Proof. We introduce an additional parameter K, which will be chosen to be large in terms of C and $\alpha $ . We will then choose $\rho =\rho (C,\alpha )$ with $0<\rho <1$ to be small in terms of K, C and $\alpha $ . We do not specify the values of K and $\rho $ ahead of time but rather assume they are sufficiently large or small to satisfy certain inequalities that arise in the proof.

Let $\delta =(m/n)^{\rho }$ and suppose for the purpose of contradiction that every set of at least m vertices fails to induce a $(\delta ,\rho ,\alpha )$ -rich subgraph. We will inductively construct a sequence of induced subgraphs $G=G[U_{0}]\supseteq G[U_{1}]\supseteq \cdots \supseteq G[U_{K}]$ and disjoint vertex sets $S_{1},\ldots ,S_{K}$ of size $|S_1|=\cdots =|S_K|=\lceil m^{\alpha }/2\rceil $ such that for each $i=1,\ldots ,K$ , we have $|U_{i}|\ge (\delta /4)|U_{i-1}|$ and $S_{i}\subseteq U_{i-1}\setminus U_i$ , as well as

$$\begin{align*}\big[e(S_{i},\{u\})\leq 4\rho\cdot |S_i|\text{ for all }u\in U_i\big]\text{ or }\big[e(S_{i},\{u\})\geq (1-4\rho)\cdot |S_i|\text{ for all }u\in U_i\big]. \end{align*}$$

This will suffice, as follows. First, note that for each $i=1,\ldots ,K$ , we have

$$ \begin{align*}&\big[d(S_{i},S_{j})\leq 4\rho\text{ for all }j\in \{i+1,\ldots,K\}\big]\\ &\qquad \text{ or }\big[d(S_{i},S_{j})\geq 1-4\rho\text{ for all }j\in \{i+1,\ldots,K\}\big].\end{align*} $$

Without loss of generality suppose that the first case holds for at least half of the indices $i=1,\ldots ,K$ , and let S be the union of the corresponding sets $S_{i}$ . Then one can compute $d(S)\leq 4\rho +1/K$ . On the other hand, $|S|\geq (K/2)\cdot m^{\alpha }/2\geq m^{\alpha }\geq n^{\alpha /2}$ and therefore $G[S]$ is a $(2C/\alpha )$ -Ramsey graph. However, now the density bound $d(S)\leq 4\rho +1/K$ contradicts Theorem 4.1 if $\rho $ is sufficiently small and K is sufficiently large (in terms of C and $\alpha $ ).

Let $U_{0}=V(G)$ . For $i=1,\ldots ,K$ we will construct the vertex sets $U_{i}$ and $S_{i}$ , assuming that $U_{0},\ldots ,U_{i-1}$ and $S_{1},\ldots ,S_{i-1}$ have already been constructed. Note that we have $|U_{i-1}|\ge (\delta /4)^{i-1}n\ge (\delta /4)^Kn=(m/n)^{\rho K}4^{-K}n\ge m$ , using that $\rho K\leq 1/3$ and $m/n\leq \rho \leq 8^{-K}$ for $\rho $ being sufficiently small with respect to K. Therefore, by our assumption, $U_{i-1}$ must contain a set W of at least $\delta |U_{i-1}|$ vertices and a set Y of more than $|U_{i-1}|^{\alpha }\geq m^{\alpha }$ vertices contradicting $(\delta ,\rho ,\alpha )$ -richness. Suppose without loss of generality that $|N(v)\cap W|\le \rho |W|$ for at least half of the vertices $v\in Y$ , and let $S_{i}\subseteq Y\subseteq U_{i-1}$ be a set of precisely $\lceil m^{\alpha }/2\rceil $ such vertices $v\in Y$ . Then, let $U=W\setminus S_{i}\subseteq U_{i-1}\setminus S_i$ , and note that we have $|U|\ge |W|/2$ since $|W|\geq \delta |U_{i-1}|\geq 4 \cdot (\delta /4)^Kn\geq 4m\geq 2|S_{i}|$ . Furthermore, let $U_{i}\subseteq U$ be the set of vertices $u\in U$ with $e(S_{i}, \{ u\} )\le 4\rho \cdot |S_i|$ . Now, we just need to show $|U_{i}|\ge (\delta /4)|U_{i-1}|$ . To this end, note that for all $v\in S_{i}$ we have $e(\{ v\} ,U)=|N(v)\cap U|\le |N(v)\cap W|\leq \rho |W|\leq 2\rho |U|$ . Hence,

$$ \begin{align*} |U\setminus U_{i}|\cdot 4\rho\cdot |S_i|&\leq \sum_{w\in U\setminus U_{i}}e(S_{i},\{ w\})=e(S_{i},U\setminus U_{i} )\\ &\le e(S_{i},U)=\sum_{v\in S_{i}}e(\{ v\} ,U)\le |S_i|\cdot 2\rho|U|, \end{align*} $$

implying that $|U\setminus U_i|\leq |U|/2$ and hence $|U_{i}|\geq |U|/2\ge |W|/4\geq (\delta /4)|U_{i-1}|$ , as desired.

Proof. We may assume that $M\leq cn/4$ (indeed, noting that $M\leq n/2$ we can otherwise just replace M by $\lfloor cn/4\rfloor $ ). The random vector $\vec {\xi }$ corresponds to a uniformly random subset $U\subseteq [n]$ of size s. Let us first expose the intersection sizes $|U\cap \{i_1,j_1\}|, \ldots , |U\cap \{i_M,j_M\}|$ , one at a time. For each $k=1,\ldots ,M$ , we have $|U\cap \{i_k,j_k\}|=1$ with probability at least $c(1-c)/4$ even when conditioning on any outcomes for the previously exposed intersection sizes $|U\cap \{i_1,j_1\}|, \ldots , |U\cap \{i_{k-1},j_{k-1}\}|$ . Hence, the number of indices $k\in [M]$ with $|U\cap \{i_k,j_k\}|=1$ stochastically dominates a binomial random variable with distribution $\mathrm {Bin}(M,c(1-c)/4)$ . Thus, by a Chernoff bound (see, e.g., Lemma 4.16), with probability at least $1-\exp (-\Omega _c(M))$ there is a set $K\subseteq [M]$ of at least $c(1-c)M/8$ indices k with $|U\cap \{i_k,j_k\}|=1$ . Let us expose and condition on all coordinates of $\vec {\xi }\in \{0,1\}^n$ except those in $\bigcup _{k\in K}\{i_k,j_k\}$ . The only remaining randomness of the vector $\vec {\xi }\in \{0,1\}^n$ is that for each $k\in K$ we have either $\xi _{i_k}=1$ or $\xi _{j_k}=1$ (each with probability $1/2$ , independently for all $k\in K$ ). Thus, after all of this conditioning, we have $\vec {a}\cdot \vec {\xi }=\sum _{k\in K} (a_{i_k}-a_{j_k})\xi _{i_k}+b$ for some $b\in \mathbb {R}$ , where $(\xi _{i_k})_{k\in K}\in \{0,1\}^K$ is uniformly random. Thus, (4.3) implies $|\mathbb E[\exp (i(\vec {a}\cdot \vec {\xi }))]|\leq \exp (-\sum _{k\in K} \|(a_{i_{k}}-a_{j_{k}})/(2\pi )\|_{\mathbb {R}/\mathbb Z}^2)\leq \exp (-\Omega _c(M\delta ^2))$ , as desired.

