ and $(af)^{*}=af^{*}$ . The real workhorse is the $n$ -ary Hardy-Littlewood Rearrangement inequality:

Lemma 8.

For $f_{1},\dots,f_{n}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}\to[0,\infty)$ we have

\int\prod_{i=1}^{n}f_{i}(x)dx\leq\int\prod_{i=1}^{n}f_{i}^{*}(x)\,dx.

Proof.

	$\displaystyle\int_{\mathbb{R}}f_{1}(x)\dots f_{n}(x)\,dx$	$\displaystyle=\int_{\mathbb{R}}\int_{0}^{\infty}\dots\int_{0}^{\infty}\mathbf{1}_{\{f_{1}>t_{1}\}}(x)\dots\mathbf{1}_{\{f_{n}>t_{n}\}}(x)\,dt_{n}\dots\,dt_{1}\,dx$
		$\displaystyle=\int_{0}^{\infty}\dots\int_{0}^{\infty}\mu(\{f_{1}>t_{1}\}\cap\dots\cap\{f_{n}>t_{n}\})\,dt_{n}\dots\,dt_{1}$
		$\displaystyle\leq\int_{0}^{\infty}\dots\int_{0}^{\infty}\mu(\{f^{}_{1}>t_{1}\}\cap\dots\cap\{f^{}_{n}>t_{n}\})\,dt_{n}\dots\,dt_{1}$
		$\displaystyle=\int_{\mathbb{R}}f^{}_{1}(x)\dots f^{}_{n}(x)\,dx.$

∎

For $n=2$ this implies $\mathinner{\!\left\lVert f*g\right\rVert}_{\infty}\leq\langle f^{*},g^{*}\rangle$ , where $f*g$ is the convolution of $f$ and $g$ . We also note the following inequalities for $0<a<b$ :

(9)		$\displaystyle\int_{a}^{b}x^{-1/2}\left(\log^{\prime}\frac{T}{x}\right)^{k}\,dx$	$\displaystyle\ll_{k}b^{1/2}\left(\log^{\prime}\frac{T}{b}\right)^{k}$
(10)		$\displaystyle\int_{a}^{b}x^{-1}\left(\log^{\prime}\frac{T}{x}\right)^{k}\,dx$	$\displaystyle\ll\log\frac{b}{a}\left(\log^{\prime}\frac{T}{a}\right)^{k},$

the first of which can be verified by applying integration by parts $k$ times.

4.3. One Large Gap

In this section we assume there exists $I$ such that $d_{I}>(1-1/(100n))d$ .

Let $d^{\prime}=d-d_{I}=\mathinner{\!\left\lvert\lambda_{1}-\lambda_{I}\right\rvert}+\mathinner{\!\left\lvert\lambda_{I+1}-\lambda_{n}\right\rvert}\leq d/100n$ . We show that $n$ and $I$ determine the positions of $\lambda_{i}$ and $\mu_{i}$ up to an $O(d^{\prime})$ error. Write $\pm C$ for an error of absolute value $\leq C$ . Then $\lambda_{i}=\lambda_{1}\pm d^{\prime}$ for $i\leq I$ and $\lambda_{i}=\lambda_{n}\pm d^{\prime}$ for $i>I$ . Writing

	$\displaystyle 0$	$\displaystyle=\sum_{i}\lambda_{i}$
		$\displaystyle=I\lambda_{1}+(n-I)\lambda_{n}\pm nd^{\prime}$

we get $\lambda_{n}=Id/n\pm d^{\prime}$ and $\lambda_{1}=(I-n)d/n\pm d^{\prime}$ . Then

(11)

\displaystyle\lambda_{i}=\begin{cases}\displaystyle{\frac{(I-n)d}{n}\pm 2d^{\prime}}&i\leq I\\ \displaystyle{\frac{Id}{n}\pm 2d^{\prime}}&i>I\end{cases}

and (using $\mathinner{\!\left\lvert\sum_{i}\mu_{i}\right\rvert}\leq 1$ to find $\mu_{I}$ )

(12)

\displaystyle\mu_{i}=\begin{cases}\displaystyle{\frac{(I-n)d}{n}\pm 2d^{\prime}}&i<I\\ \displaystyle{\frac{(I-2n)d}{n}\pm(2nd^{\prime}+1)}&i=I\\ \displaystyle{\frac{Id}{n}\pm 2d^{\prime}}&i>I\end{cases}

4.3.1. The upper bound when $(n,I)\neq(4,2)$

Lemma 9.

Assume there exists $I$ such that $d_{I}>(1-1/(100n))d$ and $(n,I)\neq(4,2)$ . Then $L_{n}(\lambda)\asymp 1$ and $J_{n}(\lambda)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ .

By Equation 11 we have $\mathinner{\!\left\lvert\lambda_{2}\right\rvert}\gg d$ so

\displaystyle\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\ll 1.

Also, if $n=4$ but $I\neq 2$ then at least one of $\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}$ and $\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}$ is $\gg d$ so

\displaystyle\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}}\ll 1.

Thus $L_{n}(\lambda)\asymp 1$ , and by Lemma 6, $J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ , so it remains to show $J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ .

If $I\geq 2$ then $\mathinner{\!\left\lvert\mu_{1}\right\rvert}\gg d$ , otherwise, if $I=1$ then $\mathinner{\!\left\lvert\mu_{n-1}\right\rvert}\gg d$ . Thus $(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}$ .

If $n=4$ and $I\neq 2$ then $\mathinner{\!\left\lvert\mu_{2}\right\rvert}\gg d$ and $L_{n-1}(\mu)\asymp 1$ .

If $n\geq 5$ , then at least one of $\mathinner{\!\left\lvert\mu_{2}\right\rvert}$ and $\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}$ must be $\gg d$ , so

\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}}\asymp 1.

If $n=5$ then the exceptional factor in $L_{n-1}(\mu)$ is also $\asymp 1$ since at least one of $\mathinner{\!\left\lvert\mu_{1}-\mu_{2}\right\rvert}$ and $\mathinner{\!\left\lvert\mu_{3}-\mu_{4}\right\rvert}$ must be $\gg d$ . Thus $L_{n-1}(\mu)\asymp 1$ for $n\geq 5$ .

Bounding $\kappa_{i,j}\leq 1$ for all $i,j$ we get

J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}\int_{\mathcal{H}}\prod_{i}F_{i}(\mu_{i})\,d\mu.

By Equation (12) we have $\mu_{I}=(I-2n)d/n\pm(2nd^{\prime}+1)$ . Then (taking $d$ sufficiently large) $\mathinner{\!\left\lvert\lambda_{I}-\mu_{I}\right\rvert}$ and $\mathinner{\!\left\lvert\lambda_{I+1}-\mu_{I}\right\rvert}$ are $\gg d$ so $F_{I}(t)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}$ . Finally

	$\displaystyle J_{n}(\lambda)$	$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\int_{\mathcal{H}}\prod_{i\neq I}F_{i}(\mu_{i})\,d\mu$
		$\displaystyle=\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\left(\mathbf{1}_{[-1,1]}\left(\operatorname{\text{\scalebox{1.5}{$\ast$}}}_{i\neq I}F_{i}\right)\right)(0)$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\prod_{i\neq I}\mathinner{\!\left\lVert F_{i}\right\rVert}_{1}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$

and we are done.

4.3.2. The case $(n,I)=(4,2)$

Lemma 10.

Assume there exists $I$ such that $d_{I}>(1-1/(100n))d$ and $(n,I)=(4,2)$ . Then

L_{n}(\lambda)\asymp\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}}

, and $J_{n}(\lambda)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n_{1}}L_{n}(\lambda)$ .

The proof proceeds similarly to Lemma 9, except that we cannot eliminate $L_{n-1}(\mu)=\log^{\prime}\mathinner{\!\left\lVert\lambda\right\rVert}/(1+\mathinner{\!\left\lvert\mu_{2}\right\rvert})$ . We write instead

J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}\int_{-\infty}^{\infty}L(\mu_{2})G(\mu_{2})\,d\mu_{2}.

where

	$\displaystyle L(t)$	$\displaystyle=\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert t\right\rvert}}$
	$\displaystyle G$	$\displaystyle=\mathbf{1}_{[-1,1]}F_{1}F_{3}.$

Using Young’s inequality and the fact that $\mathinner{\!\left\lVert F_{i}\right\rVert}_{1}\asymp 1$ we have

	$\displaystyle J_{n}(\lambda)$	$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\mathinner{\!\left\lVert L\mathbf{1}_{[-1,1]}F_{1}*F_{3}\right\rVert}_{\infty}$
		$\displaystyle\leq\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\mathinner{\!\left\lVert L*F_{1}\right\rVert}_{\infty}\mathinner{\!\left\lVert\mathbf{1}_{[-1,1]}\right\rVert}_{1}\mathinner{\!\left\lVert F_{3}\right\rVert}_{1}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\mathinner{\!\left\lVert L*F_{1}\right\rVert}_{\infty}.$

Then using the Hardy-Littlewood rearrangement inequality and equation (9):

	$\displaystyle\mathinner{\!\left\lVert L*F_{1}\right\rVert}_{\infty}$	$\displaystyle\leq\langle L^{},F^{}_{1}\rangle$
		$\displaystyle\ll\int_{0}^{d_{1}}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert x\right\rvert}}d_{1}^{-1/2}x^{-1/2}\,dx$
		$\displaystyle\ll\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+d_{1}},$

so $J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+d_{1}}$ and similarly $J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+d_{3}}$ . Together these imply

J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}}

as desired. For the lower bound, we restrict to $\mathcal{H}^{\prime}$ , then since $\kappa_{i,j}\gg 1$ for all $i<j$ , and $(1+\mathinner{\!\left\lVert\mu\right\rVert}^{-2}\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-2}$ we have

\displaystyle J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-2}\int_{\mathcal{H}^{\prime}}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{2}\right\rvert}}\prod_{1\leq i\leq 3}F_{i}(\mu_{i})\,d\mu.

By Equation (12) we have $\mathinner{\!\left\lvert\mu_{2}\right\rvert}\leq 1+4d^{\prime}$ for $\mu\in\mathcal{H}^{\prime}$ , so

\displaystyle J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-2}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{2+4d^{\prime}}\int_{\mathcal{H}^{\prime}}\prod_{1\leq i\leq 3}F_{i}(\mu_{i})\,d\mu

and using the same argument as in Lemma 6 the integral is $\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}$ . Finally since $d^{\prime}=\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}$ we have

\displaystyle J_{n}(\lambda)\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-3}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}}.

4.4. Two Large Gaps

In this section we assume no index $I$ satisfies $d_{I}>(1-1/100n)d$ . This implies there exist indices $I<J$ such that $d_{I},d_{J}\geq d/100n^{2}$ .

4.4.1. The case $(I,J)\neq(1,n-1)$

Lemma 11.

Suppose there exist $I<J$ with $(I,J)\neq(1,n-1)$ such that $d_{I},d_{J}\geq d/100n^{2}$ . Then $L_{n}(\lambda)\asymp 1$ and $J_{n}(\lambda)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ .

Proof.

If $\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}\leq d/100n^{2}$ , then $d_{i}<d/100n^{2}$ for $2\leq i\leq n-2$ , forcing $(I,J)=(1,n-1)$ . Hence $\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ and the main term in $L_{n}(\lambda)$ is $\ll 1$ . The exceptional term of $L_{n}(\lambda)$ is also $\ll 1$ since one of $\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert},\mathinner{\!\left\lvert\lambda_{3}-\lambda_{4}\right\rvert}$ must be $d_{I}$ or $d_{J}$ . It remains to prove $J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}$ .

If $n=3$ then $(I,J)=(1,2)$ is forced, so we must have $n\geq 4$ . By symmetry (replacing $\lambda_{1},\dots,\lambda_{n}$ with $-\lambda_{n},\dots,-\lambda_{1}$ ) we may assume $J\leq n-2$ .

First we rewrite $L_{n-1}(\mu)$ to depend solely on $\mu_{J}$ :

Lemma 12.

Under the assumptions of Lemma 11,

L_{n-1}(\mu)\ll\mathcal{L}(\mu_{J})\mathrel{\mathop{\ordinarycolon}}=\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{J}-\mu_{J}\right\rvert}}\right)\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{J+1}-\mu_{J}\right\rvert}}\right)^{n-2}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\mu_{J}\right\rvert}}\right)^{n-3}.

Proof.

If $J=n-2$ then bound

\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}}\right)^{n-3}\ll\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\mu_{J}\right\rvert}}\right)^{n-3}.

Otherwise, suppose $J\leq n-3$ . Since $1\leq I<J$ we must have $n\geq 5$ . If $(I,J)=(1,2)$ then $\mathinner{\!\left\lvert\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}\geq\mathinner{\!\left\lvert\mu_{2}-\mu_{3}\right\rvert}\geq\mathinner{\!\left\lvert\mu_{2}-\lambda_{3}\right\rvert}=\mathinner{\!\left\lvert\lambda_{J+1}-\mu_{J}\right\rvert}$ and

\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}}\right)^{n-3}\ll\log^{\prime}\left(\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{J+1}-\mu_{J}\right\rvert}}\right)^{n-3}.

Otherwise, if $(I,J)\neq(1,2)$ then $n\geq 6$ is forced, and $\mathinner{\!\left\lvert\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}\geq\mathinner{\!\left\lvert\mu_{2}-\mu_{n-2}\right\rvert}\geq\mathinner{\!\left\lvert\lambda_{3}-\lambda_{n-2}\right\rvert}\geq d_{J}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ , so

\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}}\right)^{n-3}\ll 1.