Proof. Note that for $\theta \le 1/(2\lVert \vec v\rVert _{\infty })$ we have that $\lVert \theta \vec {v}\rVert _{\infty }\le 1/2$ . Thus, we have that

$$\begin{align*}\operatorname{dist}(\theta \vec{v},\mathbb{Z}^{n}) = \operatorname{dist}(\theta \vec{v},\vec{0}) = \theta>L\sqrt{\log_{+}(\theta/L),}\end{align*}$$

where we have used that $x>\sqrt {\log _{+}(x)}$ for $x> 0$ . The result follows by the definition of LCD.

Proof. Choose a partition $[n] = R\cup (I_1\cup \cdots \cup I_m)$ and $\kappa _j\geq 0$ for $j=1,\ldots ,m$ with $|I_1|=\cdots =|I_m| = \lceil n^{1-2\gamma }\rceil $ such that $|d_i-\kappa _j|\le n^{1/2+4\gamma }$ for all $1\le j\le m$ and $i\in I_j$ , such that m is as large as possible. It then suffices to prove that $|R|\leq n^{1-\gamma }$ .

So let us assume for contradiction that $|R|>n^{1-\gamma }$ , and fix a subset $S\subseteq R$ of size $|S|=\lceil n^{1-\gamma }\rceil $ . Note that $D_L(\vec {d}_S/\lVert \vec {d}_S\rVert _2)\le \widehat {D}_{L,\gamma }(\vec {d})\le n^{1/2}$ by Definition 4.11. Furthermore, since $\lVert \vec {d}_S/\lVert \vec {d}_S\rVert _2\rVert _{\infty }\le Hn/n^{3/2-2\gamma }= Hn^{-1/2+2\gamma }$ , Lemma 4.10 implies $D_L(\vec {d}_S/\lVert \vec {d}_S\rVert _2)\ge (H^{-1}/2) n^{1/2-2\gamma }$ . Thus, by Definition 4.9, there is some $\theta \in [(H^{-1}/2)n^{1/2-2\gamma },2n^{1/2}]$ such that

(4.4) $$ \begin{align} \lVert(\theta/\lVert\vec{d}_S\rVert_2)\vec{d}_S-\vec{w}\rVert_2\le L\sqrt{\log_+ (\theta/L)}\le L\sqrt{\log n} \end{align} $$

for some $\vec {w}\in \mathbb {Z}^{S}$ . By choosing $\vec {w}$ to minimize the left-hand side, we may assume that $\vec {w}$ has nonnegative entries (recall that $\vec {d}$ has nonnegative entries).

Now, the number of indices $i\in S$ with $|(\theta /\lVert \vec {d}_S\rVert _2)d_i-w_i|> n^{-1/2+2\gamma }$ is at most

$$\begin{align*}\frac{\lVert(\theta/\lVert\vec{d}_S\rVert_2)\vec{d}_S-\vec{w}\rVert_2^2}{n^{-1+4\gamma}}\leq \frac{L^2 \log n}{n^{-1+4\gamma}}\le n^{1-3\gamma}.\end{align*}$$

Furthermore, note that $\theta \leq 2n^{1/2}$ and (4.4) imply $\lVert \vec {w}\rVert _2\le 3n^{1/2}$ , and hence, the number of indices $i\in S$ with $w_i\geq n^{2\gamma /3}$ is at most $9n^{1-4\gamma /3}$ . Thus, as $|S|=\lceil n^{1-\gamma }\rceil $ , there must be at least $|S|/2\geq n^{1-\gamma }/2$ indices $i\in S$ with $|(\theta /\lVert \vec {d}_S\rVert _2)d_i-w_i|\leq n^{-1/2+2\gamma }$ and $w_i\in [0,n^{2\gamma /3}]\cap \mathbb {Z}$ . Hence, by the pigeonhole principle there is some $\kappa \geq 0$ and a subset $I_{m+1}\subseteq S\subseteq R$ of size $|I_{m+1}| = \lceil n^{1-2\gamma }\rceil $ such that for all $i\in I_{m+1}$ we have $w_i=\kappa $ and

$$ \begin{align*}|(\theta/\lVert\vec{d}_S\rVert_2)d_i-\kappa |=|(\theta/\lVert\vec{d}_S\rVert_2)d_i-w_i|&\leq n^{-1/2+2\gamma}= \frac{n^{1/2-2\gamma}}{n^{(1-\gamma)/2}n}\cdot n^{1/2+(7/2)\gamma}\\ &\lesssim_H \frac{\theta}{\lVert\vec{d}_S\rVert_2}\cdot n^{1/2+(7/2)\gamma}. \end{align*} $$

Defining $\kappa _{m+1}=(\lVert \vec {d}_S\rVert _2/\theta )\kappa \geq 0$ , this implies $|d_i-\kappa _{m+1}|\leq n^{1/2+4\gamma }$ for all $i\in I_{m+1}$ . But now the partition $V(G) = (R\setminus I_{m+1})\cup (I_1\cup \cdots \cup I_{m+1})$ contradicts the maximality of m.

Proof. We sample a uniformly random vector $\vec {x}\in S$ in $\ell :=\ell _1+\cdots +\ell _m+\ell _1'+\cdots +\ell _m'$ steps, as follows. In the first $\ell _1$ steps, we pick the $\ell _1$ indices $i\in I_1$ such that $x_i=1$ (at each step, pick an index $i\in I_1$ uniformly at random among the indices where $x_i$ is not yet defined, and define $x_i=1$ ). In the next $\ell _2$ steps, we pick the $\ell _2$ indices $i\in I_2$ such that $x_i=1$ , and so on. After $\ell _1+\cdots +\ell _m$ steps, we have defined all the $1$ -entries of $\vec {x}$ . Now, we repeat the procedure (for $\ell _1'+\cdots +\ell _m'$ steps) for the $-1$ -entries.

For $t=0,\ldots ,\ell $ , define $X_t$ to be the expectation of $f(\vec {x})$ conditioned on the coordinates of $\vec {x}$ defined up to step t. Then $X_0,\ldots ,X_t$ is the Doob martingale associated to our process of sampling $\vec {x}$ . Note that $X_0=\mathbb E f(\vec {x})$ and $X_{\ell }=f(\vec {x})$ .

We claim that we always have $|X_t-X_{t-1}|\le a$ for $t=1,\ldots ,\ell $ . Indeed, let us condition on any outcomes of the first $t-1$ steps of our process of sampling $\vec {x}$ . Now, for any two possible indices i and $i'$ chosen the t-th step, we can couple the possible outcomes of $\vec {x}$ if i is chosen in the t-th step with the possible outcomes of $\vec {x}$ if $i'$ is chosen in the t-th step, simply by switching the i-th and the $i'$ -th coordinate. Using our assumption on f, this shows that for any two possible outcomes in the t-th step the corresponding conditional expectations differ by at most a. This implies $|X_t-X_{t-1}|\le a$ , as claimed.

Now, the inequality in the lemma follows from the Azuma–Hoeffding inequality for martingales (see, for example, [Reference Janson, Łuczak and Rucinski59, Theorem 2.25]).

Proof. If $\lambda =0$ , then $\varphi _{X}(t)=\varphi _{aW}(t)=\varphi _W(at)=\exp (-a^2t^2/2)$ , as desired. So let us assume $\lambda \neq 0$ . Note that $X=a W+\lambda W^2 = \lambda (W + a/(2\lambda ))^2-a^2/(4\lambda )$ , and thus,

$$\begin{align*}|\varphi_X(t)| = |\varphi_{\lambda(W + a/(2\lambda))^2}(t)| = |\varphi_{(W + a/(2\lambda))^2}(\lambda t)|. \end{align*}$$

Using the formula for the characteristic function of a noncentral chi-squared distribution with $1$ degree of freedom and noncentrality parameter $(a/(2\lambda ))^2$ , we obtain

$$\begin{align*}|\varphi_{(W + a/(2\lambda))^2}(\lambda t)| =\frac{\left|\exp\left(\frac{i\cdot a^2/(4\lambda^2)\cdot \lambda t}{1-2i\lambda t}\right)\right|}{|1-2i\lambda t|^{1/2}} =\frac{\left|\exp\left(\frac{i\cdot a^2/(4\lambda^2)\cdot \lambda t\cdot (1+2i\lambda t)}{1+4\lambda^2 t^2}\right)\right|}{(1+4\lambda^2 t^2)^{1/4}} =\frac{\exp\left(\frac{-a^2t^2}{2(1+4\lambda^2t^2)}\right)}{(1+4\lambda^2t^2)^{1/4}}. \end{align*}$$

Proof of Lemma 5.5(a)

First, note that we may assume that there are no indices i with $\sigma _i=0$ (indeed, if $\sigma _i=0$ , then $\lambda _i=a_i=0$ and we can just omit all such indices). By rescaling, we may assume that $\sigma ^2 = 1$ . Note that $\varphi _{X}(t)=\prod _{i=1}^{n}\varphi _{X_i}(t)$ , and hence $\int _{-\infty }^{\infty }|\varphi _{X}(t)|\,dt<\infty $ . Also recall that the standard Gaussian distribution has density $u\mapsto e^{-u^2/2}/\sqrt {2\pi }$ and characteristic function $t\mapsto e^{-t^2/2}$ . Thus, by the inversion formula (4.1), it suffices to show that

(5.1) $$ \begin{align} \frac{1}{2\pi}\int_{-\infty}^{\infty}\bigg|\prod_{i=1}^{n}\varphi_{X_i}(t) - e^{-t^2/2}\bigg|\,dt\lesssim \frac1\Gamma + \int_{|t|\ge \Gamma/32}\;\prod_{i=1}^{n}|\varphi_{X_i}(t)|\,dt. \end{align} $$

Note that $\mathbb E [X_i]=0$ for $i=1,\ldots ,n$ , and let us write $L=(\sum _{i=1}^{n}\mathbb {E}[|X_i|^3])/(\sum _{i=1}^n \sigma _i^{2})^{3/2}=\sum _{i=1}^{n}\mathbb {E}[|X_i|^3]$ . Then for $|t|\le 1/(4L)$ , by [Reference Petrov and Brown83, Chapter V, Lemma 1] (which is a standard estimate in proofs of central limit theorems) we have

$$\begin{align*}\bigg|\prod_{i=1}^{n}\varphi_{X_i}(t) - e^{-t^2/2}\bigg|=\left|\varphi_{X}(t) - e^{-t^2/2}\right|\le 16L\cdot |t|^3e^{-t^2/3}.\end{align*}$$

By Hölder’s inequality and Theorem 4.14 (hypercontractivity), we have $\sigma _i^{3}\le \mathbb {E}[|X_i|^{3}]\le 8\sigma _i^{3}$ for $i=1,\ldots ,n$ , so we obtain $1/\Gamma \leq L\leq 8/\Gamma $ . Thus, the interval $|t|\le \Gamma /32$ contributes at most $\int _{-\Gamma /32}^{\Gamma /32}16L\cdot |t|^3e^{-t^2/3}\,dt\lesssim L\int _{-\infty }^{\infty }|t|^3e^{-t^2/3}\,dt \lesssim L\lesssim 1/\Gamma $ to the integral in (5.1). Therefore, we obtain

$$ \begin{align*} \int_{|t|\ge {\Gamma}/{32}}\bigg|\prod_{i=1}^{n}\varphi_{X_i}(t) - e^{-t^2/2}\bigg|\,dt&\le \int_{|t|\ge {\Gamma}/{32}}e^{-t^2/2} + \bigg|\prod_{i=1}^{n}\varphi_{X_i}(t)\bigg|\,dt\\ &\lesssim \frac{1}{\Gamma} + \int_{|t|\ge {\Gamma}/{32}} \bigg|\prod_{i=1}^{n}\varphi_{X_i}(t)\bigg|\,dt. \\[-45pt] \end{align*} $$