As for the exceptional factor in $L_{n-1}(\mu)$ that appears when $n=5$ , if $J=3$ we bound

\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{1}-\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{3}-\mu_{4}\right\rvert}}\ll\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{J+1}-\mu_{J}\right\rvert}},

and if $J=2$ we bound

\ll\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{J}-\mu_{J}\right\rvert}}.

∎

Since $\mathinner{\!\left\lVert\mu\right\rVert}\gg\mathinner{\!\left\lvert\mu_{1}-\mu_{n-1}\right\rvert}\gg d_{J}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ , we have $(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}$ .

Since $\mathinner{\!\left\lvert\lambda_{J+1}-\mu_{I}\right\rvert}\geq d_{J}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ we have

	$\displaystyle\kappa_{I,J}$	$\displaystyle=\mathinner{\!\left\lvert\lambda_{I}-\mu_{J}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{J+1}-\mu_{I}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\mu_{I}-\mu_{J}\right\rvert}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1/2}\mathinner{\!\left\lvert\mu_{I}-\mu_{J}\right\rvert}^{1/2},$

and since $\mathinner{\!\left\lvert\lambda_{J}-\mu_{n-1}\right\rvert}\geq d_{J}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ we have

	$\displaystyle\kappa_{J,n-1}$	$\displaystyle=\mathinner{\!\left\lvert\lambda_{J}-\mu_{n-1}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{n}-\mu_{J}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\mu_{J}-\mu_{n-1}\right\rvert}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1/2}\mathinner{\!\left\lvert\mu_{J}-\mu_{n-1}\right\rvert}^{1/2}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1/2}\mathinner{\!\left\lvert\lambda_{n}-\mu_{J}\right\rvert}^{1/2}.$

Applying all the above inequalities, and $\kappa_{i,j}\leq 1$ for all other pairs $i,j$ , to Equation (8) and converting to a convolution we have

J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\int F(t)G(-t)\,dt

where

	$\displaystyle F(t)$	$\displaystyle=\int_{\mu_{I}+\mu_{J}=t}F_{I}(\mu_{I})F_{J}(\mu_{J})\mathinner{\!\left\lvert\mu_{I}-\mu_{J}\right\rvert}^{1/2}\mathinner{\!\left\lvert\mu_{J}-\lambda_{n}\right\rvert}^{1/2}\mathcal{L}(\mu_{J})\,d\mu_{J}$
	$\displaystyle G(s)$	$\displaystyle=\left(\mathbf{1}_{[-1,1]}\left(\operatorname{\text{\scalebox{1.5}{$\ast$}}}_{i\neq I,J}F_{i}\right)\right)(s).$

The estimation of $F(t)$ is contained in Lemma 14, with

\displaystyle A=\lambda_{I}

\displaystyle B=\lambda_{I+1}

\displaystyle C=\lambda_{J}

\displaystyle D=\lambda_{J+1}

\displaystyle E=\lambda_{n}

\displaystyle T=d,

\displaystyle k=2n-4

giving

	$\displaystyle F(t)\ll\left(\log^{\prime}\frac{\mathinner{\!\left\lvert\lambda_{J}-\lambda_{I+1}\right\rvert}}{\mathinner{\!\left\lvert\lambda_{I+1}+\lambda_{J}-t\right\rvert}}\right)^{k+1}+\left(\log^{\prime}\frac{\mathinner{\!\left\lvert\lambda_{n}-\lambda_{J+1}\right\rvert}}{\mathinner{\!\left\lvert\lambda_{I}+\lambda_{J+1}-t\right\rvert}}\right)^{k+1}+$
	$\displaystyle\left(\log^{\prime}\frac{d}{\mathinner{\!\left\lvert\lambda_{I}+\lambda_{J}-t\right\rvert}}\right)^{k}+\left(\log^{\prime}\frac{d}{\mathinner{\!\left\lvert\lambda_{I+1}+\lambda_{J+1}-t\right\rvert}}\right)^{k}.$

Let $d^{\prime}=d-d_{I}-d_{J}$ . Let $H_{1}(t)$ be the sum of the first two log terms and $H_{2}(t)$ be the sum of the second two log terms. We have $\mathinner{\!\left\lvert\lambda_{J}-\lambda_{I+1}\right\rvert},\mathinner{\!\left\lvert\lambda_{n}-\lambda_{J+1}\right\rvert}<d^{\prime}$ so $H^{*}_{1}(t)\ll(\log^{\prime}\frac{d^{\prime}}{t})^{k+1}$ . Let $d_{K}$ be the next largest gap after $d_{I}$ and $d_{J}$ , we have $d_{K}\geq d^{\prime}/(n-3)$ . Then

	$\displaystyle\int H_{1}(t)G(-t)\,dt$	$\displaystyle\ll\mathinner{\!\left\lVert H_{1}*F_{K}\right\rVert}_{\infty}\prod_{i\neq I,J,K}\mathinner{\!\left\lVert F_{i}\right\rVert}_{1}$
		$\displaystyle\ll\langle H^{}_{1},F^{}_{K}\rangle$
		$\displaystyle\ll\int_{0}^{d_{K}}\left(\log^{\prime}\frac{d^{\prime}}{x}\right)^{k+1}d_{K}^{-1/2}x^{-1/2}\,dx$
		$\displaystyle\ll\left(\log^{\prime}\frac{d^{\prime}}{d_{K}}\right)^{k+1}$
		$\displaystyle\ll 1,$

as desired. To bound $\int H_{2}(t)G(-t)dt$ there are two cases. First suppose $d^{\prime}\geq d/(100n)$ . Then

	$\displaystyle\int H_{2}(t)G(-t)\,dt$	$\displaystyle\ll\langle H^{}_{2},F^{}_{K}\rangle$
		$\displaystyle\ll\int_{0}^{d_{K}}\left(\log^{\prime}\frac{d}{x}\right)^{k}d^{-1/2}_{K}x^{-1/2}\,dx$
		$\displaystyle\ll\left(\log^{\prime}\frac{d}{d_{K}}\right)^{k}$
		$\displaystyle\ll 1,$

since $d_{K}\gg d$ . Otherwise $d^{\prime}<d/(100n)$ . If $G(-t)\neq 0$ then

	$\displaystyle t$	$\displaystyle\in-\operatorname{supp}G$
		$\displaystyle\subset[-1,1]+\sum_{\begin{subarray}{c}1\leq i\leq n-1\\ i\neq I,J\end{subarray}}[-\lambda_{i+1},-\lambda_{i}]$
		$\displaystyle=[-1+\lambda_{1}+\lambda_{I+1}+\lambda_{J+1},\,\,1+\lambda_{n}+\lambda_{I}+\lambda_{J}].$

Recall that $\lambda_{1}<-d/n$ . Subtracting $\lambda_{I}+\lambda_{J}$ from the lower bound for $t$ we get

	$\displaystyle t-\lambda_{I}-\lambda_{J}$	$\displaystyle\geq-1+\lambda_{1}+d_{I}+d_{J}$
		$\displaystyle\geq-1-d/n+d_{I}+d_{J}$
		$\displaystyle\geq-1+(1-1/100n-1/n)d$

so for sufficiently large $\lambda$ , $\mathinner{\!\left\lvert t-\lambda_{I}-\lambda_{J}\right\rvert}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ , and similarly $\mathinner{\!\left\lvert t-\lambda_{I+1}+\lambda_{J+1}\right\rvert}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ . Hence $H_{2}(t)\ll 1$ on the support of $G(-t)$ and $\int H_{2}(t)G(-t)\,dt\ll 1$ as desired. ∎

4.4.2. The case $(I,J)=(1,n-1)$

Lemma 13.

Let $(I,J)=(1,n-1)$ and suppose $d_{I},d_{J}\geq d/100n^{2}$ . Then

L_{n}(\lambda)\asymp\left(\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\right)^{n-2},

and $J_{n}(\lambda)\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}L_{n}(\lambda)$ .

Proof.

Clearly the exceptional term in $L_{4}(\lambda)$ is trivial when $d_{1},d_{3}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ . We split into three cases: the upper bound for $n=3$ , the upper bound for $n\geq 4$ , and the lower bound.

4.4.3. The upper bound when $n=3$

Since $\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert},\mathinner{\!\left\lvert\lambda_{2}-\lambda_{3}\right\rvert}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ we have

	$\displaystyle(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-1}\kappa_{1,2}$	$\displaystyle\ll\mathinner{\!\left\lvert\lambda_{1}-\mu_{2}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{3}-\mu_{1}\right\rvert}^{-1/2}$
		$\displaystyle\ll\mathinner{\!\left\lvert\lambda_{1}-\lambda_{2}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{3}-\lambda_{2}\right\rvert}^{-1/2}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}.$

Then

\displaystyle J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\int_{-1}^{1}\int_{\begin{subarray}{c}\mu_{1}+\mu_{2}=t\\ \lambda_{1}\leq\mu_{1}\leq\lambda_{2}\leq\mu_{2}\leq\lambda_{3}\end{subarray}}\mathinner{\!\left\lvert\lambda_{1}-\mu_{1}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{2}-\mu_{1}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{2}-\mu_{2}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{3}-\mu_{2}\right\rvert}^{-1/2}\,d\mu_{1}\,dt.

By Lemma 15 the inner integral is

\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert 2\lambda_{2}-t\right\rvert}}+\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{1}+\lambda_{3}-t\right\rvert}}\right)=\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert 2\lambda_{2}-t\right\rvert}}+\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{2}+t\right\rvert}}\right),

and

	$\displaystyle J_{n}(\lambda)$	$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-2}\int_{-1}^{1}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{2}+t\right\rvert}}+\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert 2\lambda_{2}-t\right\rvert}}\,dt$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-2}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}}$

as desired.

For $2\leq i\leq n-2$ we have

	$\displaystyle\kappa_{1,i}$	$\displaystyle=\mathinner{\!\left\lvert\mu_{1}-\mu_{i}\right\rvert}\mathinner{\!\left\lvert\lambda_{i+1}-\mu_{1}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{1}-\mu_{i}\right\rvert}^{-1/2}$
		$\displaystyle\ll\mathinner{\!\left\lvert\mu_{1}-\mu_{i}\right\rvert}\mathinner{\!\left\lvert\lambda_{i+1}-\mu_{1}\right\rvert}^{-1/2}\mathinner{\!\left\lVert\lambda\right\rVert}^{-1/2}$
		$\displaystyle\ll\mathinner{\!\left\lvert\mu_{1}-\mu_{i}\right\rvert}^{1/2}\mathinner{\!\left\lVert\lambda\right\rVert}^{-1/2}$

and similarly $\kappa_{i,n}\ll\mathinner{\!\left\lvert\mu_{i}-\mu_{n}\right\rvert}^{1/2}\mathinner{\!\left\lVert\lambda\right\rVert}^{-1/2}$ so by AM-GM, $\kappa_{1,i}\kappa_{i,n}\ll\mathinner{\!\left\lvert\mu_{1}-\mu_{n-1}\right\rvert}\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}$ . Also, $\kappa_{1,n}\ll\mathinner{\!\left\lvert\mu_{1}-\mu_{n-1}\right\rvert}\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}$ . We can use these factors to cancel $(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}$ , and the extra factor in $L_{n-1}(\mu)$ if $n=5$ :

	$\displaystyle\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{1}-\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{3}-\mu_{4}\right\rvert}}\right)$	$\displaystyle(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}\kappa_{1,n-1}\prod_{2\leq i\leq n-2}\kappa_{1,i}\kappa_{i,n-1}$
		$\displaystyle\ll\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{\mathinner{\!\left\lvert\mu_{1}-\mu_{2}\right\rvert}}\mathinner{\!\left\lVert\mu\right\rVert}^{-1/2}\mathinner{\!\left\lvert\mu_{1}-\mu_{2}\right\rvert}^{1/2}\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2},$

where in the last line we use the fact that $x\log(1/x)\leq 1$ for $x\leq 1$ . Bound $\kappa_{i,j}\leq 1$ for all remaining $i,j$ .

Bound

\displaystyle\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{2}\right\rvert}+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}}\right)^{n-3}

\displaystyle\ll\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert\mu_{n-2}\right\rvert}}\right)^{n-3}.

Write

\displaystyle J_{n}(\lambda)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}\int F(t)G(-t)\,dt

where

	$\displaystyle F(t)$	$\displaystyle=\int_{\begin{subarray}{c}\mu_{1}+\mu_{n-1}=t\\ \lambda_{1}\leq\mu_{1}\leq\lambda_{2}\\ \lambda_{n-1}\leq\mu_{n-1}\leq\lambda_{n}\end{subarray}}F_{1}(\mu_{1})F_{n-1}(\mu_{n-1})\,d\mu_{1}$
	$\displaystyle G$	$\displaystyle=\mathbf{1}_{[-1,1]}F^{L}_{n-2}\left(\operatorname*{\text{\scalebox{1.5}{$\ast$}}}_{i\neq 1,n-2,n-1}F_{i}\right)$
	$\displaystyle F^{L}_{n-2}(x)$	$\displaystyle=F_{n-2}(x)\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\mu\right\rVert}}{1+\mathinner{\!\left\lvert x\right\rvert}}\right)^{n-3}.$

Applying Lemma 15 and using $d_{1},d_{n-1}\gg\mathinner{\!\left\lVert\lambda\right\rVert}$ gives

\displaystyle F(t)\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{1}+\lambda_{n}-t\right\rvert}}+\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{2}+\lambda_{n-1}-t\right\rvert}}\right).