Proof of Lemma 5.5(b)

As before, we may assume that there are no indices i with $\sigma _i=0$ , and by rescaling we may assume that $\sigma ^2 = 1$ . Via Lemma 5.4, we estimate

$$ \begin{align*} \int_{|t|\ge K}\bigg|\prod_{i=1}^{n}\varphi_{X_i}(t)\bigg|\,dt &\le \prod_{i=1}^{n}\bigg(\int_{|t|\ge K}|\varphi_{X_i}(t)|^{1/\sigma_i^2}\,dt\bigg)^{\sigma_i^2}\\ &=\prod_{i=1}^{n}\Bigg(\int_{|t|\ge K}\frac{\exp\Big(-{a_i^2t^2}/\left({(2+8\lambda_i^2t^2)\sigma_i^2}\right)\Big)}{(1+4\lambda_i^2t^2)^{1/(4\sigma_i^2)}}\,dt\Bigg)^{\sigma_i^2}\\ &\le\prod_{i=1}^{n}\Bigg(\int_{|t|\ge K}\frac{\exp\Big(-{a_i^2t^2}/\left({(2+8\lambda_i^2t^2)\sigma_i^2}\right)\Big)}{1+\lambda_i^2t^2/\sigma_i^2}\,dt\Bigg)^{\sigma_i^2}. \end{align*} $$

In the first step, we have used Hölder’s inequality with weights $\sigma _1^2,\ldots ,\sigma _n^2$ (which sum to 1) and in the final step we have used Bernoulli’s inequality (which says that $(1+x)^r\ge 1+rx$ for $x\ge 0$ and $r\ge 1$ ; recall that we are assuming that $4(a_i^2+2\lambda _i^2)=4\sigma _i^2\le 1$ for each i).

Since $\sum _{i=1}^n\sigma _i^2=1$ , it now suffices to prove that for each $i=1,\ldots ,n$ we have

$$\begin{align*}\int_{|t|\ge K}\frac{\exp\Big(-{a_i^2t^2}/\left({(2+8\lambda_i^2t^2)\sigma_i^2}\right)\Big)}{1+\lambda_i^2t^2/\sigma_i^2}\,dt\lesssim \frac{1}{K}. \end{align*}$$

Fix some i. If $|\lambda _i|\ge |a_i|$ , then $\lambda _i^2\geq \sigma _i^2/3$ and

$$\begin{align*}\int_{|t|\ge K}\frac{\exp\Big(-{a_i^2t^2}/\left({(2+8\lambda_i^2t^2)\sigma_i^2}\right)\Big)}{1+\lambda_i^2t^2/\sigma_i^2}\,dt\le\int_{|t|\ge K}\frac{1}{1+t^2/3}\,dt\lesssim \frac{1}{K}.\end{align*}$$

Otherwise, if $|a_i|\ge |\lambda _i|$ , we have $a_i^2\geq \sigma _i^2/3$ , $\sigma _i^2\le 1$ , and therefore

$$ \begin{align*} \frac{\exp\Big(-{a_i^2t^2}/\left({(2+8\lambda_i^2t^2)\sigma_i^2}\right)\Big)}{1+\lambda_i^2t^2/\sigma_i^2}&\le \frac{\Big(1+{a_i^2t^2}/\left({(2+8\lambda_i^2t^2)\sigma_i^2}\right)\Big)^{-1}}{1+\lambda_i^2t^2/\sigma_i^2}\\ &\lesssim \frac{\big(1+{t^2}/({1+\lambda_i^2t^2})\big)^{-1}}{1+\lambda_i^2t^2/\sigma_i^2}\\ &\le \frac{\big(1+{t^2}/{(1+\lambda_i^2t^2)}\big)^{-1}}{1+\lambda_i^2t^2} = \frac{1}{1+(1+\lambda_i^2)t^2}. \end{align*} $$

It follows that

$$ \begin{align*} &\int_{|t|\ge K}\frac{\exp\Big(-{a_i^2t^2}/\left({(2+8\lambda_i^2t^2)\sigma_i^2}\right)\Big)}{1+\lambda_i^2t^2/\sigma_i^2}\,dt \lesssim \int_{|t|\ge K}\frac{1}{1+(1+\lambda_i^2)t^2}\,dt \lesssim \frac{1}{K}.\\[-46pt] \end{align*} $$

Proof. We may assume without loss of generality that $|\lambda _1|\geq \cdots \geq |\lambda _n|$ . By adding at most two terms with $a_i=\lambda _i=0$ , we may assume n is divisible by $3$ . Note that if $\sigma _i^2\le \sigma ^2/4$ for all $i\in [n]$ , the result follows immediately from Lemma 5.5(b). Therefore, it suffices to consider the case when there is an index j such that $\sigma _j^2\ge \sigma ^2/4$ .

Note that the given condition implies $\sum _{k=1}^{n/3}\lambda _{3k}^2\ge \frac {1}{3}\sum _{k= 3}^{n}\lambda _k^2\ge \eta \lambda _j^2/3$ . Now, Lemma 5.4 yields

$$ \begin{align*} \prod_{i=1}^{n}|\varphi_{X_i}(t)|&\le \exp\bigg(\frac{-a_j^2t^2}{2+8\lambda_j^2t^2}\bigg)\prod_{i=1}^{n}\frac{1}{(1+4\lambda_i^2t^2)^{1/4}}\le \exp\bigg(\frac{-a_j^2t^2}{2+8\lambda_j^2t^2}\bigg)\prod_{i=1}^{n/3}\frac{1}{(1+4\lambda_{3i}^2t^2)^{3/4}}\\ &\le \exp\bigg(\frac{-a_j^2t^2}{2+8\lambda_j^2t^2}\bigg)\bigg(1+4\sum_{i=1}^{n/3}\lambda_{3i}^2t^2\bigg)^{-3/4}\le \exp\bigg(\frac{-a_j^2t^2}{2+8\lambda_j^2t^2}\bigg)(1+\eta \lambda_j^2t^2)^{-3/4}\\ &\le \bigg(1 + \frac{a_j^2t^2}{2+8\lambda_j^2t^2}\bigg)^{-3/4}(1+\eta\lambda_j^2t^2)^{-3/4}\\ &\lesssim_{\eta} (\lambda_j^2t^2+a_j^2t^2)^{-3/4}\lesssim (\sigma_j |t|)^{-3/2}\lesssim (\sigma |t|)^{-3/2}. \end{align*} $$

Thus, we have

$$\begin{align*}\int_{|t|\ge 1/(32\sigma)}\prod_{i=1}^{n}|\varphi_{X_i}(t)|\,dt\lesssim_{\eta}\int_{|t|\ge 1/(32\sigma)}(\sigma |t|)^{-3/2}\,dt\lesssim 1/\sigma.\\[-47pt] \end{align*}$$

Proof of Theorem 1.6

By rescaling, we may assume $\sigma (f)=1$ . It suffices to show that the probability density function $p_{f-\mathbb E f}$ of $f-\mathbb Ef$ satisfies $p_{f-\mathbb E f}(u)\lesssim _{\eta } 1$ for all u.