By symmetry (replacing $\lambda_{1},\dots,\lambda_{n}$ with $-\lambda_{n},\dots,-\lambda_{1}$ ) we may assume $\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}\geq\mathinner{\!\left\lvert\lambda_{2}\right\rvert}$ , which also implies $\lambda_{n-1}\geq 0$ . Now we split into three cases:

•

If $\lambda_{n-1}\leq 100$ then bound

\displaystyle F^{L}_{n-2}(x)

\displaystyle\ll(\log^{\prime}\mathinner{\!\left\lVert\lambda\right\rVert})^{n-3}F_{n-2}(x)

and

	$\displaystyle\int F(t)G(-t)\,dt$	$\displaystyle\ll(\log^{\prime}\mathinner{\!\left\lVert\lambda\right\rVert})^{n-3}\mathinner{\!\left\lVert F*\mathbf{1}_{[-1,1]}\right\rVert}_{\infty}\prod_{i\neq 1,n-1}\mathinner{\!\left\lVert F_{i}\right\rVert}_{1}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}(\log^{\prime}\mathinner{\!\left\lVert\lambda\right\rVert})^{n-3}\int_{0}^{2}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{x}\,dx$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}(\log^{\prime}\mathinner{\!\left\lVert\lambda\right\rVert})^{n-2}$
		$\displaystyle\asymp\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\right)^{n-2}$

and we are done.

•

Suppose $\lambda_{n-1}>100$ and $\mathinner{\!\left\lvert\lambda_{2}-\lambda_{n-1}\right\rvert}\leq\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}/100$ . Then $\lambda_{2},\dots,\lambda_{n-1}>0$ and $\lambda_{2}>10$ . Since $F^{L}_{n-2}$ is supported on $[\lambda_{n-2},\lambda_{n-1}]$ we can bound

	$\displaystyle F^{L}_{n-2}(x)$	$\displaystyle\ll\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}}\right)^{n-3}F_{n-2}(x)$
		$\displaystyle\ll\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\right)^{n-3}F_{n-2}(x).$

For $t$ satisfying $G(-t)\neq 0$ we have

	$\displaystyle t$	$\displaystyle\in-\operatorname{supp}G$
		$\displaystyle\subset[-1-\sum_{2\leq i\leq n-2}\lambda_{i+1},1-\sum_{2\leq i\leq n-2}\lambda_{i}]$
		$\displaystyle=[-1+\lambda_{1}+\lambda_{2}+\lambda_{n},1+\lambda_{1}+\lambda_{n-1}+\lambda_{n}]$

so $t-\lambda_{1}-\lambda_{n}\geq\lambda_{2}-1\gg\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}$ . On the other hand,

	$\displaystyle\lambda_{2}+\lambda_{n-1}-t$	$\displaystyle\geq\lambda_{2}+\lambda_{n-1}-(1+\lambda_{1}+\lambda_{n-1}+\lambda_{n})$
		$\displaystyle\geq\lambda_{2}-1-\lambda_{1}-\lambda_{n}$
		$\displaystyle\geq\lambda_{2}-1+(\lambda_{2}+\dots+\lambda_{n-1})$
		$\displaystyle\geq(n-1)\lambda_{2}-1$
		$\displaystyle\gg 1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}.$

Thus

\displaystyle F(t)

\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}},

and the result follows.

•

Finally suppose $\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}>100$ and $\mathinner{\!\left\lvert\lambda_{2}-\lambda_{n-1}\right\rvert}\geq\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}/100$ . Then $\mathinner{\!\left\lvert\lambda_{2}-\lambda_{n-1}\right\rvert}\asymp 1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}$ . If $\lambda_{n-2}<\lambda_{n-1}/2$ , then $d_{n-2}>\mathinner{\!\left\lvert\lambda_{2}-\lambda_{n-1}\right\rvert}/4\gg\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}$ and by monotone rearrangement,

	$\displaystyle\int F(t)G(-t)\,dt$	$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\mathinner{\!\left\lVert F*F^{L}_{n-2}\right\rVert}_{\infty}\prod_{i\neq 1,n-2,n-1}\mathinner{\!\left\lVert F_{i}\right\rVert}_{1}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\int_{0}^{d_{n-2}}d_{n-2}^{-1/2}x^{-1/2}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{x}\right)^{n-2}\,dx$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{d_{n-2}}\right)^{n-2}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}}\right)^{n-2}$

and we are done. Otherwise if $\lambda_{n-2}>\lambda_{n-1}/2$ then

	$\displaystyle F^{L}_{n-2}(x)$	$\displaystyle\ll\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}}\right)^{n-3}F_{n-2}(x)$
		$\displaystyle\ll\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\right)^{n-3}F_{n-2}(x).$

Choose $2\leq K\leq n-2$ such that $d_{K}$ is the next largest after $d_{J}$ , then

	$\displaystyle d_{K}$	$\displaystyle>(d-d_{I}-d_{J})/(n-3)$
		$\displaystyle=\mathinner{\!\left\lvert\lambda_{2}-\lambda_{n-1}\right\rvert}/(n-3)$
		$\displaystyle\gg 1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}$

and

	$\displaystyle\int F(t)G(-t)\,dt$	$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}}\right)^{n-3}\mathinner{\!\left\lVert F*K\right\rVert}_{\infty}\prod_{i\neq 1,K,n-1}\mathinner{\!\left\lVert F_{i}\right\rVert}_{1}$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}}\right)^{n-3}\int_{0}^{d_{K}}d_{K}^{-1/2}x^{-1/2}\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{x}\,dx$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}}\right)^{n-3}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{d_{K}}\right)$
		$\displaystyle\ll\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-2}\right\rvert}}\right)^{n-2}.$

4.4.4. Lower bound when $n\geq 3$ and $(I,J)=(1,n-1)$

By assuming $L_{n}(\lambda)$ is larger than some constant, we may assume $1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}<d/(100n)$ . This further implies $\mathinner{\!\left\lvert\lambda_{1}+\lambda_{n}\right\rvert}=\mathinner{\!\left\lvert\lambda_{2}+\dots+\lambda_{n-1}\right\rvert}\leq d/100$ , and since $d=\lambda_{n}-\lambda_{1}$ , we have $\lambda_{1}=(\lambda_{1}+\lambda_{n}-d)/2\leq-d/2+d/200\leq-d/3$ and $\lambda_{n}\geq d/3$ .