Since F is a real symmetric matrix, we can write $F = QD Q^{\intercal }$ , where D is a diagonal matrix with entries $\lambda _1,\ldots ,\lambda _n$ and Q is an orthogonal matrix. Let $\vec W=Q^{\intercal }\vec Z$ , and note that $\vec W$ is also distributed as $\mathcal {N}(0,1)^{\otimes n}$ (since the distribution $\mathcal {N}(0,1)^{\otimes n}$ is invariant under orthogonal transformations). We have

$$\begin{align*}f(\vec Z) = f_0+\vec{f}\cdot \vec{Z} + \vec{Z}^\intercal F\vec{Z} = f_0+\vec{f}\cdot (Q\vec{W}) + \vec{W}^\intercal Q^\intercal FQ\vec{W} = f_0+(Q^\intercal\vec{f})\cdot \vec{W} + \vec{W}^\intercal D \vec{W}.\end{align*}$$

Let $\vec {a} = (a_1,\ldots ,a_n)=Q^\intercal \vec {f}$ . We have

$$\begin{align*}f-\mathbb E f= \sum_{i=1}^n(a_iW_i+\lambda_i(W_i^2-1)). \end{align*}$$

Let $\sigma _1,\ldots ,\sigma _n\geq 0$ be such that $\sigma _i^2=a_i^2+2\lambda _i^2$ , so $1=\sigma (f)^2 = \sigma _1^2+\cdots +\sigma _n^2$ . Note that the assumption in the theorem statement implies $n\ge 3$ , and combining the assumption with Theorem 4.13 yields

$$\begin{align*}\eta\le \min_{\substack{\widetilde F\in\mathbb{R}^{n\times n}\\\operatorname{rank}(\widetilde F)\le 2}}\frac{\|F-\widetilde F\|^2_{\mathrm F}}{\|F\|^2_{\mathrm{F}}}=\min_{\substack{I\subseteq [n]\\|I|=n-2}} \frac{\sum_{i\in I}\lambda_i^2}{\lambda_1^2+\cdots+\lambda_n^2}.\end{align*}$$

Hence, for any subset $I\subseteq [n]$ with $|I|= n-2$ and any $j\in [n]$ we obtain $\sum _{i\in I}\lambda _i^2\ge \eta (\lambda _1^2+\,\cdots\, +\lambda _n^2)\geq \eta \lambda _j^2$ . Let $\Gamma $ be as in Lemma 5.5, and recall that $\Gamma \ge 1$ .

Now, by combining Lemma 5.5(a) and Lemma 5.7, we have that

$$ \begin{align*} \sup_{u\in\mathbb{R}}p_{f}(u)=\sup_{u\in\mathbb{R}}p_{f-\mathbb E f}(u)\lesssim \frac{1}{\sqrt{2\pi}} + \frac{1}{\Gamma} + \int_{|t|\ge \Gamma/32}\prod_{i=1}^{n}|\varphi_{X_i}(t)|\,dt\lesssim_{\eta} 1. \end{align*} $$

By integrating over the desired interval, we obtain the bound in Theorem 1.6.

Proof. We may assume $a\ge 0$ (changing a to $-a$ does not change the distribution of X). First, note that the case $\lambda =0$ is easy since then we have $\sigma (X)=a$ and $p_X(u)=e^{-(u/a)^2/2}/(\sqrt {2\pi }a)\gtrsim _{A'} 1/\sigma (X)$ . So let us assume $\lambda \neq 0$ and define $g\colon \mathbb {R}\to \mathbb {R}$ by

$$\begin{align*}g(t)=at+\lambda(t^2-1)=\lambda\cdot \left(t+\frac{a}{2\lambda}\right)^2-\lambda-\frac{a^2}{4\lambda}.\end{align*}$$

Then for all $t\in \mathbb {R}$ , we have

$$\begin{align*}(g'(t))^2=4\lambda^2\cdot \left(t+\frac{a}{2\lambda}\right)^2=4\lambda\cdot g(t)+4\lambda^2+a^2\leq 4\sigma(X)\cdot |g(t)|+4\sigma(X)^2. \end{align*}$$

Hence, for any $t\in \mathbb {R}$ with $g(t)=u$ , recalling $0\le u\le A'\sigma (X)$ , we obtain $|g'(t)|\leq \sqrt {4(A'+1)\sigma (X)^2}\lesssim _{A'} \sigma (X)$ .

We claim that we can find $t\in [-3A',3A']$ with $g(t)=u$ . Indeed, in case (1), we have $g(0)=-\lambda \leq u$ and $g(2A'+1)\geq 2A'a+2A'\lambda \geq A'\sigma (X)\geq u$ , and hence by the intermediate value theorem there exists $t\in [0,2A'+1]\subseteq [-3A',3A']$ with $g(t)=u$ . In case (2), observe that $a^2=\sigma (X)^2-2\lambda ^2\geq 100 A^{\prime 2}\cdot \lambda ^2-2\lambda ^2\geq 81A^{\prime 2}\cdot \lambda ^2$ , so $a\geq 9A'\cdot |\lambda |$ and therefore $|\lambda (9A^{\prime 2}-1)|\leq A'a$ and $\sigma (X)^2=a^2+2\lambda ^2\leq 4a^2$ . Hence, $g(-3A')=-3A'a+\lambda (9A^{\prime 2}-1)\leq -2A'a\leq 0\leq u$ and $g(3A')=3A'a+\lambda (9A^{\prime 2}-1)\geq 2A'a\geq A'\sigma (X)\geq u$ and we can again conclude that there exists $t\in [-3A',3A']$ with $g(t)=u$ .

Now, we have

$$\begin{align*}p_X(u)=p_{g(W)}(g(t))\geq \frac{p_W(t)}{|g'(t)|}\gtrsim_{A'} \frac{e^{-(3A')^2/2}}{\sigma(X)}\gtrsim_{A'} \frac{1}{\sigma(X)}.\\[-42pt] \end{align*}$$

Proof. By rescaling, we may assume that $\sigma (X) = 1$ . Note that then

where we have used that

for all $x\in \mathbb {R}$ . The result follows.

Proof of Theorem 5.2(2)

We may assume $\sigma (f)=1$ . Borrowing the notation from the proof of Theorem 1.6, we write

$$\begin{align*}f-\mathbb E f= \sum_{i=1}^n(a_iW_i+\lambda_i(W_i^2-1)), \end{align*}$$

with $(W_1,\ldots ,W_n)\sim \mathcal {N}(0,1)^{\otimes n}$ , and $\sigma _i^2=a_i^2+2\lambda _i^2$ (then we have $1=\sigma ^2=\sigma _1^2+\cdots +\sigma _n^2$ ). It now suffices to prove that for all $u\in [0,A+1]$ we have $p_{f-\mathbb E f}(u) \gtrsim _A 1$ . Let L be a large integer depending only on A (such that $L\geq 2$ and $L\geq C_{5.5}(1+32C_{5.5}')\cdot 2\sqrt {2\pi }\cdot e^{(A+1)^2/2}$ for the constants $C_{5.5}$ and $C_{5.5}'$ in Lemma 5.5). We break into cases.

First, suppose $\max _i\sigma _i\le 1/L$ . In this case, we define $\Gamma =\sigma (f)^{3}/\sum _{i=1}^n\sigma _i^3=1/\sum _{i=1}^n\sigma _i^3$ and note that $\sum _{i=1}^{n}\sigma _i^3\le (\max _i\sigma _i)(\sum _{i=1}^{n} \sigma _i^2)\le 1/L$ , so $\Gamma \ge L$ . We also have $\sigma _i^2\leq 1/L^2\leq 1/4$ , so Lemma 5.5(b) applies. So by combining parts (a) and (b) of Lemma 5.5, for all $u\in [0,A+1]$ we obtain, as desired,

$$\begin{align*}p_{f-\mathbb E f}(u)\ge \frac{e^{-u^2/2}}{\sqrt{2\pi}}-\frac{C_{5.5}(1+32C_{5.5}')}{\Gamma}\ge \frac{e^{-(A+1)^2/2}}{\sqrt{2\pi}}-\frac{C_{5.5}(1+32C_{5.5}')}{L}\ge \frac12\cdot \frac{e^{-(A+1)^2/2}}{\sqrt{2\pi}}\gtrsim_A 1.\end{align*}$$

Otherwise, there is $i^*\in [n]$ such that $\sigma _{i^*} \ge 1/L$ . We claim that then there is an index $j\in [n]$ satisfying at least one of the following two conditions:

1. (1) $\sigma _{j} \ge 1/(10(A+19)L^2)$ and $\lambda _{j}\geq 0$ , or

2. (2) $\sigma _{j} \ge 1/L$ and $10(A+19)L\cdot |\lambda _{j}|\le \sigma _{j}$ .

Indeed, if $10(A+19)L\cdot |\lambda _{i^*}|\le \sigma _{i^*}$ we can simply take $j=i^*$ and (2) is satisfied. Otherwise, we have $|\lambda _{i^*}|> \sigma _{i^*}/(10(A+19)L)\geq 1/(10(A+19)L^2)$ and the assumption in Theorem 5.2(2) yields $\lambda _1\geq |\lambda _{i^*}|\geq 1/(10(A+19)L^2)$ . So in particular $\lambda _1\geq 0$ and $\sigma _1\geq \lambda _1\geq 1/(10(A+19)L^2)$ , and we can take $j=1$ and (1) is satisfied.

Now, let $X_j=a_{j} W_{j}+\lambda _{j}(W_{j}^2-1)$ and let $X'=f-\mathbb E f-X_j=\sum _{i\neq j} (a_i W_i+\lambda _i(W_i^2-1))$ contain all terms of $f-\mathbb E f$ except the term $X_j$ . By Theorem 4.14 (hypercontractivity), we have $\mathbb E[(X')^4]\leq 81\sigma (X')^4$ , and therefore, Lemma 5.9 shows that $-18\le -18\sigma (X')\le X'\le 0$ with probability at least $1/405$ .

We claim that we can apply Lemma 5.8 to $X_j$ and $u\in [0,A+19]$ , showing that $p_{X_j}(u)\gtrsim _A 1/\sigma _{j}\geq 1$ . Indeed, in case (1) we have $0\leq u\leq 10(A+19)^2L^2\sigma _j$ and can apply case (1) of Lemma 5.8 with $A'=10(A+19)^2L^2$ , while in case (2) we have $0\leq u\leq (A+19)L\sigma _j$ and can apply case (2) of Lemma 5.8 with $A'=(A+19)L$ .

Therefore, for any $u\in [0,A+1]$ we obtain

$$ \begin{align*} p_{f-\mathbb E f}(u)=p_{X'+X_j}(u)&\geq \int_{-18}^{0} p_{X'}(y)p_{X_j}(u-y)\,dy\\ &\gtrsim_A \int_{-18}^{0} p_{X'}(y)\,dy =\Pr[-18\leq X'\leq 0]\gtrsim 1.\\[-45pt] \end{align*} $$

Proof. Define the function $g\colon \mathbb {R}\to \mathbb {R}$ by $g(t)=at+\lambda (t^2-1)$ . As in the proof of Lemma 5.8, we can calculate $(g'(t))^2=4\lambda \cdot g(t)+4\lambda ^2+a^2$ for all $t\in \mathbb {R}$ . Now, consider some $u\in \mathbb {R}$ with $x/10\leq |u|\leq 2x$ . There are at most two different $t\in \mathbb {R}$ with $g(t)=u$ . For any such t, we have (using the assumption that $|\lambda |\cdot x\leq a^2/10$ )

$$\begin{align*}(g'(t))^2\geq 4\lambda^2+a^2-4|\lambda|\cdot 2x\geq a^2/5\geq a^2/9.\end{align*}$$