Similarly to Section 4.1, we restrict the integral to the region

\displaystyle\mathcal{H}^{\prime\prime}=\{\mu\in\mathcal{H}\mathrel{\mathop{\ordinarycolon}}\mu_{i}\in M_{i}\text{ for }2\leq i\leq n-2,\,\,\mu_{1}<-d/4,\,\,\mu_{n-1}>d/4\},

on which we have $\kappa_{i,j}\geq 1/4$ for all $i<j$ , and $F_{i}\gg d_{i}^{-1}\mathbf{1}_{M_{i}}$ for $2\leq i\leq n-2$ . Since $\lambda_{2}<\mu_{2}<\mu_{n-2}<\lambda_{n-1}$ we have

\displaystyle L_{n-1}(\mu)\gg\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\right)^{n-3}.

Bound $\mathinner{\!\left\lvert\lambda_{2}-\mu_{1}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{n-1}-\mu_{n-1}\right\rvert}^{-1/2}\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-1}$ and $(1+\mathinner{\!\left\lVert\mu\right\rVert})^{-n+2}\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+2}$ . Using all the above inequalities gives

	$\displaystyle J_{n}(\lambda)\gg$	$\displaystyle\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\left(\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}\right)^{n-3}\times$
		$\displaystyle\int_{\mathcal{H}^{\prime\prime}}\mathinner{\!\left\lvert\lambda_{1}-\mu_{1}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{n}-\mu_{n-1}\right\rvert}^{-1/2}\prod_{2\leq i\leq n-2}F_{i}(\mu_{i})J(\mu)d\mu,$

which we rewrite as a convolution

	$\displaystyle J_{n}(\lambda)$	$\displaystyle\gg\mathinner{\!\left\lVert\lambda\right\rVert}^{-n+1}\int F(t)G(-t)\,dt$
	$\displaystyle F(t)$	$\displaystyle=\int_{\begin{subarray}{c}\mu_{1}+\mu_{n-1}=t\\ \lambda_{1}\leq\mu_{1}\leq-d/4\\ d/4\leq\mu_{n-1}\leq\lambda_{n}\end{subarray}}\mathinner{\!\left\lvert\lambda_{1}-\mu_{1}\right\rvert}^{-1/2}\mathinner{\!\left\lvert\lambda_{n}-\mu_{n-1}\right\rvert}^{-1/2}d\mu$
	$\displaystyle G$	$\displaystyle=\mathbb{1}_{[-1,1]}\left(\operatorname{\text{\scalebox{1.5}{$\ast$}}}_{2\leq i\leq n-2}d_{i}^{-1}\mathbb{1}_{M_{i}}\right).$

Since $M_{i}\subset[\lambda_{2},\lambda_{n-1}]$ for $2\leq i\leq n-2$ , we have $G(-t)\neq 0\implies\mathinner{\!\left\lvert t\right\rvert}\leq d/100$ . If $t\leq\lambda_{1}+\lambda_{n}$ , then substituting $y=\mu_{1}-\lambda_{1}$ in the equation for $F$ gives

	$\displaystyle F(t)$	$\displaystyle=\int_{0}^{t-d/4-\lambda_{1}}y^{-1/2}(\lambda_{n}-(t-(y+\lambda_{1})))^{-1/2}\,dy$
		$\displaystyle=\int_{0}^{t-d/4-\lambda_{1}}y^{-1/2}(y+\lambda_{1}+\lambda_{n}-t)^{-1/2}\,dy.$

Apply Lemma 17 with $a=\lambda_{1}+\lambda_{n}-t$ and $T=t-d/4-\lambda_{1}$ . The preceding inequalities for $\mathinner{\!\left\lvert t\right\rvert},\lambda_{1}$ , and $\mathinner{\!\left\lvert\lambda_{1}+\lambda_{n}\right\rvert}$ ensure $a\leq T$ . This gives

\displaystyle F(t)\gg\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\lambda_{1}+\lambda_{n}-t}.

Likewise if $t\geq\lambda_{1}+\lambda_{n}$ then

\displaystyle F(t)\gg\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\lambda_{1}+\lambda_{n}-t},

thus

\displaystyle F(t)\gg\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert t-(\lambda_{1}+\lambda_{n})\right\rvert}}.

Since $\mathinner{\!\left\lvert\lambda_{1}+\lambda_{n}\right\rvert}\ll\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}$ we have $\mathinner{\!\left\lvert t-(\lambda_{1}+\lambda_{n})\right\rvert}\ll\mathinner{\!\left\lvert t\right\rvert}+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}$ and

\displaystyle F(t)\gg\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{\mathinner{\!\left\lvert t\right\rvert}+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}}.

Meanwhile, we have $\int G(t)\,dt\gg 1$ , and the support of $G$ is contained in an interval of radius $\ll 1+\mathinner{\!\left\lvert\lambda_{2}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n-1}\right\rvert}$ around 0, so

\displaystyle\int F(t)G(-t)\,dt\gg\log^{\prime}\frac{\mathinner{\!\left\lVert\lambda\right\rVert}}{1+\mathinner{\!\left\lvert\lambda_{1}\right\rvert}+\mathinner{\!\left\lvert\lambda_{n}\right\rvert}}

as desired. ∎

5. Lemmas

Here we have separated out the main calculation for the two large gaps case.

Lemma 14.

Let $A\leq B\leq C\leq D\leq E$ , $T$ , and $(L_{i})_{1\leq i\leq k}$ be given with $\mathinner{\!\left\lvert A-E\right\rvert}\leq T$ . Then for $A+C\leq t\leq B+D$ ,

\displaystyle(*)=\int_{\begin{subarray}{c}x+y=t\\ A\leq x\leq B\\ C\leq y\leq D\end{subarray}}\left(\mathinner{\!\left\lvert x-A\right\rvert}\mathinner{\!\left\lvert x-B\right\rvert}\mathinner{\!\left\lvert y-C\right\rvert}\mathinner{\!\left\lvert y-D\right\rvert}\right)^{-1/2}\mathinner{\!\left\lvert x-y\right\rvert}^{1/2}\mathinner{\!\left\lvert y-E\right\rvert}^{1/2}\prod_{1\leq i\leq k}\log^{\prime}\frac{T}{\mathinner{\!\left\lvert L_{i}-y\right\rvert}}\,dx

is bounded above by

	$\displaystyle\ll_{k}T\mathinner{\!\left\lvert A-B\right\rvert}^{-1/2}\mathinner{\!\left\lvert C-D\right\rvert}^{-1/2}\times$
	$\displaystyle\left[\left(\log^{\prime}\frac{\mathinner{\!\left\lvert B-C\right\rvert}}{\mathinner{\!\left\lvert B+C-t\right\rvert}}\right)^{k+1}+\left(\log^{\prime}\frac{\mathinner{\!\left\lvert D-E\right\rvert}}{\mathinner{\!\left\lvert A+D-t\right\rvert}}\right)^{k+1}+\left(\log^{\prime}\frac{T}{\mathinner{\!\left\lvert A+C-t\right\rvert}}\right)^{k}+\left(\log^{\prime}\frac{T}{\mathinner{\!\left\lvert B+D-t\right\rvert}}\right)^{k}\right].$

Proof.