We furthermore claim that any such t must satisfy $|t|\geq x/(20|a|)$ . Indeed, if $|t|< x/(20|a|)$ , then (using that $x\geq 10^3\sigma \geq 10^3a$ and the assumption $|\lambda |\cdot x\leq a^2/10$ )

$$\begin{align*}|g(t)|=|at+\lambda(t^2-1)|\leq |a|\cdot \frac{x}{20|a|}+|\lambda|\cdot \max\left\{\frac{x^2}{400a^2},1\right\}\leq \frac{x}{20}+|\lambda|\cdot\frac{x^2}{400a^2}\leq \frac{x}{20}+\frac{x}{4000}<\frac{x}{10}.\end{align*}$$

As $|u|\geq x/10$ , this contradicts $g(t)=u$ . Thus, any $t\in \mathbb {R}$ with $g(t)=u$ must indeed also satisfy $|t|\geq x/(20|a|)$ . Now, we obtain (using again that $x\geq 10^3\sigma \geq 10^3a$ )

$$\begin{align*}p_X(u)=\sum_{\substack{t\in\mathbb{R}\\g(t)=u}} \frac{p_W(t)}{|g'(t)|}\leq 2\cdot \frac{1}{|a|/3}\cdot \exp\left(-\frac{x^2}{800a^2}\right)\lesssim \frac{1}{|a|}\exp\left(-\frac{x}{\sigma}\right). \\[-55pt] \end{align*}$$

Proof of Theorem 5.2(1)

By rescaling, we may assume $\sigma :=\sigma (f)=1$ . If $|x|\leq 10^3=10^3\sigma (f)$ , the desired bound follows from Theorem 1.6. So we may assume that $|x|\geq 10^3\sigma (f)$ . Also, note that the assumption in Theorem 5.2(1) implies that $\eta \le 1$ . Borrowing the notation from the proof of Theorem 1.6, we write

$$\begin{align*}f-\mathbb E f= \sum_{i=1}^n(a_iW_i+\lambda_i(W_i^2-1)), \end{align*}$$

with $(W_1,\ldots ,W_n)\sim \mathcal N(0,1)^{\otimes n}$ and $\sigma _i^2=a_i^2+2\lambda _i^2$ (then we have $1=\sigma ^2=\sigma _1^2+\cdots +\sigma _n^2$ ). We may assume that $|\lambda _1|\geq \cdots \geq |\lambda _n|$ . Note that using Theorem 4.13 the assumption in Theorem 5.2(1) implies that for every subset $I\subseteq [n]$ of size $|I|=n-3$ we have $\sum _{i\in I}\lambda _i^2\geq \eta (\lambda _1^2+\cdots +\lambda _n^2)$ . In particular, $\sum _{i=4}^{n}\lambda _i^2\geq \eta (\lambda _1^2+\cdots +\lambda _n^2)$ .

By adding at most three terms with $a_i=\lambda _i=0$ , we may assume that $n\equiv 1 \pmod {4}$ . For a subset $J\subseteq [n]$ , let $X_{J}=\sum _{i\in J}(a_iW_i+\lambda _i(W_i^2-1))$ and $\sigma _{J}^{2}=\sum _{i\in J}\sigma _{i}^2=\sigma (X_{J})^2$ .

Let $i^*\in [n]$ be chosen such that $\sigma _{i^*}^2$ is maximal, and define $J_0=\{i^*\}$ . We claim that we can find a partition of $[n]\setminus J_0=[n]\setminus \{i^*\}$ into four subsets $J_1,J_2,J_3,J_4$ satisfying the following conditions.

1. (a) For $h=1,2,3,4$ , we have $\sigma _{[n]\setminus J_h}^2\geq \eta /2$ .

2. (b) For any $h=0,\ldots ,4$ and any subset $I\subseteq [n]\setminus J_h$ of size $|I|=n-|J_h|-2$ , we have $\sum _{i\in I}\lambda _i^2\geq (\eta /4)\cdot (\lambda _1^2+\cdots +\lambda _n^2)$ .

Indeed, we can build such a partition iteratively: Let us divide $[n]\setminus \{i^*\}$ into $n/4$ quadruplets (starting with the four smallest indices, then the next four, and so on). Iteratively, for each quadruplet, distribute one element to each of $J_1,J_2,J_3,J_4$ in the following way. We assign the index i in the quadruplet with the largest $\sigma _i^2$ to the set $J_h$ which had the smallest value of $\sigma _{J_h}^{2}$ at the end of the last step, we assign the index i with the second-largest $\sigma _i^2$ to the set $J_h$ which had the second-smallest value of $\sigma _{J_h}^{2}$ , and so on. One can check that this assignment process maintains the property that at the end of any step, the values $\sigma _{J_h}^{2}$ for $h=1,2,3,4$ differ by at most $\max _{i}\sigma _{i}^{2}=\sigma _{i^*}^2$ . Hence, $\sigma _{[n]\setminus J_1}^2\geq \sigma _{J_2}^2+\sigma _{i^*}^2\geq \sigma _{J_1}^2=1-\sigma _{[n]\setminus J_1}^2$ , so $\sigma _{[n]\setminus J_1}^2\geq 1/2\geq \eta /2$ . Analogously, one can show $\sigma _{[n]\setminus J_h}^2\geq \eta /2$ for $h=2,3,4$ , so (a) is satisfied. To check (b), note that for each $h=0,\ldots ,4$ the set $[n]\setminus J_h$ is missing either one element from each of the quadruplets considered during the construction (if $1\leq h\leq 4$ ) or is missing one element in total (if $h=0$ ). For a subset $I\subseteq [n]\setminus J_h$ of size $|I|=n-|J_h|-2$ , two additional elements are missing. Thus, for every $k=1,\ldots ,n/4$ the set $I\subseteq [n]$ is missing at most $k+2$ of the elements in $[4k]$ . Thus, recalling that $|\lambda _1|\geq \cdots \geq |\lambda _n|$ , we obtain

$$ \begin{align*} \sum_{i\in I}\lambda_i^2 &\ge \lambda_4^2+(\lambda_6^2+\lambda_7^2+\lambda_8^2)+(\lambda_{10}^2+\lambda_{11}^2+\lambda_{12}^2)+\cdots \\[-2pt] &\geq \lambda_4^2+\lambda_8^2+\lambda_{12}^2+\cdots\ge \frac{1}{4}\sum_{i=4}^{n}\lambda_i^2\geq (\eta/4)\cdot (\lambda_1^2+\cdots+\lambda_n^2). \end{align*} $$

This establishes (b). Thus, the sets $J_1,\ldots ,J_4$ indeed satisfy the desired conditions.