For $x\in[P,Q]$ we have

(13)

\displaystyle\mathinner{\!\left\lvert x-P\right\rvert}^{-1/2}\mathinner{\!\left\lvert x-Q\right\rvert}^{-1/2}\ll\mathinner{\!\left\lvert P-Q\right\rvert}^{-1/2}\left(\mathinner{\!\left\lvert x-P\right\rvert}^{-1/2}+\mathinner{\!\left\lvert x-Q\right\rvert}^{-1/2}\right).

Using this identity twice with $(P,Q)=(A,B)$ and $(P,Q)=(C,D)$ and substituting $y=t-x$ gives

\displaystyle(*)\ll\mathinner{\!\left\lvert A-B\right\rvert}^{-1/2}\mathinner{\!\left\lvert C-D\right\rvert}^{-1/2}\sum_{\begin{subarray}{c}X\in\{A,B\}\\ Y\in\{C,D\}\end{subarray}}I(X,Y)

where $I(X,Y)=$

\displaystyle\int\limits_{\max(A,t-D)}^{\min(B,t-C)}\mathinner{\!\left\lvert x-X\right\rvert}^{-1/2}\mathinner{\!\left\lvert x-t+Y\right\rvert}^{-1/2}\mathinner{\!\left\lvert 2x-t\right\rvert}^{1/2}\mathinner{\!\left\lvert x-t+E\right\rvert}^{1/2}\prod_{1\leq i\leq k}\log^{\prime}\frac{T}{\mathinner{\!\left\lvert L_{i}-t+x\right\rvert}}dx.

First consider $I(A,C)$ . Extend the bounds of the integral to $[A,t-C]$ . Bound $\mathinner{\!\left\lvert 2x-t\right\rvert}^{1/2}\mathinner{\!\left\lvert x-t+E\right\rvert}^{1/2}\ll T$ . Then

I(A,C)\ll T\int_{A}^{t-C}\mathinner{\!\left\lvert x-A\right\rvert}^{-1/2}\mathinner{\!\left\lvert x-(t-C)\right\rvert}^{-1/2}\prod_{1\leq i\leq k}\log^{\prime}\frac{T}{\mathinner{\!\left\lvert L_{i}-t+x\right\rvert}}\,dx

and using Equation (13) with $(P,Q)=(A,t-C)$ and then using monotone rearrangement on each factor we obtain

I(A,C)\ll T\mathinner{\!\left\lvert t-C-A\right\rvert}^{-1/2}\int_{0}^{t-C-A}x^{-1/2}\left(\log^{\prime}\frac{T}{x}\right)^{k}\,dx\ll_{k}T\left(\log^{\prime}\frac{T}{\mathinner{\!\left\lvert A+C-t\right\rvert}}\right)^{k}.

The calculation for $I(B,D)$ is similar to $I(A,C)$ .

Now consider $I(A,D)$ . Bound $\mathinner{\!\left\lvert 2x-t\right\rvert}^{1/2}\ll T^{1/2}$ . Suppose $t\leq A+D$ . Substitute $x^{\prime}=x-A$ . The integral now runs from $0$ to $B-A$ , extend it to $[0,T]$ . The integral now reads

T^{1/2}\int_{0}^{T}\mathinner{\!\left\lvert x^{\prime}\right\rvert}^{-1/2}\mathinner{\!\left\lvert x^{\prime}+A+D-t\right\rvert}^{-1/2}\mathinner{\!\left\lvert x^{\prime}+A+E-t\right\rvert}^{1/2}\prod_{1\leq i\leq k}\log^{\prime}\frac{T}{\mathinner{\!\left\lvert L_{i}-t+x^{\prime}+A\right\rvert}}\,dx.

Applying Lemma 16 with $a=A+D-t$ and $b=A+E-t$ gives

	$\displaystyle I(A,D)$	$\displaystyle\ll_{k}T\left(\log^{\prime}\frac{A+E-t}{A+D-t}\right)^{k+1}$
		$\displaystyle=T\left(\log^{\prime}\left(1+\frac{E-D}{A+D-t}\right)\right)^{k+1}$
		$\displaystyle\ll T\left(\log^{\prime}\frac{\mathinner{\!\left\lvert D-E\right\rvert}}{\mathinner{\!\left\lvert A+D-t\right\rvert}}\right)^{k+1}.$

If $t\geq A+D$ , then substitute $x^{\prime}=x-t+D$ , extend the bounds of the integral to $[0,T]$ , and apply Lemma 16 with $a=t-A-D,b=E-D$ . The calculation for $I(B,C)$ is similar. ∎

Lemma 15.

Let $A\leq B\leq C\leq D$ be given with $\mathinner{\!\left\lvert A-D\right\rvert}\leq T$ . Then for $A+B\leq t\leq C+D$ ,

\displaystyle\int_{\begin{subarray}{c}x+y=t\\ A\leq x\leq B\\ C\leq y\leq D\end{subarray}}\left(\mathinner{\!\left\lvert x-A\right\rvert}\mathinner{\!\left\lvert x-B\right\rvert}\mathinner{\!\left\lvert y-C\right\rvert}\mathinner{\!\left\lvert y-D\right\rvert}\right)^{-1/2}\,dx

is bounded above by

\displaystyle\ll\mathinner{\!\left\lvert A-B\right\rvert}^{-1/2}\mathinner{\!\left\lvert C-D\right\rvert}^{-1/2}\left(\log^{\prime}\frac{T}{\mathinner{\!\left\lvert B+C-t\right\rvert}}+\log^{\prime}\frac{T}{\mathinner{\!\left\lvert A+D-t\right\rvert}}\right).

Proof.

Substitute $A,B,C+T,D+T,D+2T,3T,t+T$ for $A,B,C,D,E,T,t$ in Lemma 14. We obtain that

\displaystyle\int_{\begin{subarray}{c}x+y=t+T\\ A\leq x\leq B\\ C+T\leq y\leq D+T\end{subarray}}\left(\mathinner{\!\left\lvert x-A\right\rvert}\mathinner{\!\left\lvert x-B\right\rvert}\mathinner{\!\left\lvert y-C-T\right\rvert}\mathinner{\!\left\lvert y-D-T\right\rvert}\right)^{-1/2}\mathinner{\!\left\lvert x-y\right\rvert}^{1/2}\mathinner{\!\left\lvert y-E-2T\right\rvert}^{1/2}\,dx

is $\ll$ than

\displaystyle T\mathinner{\!\left\lvert A-B\right\rvert}^{-1/2}\mathinner{\!\left\lvert C-D\right\rvert}^{-1/2}\left(\log^{\prime}\frac{T+\mathinner{\!\left\lvert B-C\right\rvert}}{\mathinner{\!\left\lvert B+C-t\right\rvert}}+\log^{\prime}\frac{T+\mathinner{\!\left\lvert D-E\right\rvert}}{\mathinner{\!\left\lvert A+D-t\right\rvert}}\right).

Note that $\mathinner{\!\left\lvert x-y\right\rvert},\mathinner{\!\left\lvert y-E-2T\right\rvert}$ in the integrand and $T+\mathinner{\!\left\lvert B-C\right\rvert},T+\mathinner{\!\left\lvert D-E\right\rvert}$ on the right hand side are all $\asymp T$ , so dividing both sides by $T$ gives the desired inequality. ∎

Lemma 16.

For $0<a,b<T$ and all $L_{1},\dots,L_{k}$ we have

\displaystyle\int_{0}^{T}

\displaystyle x^{-1/2}(x+a)^{-1/2}(x+b)^{1/2}\prod_{1\leq i\leq k}\log^{\prime}\frac{T}{\mathinner{\!\left\lvert L_{i}+x\right\rvert}}\,dx\ll_{k}T^{1/2}\left(\log^{\prime}\frac{b}{a}\right)^{k+1}.

Proof.

Since $x^{-1/2}(x+a)^{-1/2}(x+b)^{1/2}$ is decreasing on $[0,T]$ , we may use monotone rearrangement and assume $L_{i}=0$ for all $i$ . If $0<a<b<T$ then the integral is

	$\displaystyle\ll\left[a^{-1/2}b^{1/2}\int_{0}^{a}x^{-1/2}+b^{1/2}\int_{a}^{b}x^{-1}+\int_{b}^{T}x^{-1/2}\right]\left(\log^{\prime}\frac{T}{x}\right)^{k}\,dx$
	$\displaystyle\ll_{k}b^{1/2}\left(\log^{\prime}\frac{T}{a}\right)^{k}+b^{1/2}\left(\log^{\prime}\frac{b}{a}\right)\left(\log^{\prime}\frac{T}{a}\right)^{k}+T^{1/2}$
	$\displaystyle\ll_{k}T^{1/2}\frac{b^{1/2}}{T^{1/2}}\left(\log^{\prime}\frac{T}{b}+\log^{\prime}\frac{b}{a}\right)^{k}+T^{1/2}\frac{b^{1/2}}{T^{1/2}}\left(\log\frac{b}{a}\right)\left(\log^{\prime}\frac{T}{b}+\log^{\prime}\frac{b}{a}\right)^{k}+T^{1/2}$
	$\displaystyle\ll_{k}T^{1/2}\left(\log^{\prime}\frac{b}{a}\right)^{k+1}$

where on the third line, we use the fact that $(b/T)^{1/2}(\log^{\prime}(T/b))^{k}\ll_{k}1$ . Otherwise if $0<b<a<T$ then $x^{-1/2}(x+a)^{-1/2}(x+b)^{1/2}\leq x^{-1/2}$ and the integral is

\displaystyle\leq\int_{0}^{T}x^{-1/2}\left(\log^{\prime}\frac{T}{x}\right)^{k}\,dx\ll_{k}T^{1/2}.

∎

Lemma 17.

For $0<a\leq T$ ,

\displaystyle\int_{0}^{T}x^{-1/2}(x+a)^{-1/2}\,dx\asymp\log^{\prime}(T/a).

Proof.

	$\displaystyle\int_{0}^{T}x^{-1/2}(x+a)^{-1/2}\,dx$	$\displaystyle=\int_{0}^{a}x^{-1/2}(x+a)^{-1/2}\,dx+\int_{a}^{T}x^{-1/2}(x+a)^{-1/2}\,dx$
		$\displaystyle\asymp a^{-1/2}\int_{0}^{a}x^{-1/2}\,dx+\int_{a}^{T}x^{-1}\,dx$
		$\displaystyle\asymp 1+\log(T/a).$

∎

References

[1] Farrell Brumley and Simon Marshall. Lower bounds for maass forms on semisimple groups. Compositio Mathematica, 156(5):959–1003, 2020.
[2] Almut Burchard. A short course on rearrangement inequalities. Lecture notes, IMDEA Winter School, Madrid, 2009.
[3] JJ Duistermaat. On the similarity between the Iwasawa projection and the diagonal part. Mem. Soc. Math. France, (15):129–138, 1984.
[4] Ky Fan and Gordon Pall. Imbedding conditions for Hermitian and normal matrices. Canadian Journal of Mathematics, 9:298–304, 1957.
[5] Bart Michels. The maximal growth of automorphic periods and oscillatory integrals for maximal flat submanifold. PhD thesis, Paris 13, 2022.
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Uniform Bounds for Maximal Flat Periods on S​L​(n,ℝ)SL(n,\mathbb{R})

Abstract.

1. Introduction

Theorem 1.

1.1. Outline of the proof

2. Proof of theorem 1

2.1. Preliminaries and Notation

2.2. Spherical Functions and the Trace Formula

2.3. Euclidean Approximation of Spherical Functions

Lemma 1.

3. The coadjoint orbit K.λK.\lambda

3.1. Preliminaries

Lemma 2.

Proof.

3.2. Inductive step

Lemma 3.

Proof.

Lemma 4.

Proof.

Lemma 5 (Fan-Pall).

4. Convolution Bounds

4.1. The general lower bound

Lemma 6.

Proof.

Lemma 7.

Proof.

4.2. Monotone rearrangement

Lemma 8.

Proof.

4.3. One Large Gap

4.3.1. The upper bound when (n,I)≠(4,2)(n,I)\neq(4,2)

Lemma 9.

4.3.2. The case (n,I)=(4,2)(n,I)=(4,2)

Lemma 10.

4.4. Two Large Gaps

4.4.1. The case (I,J)≠(1,n−1)(I,J)\neq(1,n-1)

Lemma 11.

Proof.

Lemma 12.

Proof.

4.4.2. The case (I,J)=(1,n−1)(I,J)=(1,n-1)

Lemma 13.

Proof.

4.4.3. The upper bound when n=3n=3

4.4.4. Lower bound when n≥3n\geq 3 and (I,J)=(1,n−1)(I,J)=(1,n-1)

5. Lemmas

Lemma 14.

Proof.

Lemma 15.

Proof.

Lemma 16.

Proof.

Lemma 17.

Proof.

References

Uniform Bounds for Maximal Flat Periods on $SL(n,\mathbb{R})$

3. The coadjoint orbit $K.\lambda$

4.3.1. The upper bound when $(n,I)\neq(4,2)$

4.3.2. The case $(n,I)=(4,2)$

4.4.1. The case $(I,J)\neq(1,n-1)$

4.4.2. The case $(I,J)=(1,n-1)$

4.4.3. The upper bound when $n=3$

4.4.4. Lower bound when $n\geq 3$ and $(I,J)=(1,n-1)$