By our assumption $|x|\geq 10^3\sigma (f)$ and by $0\leq \varepsilon \leq \sigma (f)$ , we have $|y|\geq 0.999|x|\geq (5/6)\cdot |x|$ for all $y\in [x,x+\varepsilon ]$ . Thus, whenever $f-\mathbb {E} f=\sum _{i=1}^n (a_iW_i+\lambda _i(W_i^2-1))=X_{J_0}+\cdots +X_{J_4}$ is contained in the interval $[x,x+\varepsilon ]$ , we must have $|X_{J_h}|\geq |x|/6$ for at least one $h\in \{0,\ldots ,4\}$ . So, we have

(5.2) $$ \begin{align} \Pr[f-\mathbb{E} f\in [x,x+\varepsilon]] \leq \sum_{h=0}^{4}\Pr\Big[|X_{J_h}|\geq |x|/6\text{ and }X_{[n]\setminus J_h}\in [x-X_{J_h},x-X_{J_h}+\varepsilon]\Big]. \end{align} $$

For $h= 1,\ldots ,4$ , note that

(5.3) $$ \begin{align} &\Pr\Big[|X_{J_h}|\geq |x|/6\text{ and }X_{[n]\setminus J_h}\in [x-X_{J_h},x-X_{J_h}+\varepsilon]\Big]\notag\\ &\quad\le \Pr[|X_{J_h}|\ge|x|/6]\cdot \mathcal L(X_{[n]\setminus J_h},\varepsilon)\lesssim_{\eta} \exp\left(-\frac{2}{2e}\cdot \frac{|x|}{6\sigma_{J_h}}\right)\frac{\varepsilon}{\sigma_{[n]\setminus J_h}} \lesssim_{\eta}\frac{\varepsilon}{\sigma}\exp\left(-\Omega\left(\frac{|x|}{\sigma}\right)\right), \end{align} $$

where in the second step we applied Theorem 4.15 to $X_{J_h}$ with $t=|x|/(6\sigma _{J_h})\geq |x|/(6\sigma )$ and Theorem 1.6 to $X_{[n]\setminus J_h}$ (noting that the assumption of Theorem 1.6 is satisfied by condition (b); see also Remark 5.1), and in the last step we used that $\sigma _{[n]\setminus J_h}^2\geq \eta /2$ by condition (a).

We now distinguish two cases. First, let us assume that $\sigma _{[n]\setminus J_0}\ge \eta ^2/(100|x|)$ . In this case, similarly to (5.3), we can bound (recalling that $\sigma =1$ )

$$ \begin{align*} &\Pr\Big[|X_{J_0}|\geq |x|/6\text{ and }X_{[n]\setminus J_0}\in [x-X_{J_0},x-X_{J_0}+\varepsilon]\Big]\\ &\qquad\lesssim_{\eta} \exp\left(-\frac{2}{2e}\cdot \frac{|x|}{6\sigma_{J_0}}\right)\cdot \frac{\varepsilon}{\sigma_{[n]\setminus J_0}} \lesssim_{\eta}\frac{\varepsilon}{\sigma}\cdot \frac{|x|}{40\sigma}\cdot \exp\left(-\frac{|x|}{20\sigma}\right)\le \frac{\varepsilon}{\sigma}\exp\left(-\frac{|x|}{40\sigma}\right), \end{align*} $$

where in the last step we used that $te^{-t}\le 1/e\le 1$ for all $t\in \mathbb {R}$ (specifically, we used this for $t=|x|/(40\sigma )$ ). Together with (5.3), this enables us to bound all five summands on the right-hand side of (5.2), implying the desired bound for $\Pr [f-\mathbb {E} f\in [x,x+\varepsilon ]]$ .

It remains to consider the case that $\sigma _{[n]\setminus J_0}< \eta ^2/(100|x|)$ . Then we in particular have $\sigma _{i^*}^2=1-\sigma _{[n]\setminus J_0}^2\ge 1-\eta ^4/(10^4|x|^2)\ge 1-\eta /2$ . Furthermore, the assumption in Theorem 5.2(1) implies $\sum _{i\in [n]\setminus \{i^*\}} \lambda _i^2\geq \eta (\lambda _1^2+\cdots +\lambda _n^2)$ , and therefore $\lambda _{i^*}^2\leq (1-\eta )(\lambda _1^2+\cdots +\lambda _n^2)\leq (1-\eta )/2$ (recalling that $1=\sigma ^2=\sum _{i=1}^{n}(a_i^2+2\lambda _i^2)$ ). Thus, we obtain $a_{i^*}^2=\sigma _{i^*}^2-2\lambda _{i^*}^2\geq (1-\eta /2)-(1-\eta )= \eta /2$ . Our assumption also implies $\eta ^4/(10^4|x|^2)> \sigma _{[n]\setminus J_0}^2 \ge \sum _{i\in [n]\setminus \{i^*\}} \lambda _i^2 \ge \eta \lambda _{i*}^2$ , meaning that $|\lambda _{i^*}|\cdot |x|\le \eta /100 \le a^2_{i^*}/10$ .

Now, we observe

$$ \begin{align*} &\Pr[f-\mathbb{E} f\in [x,x+\varepsilon]]\\ &\qquad \leq \sum_{h=1}^{4}\Pr\Big[|X_{J_h}|\geq |x|/6\text{ and }X_{[n]\setminus J_h}\in [x-X_{J_h},x-X_{J_h}+\varepsilon]\Big]\\ &\qquad\qquad\qquad+\Pr\Big[|X_{[n]\setminus J_0}|\leq (4/6)|x|\text{ and }X_{J_0}\in [x-X_{[n]\setminus J_0},x-X_{[n]\setminus J_0}+\varepsilon]\Big]. \end{align*} $$

Again, (5.3) gives an upper bound for the summands for $h=1,\ldots ,4$ . To bound the last summand, let us fix any outcome of $X_{[n]\setminus J_0}$ with $|X_{[n]\setminus J_0}|\leq (4/6)|x|$ . Then the probability that $X_{J_0}=a_{i^*}W_{i^*}+\lambda _{i^*}(W_{i^*}^2-1)$ lies in the interval $[x-X_{[n]\setminus J_0},x-X_{[n]\setminus J_0}+\varepsilon ]$ (which has length $\varepsilon $ and is somewhere between $x/10$ and $2x$ ) is by Lemma 5.10 bounded by

$$\begin{align*}\Pr[X_{J_0}\in [x-X_{[n]\setminus J_0},x-X_{[n]\setminus J_0}+\varepsilon]]\lesssim \frac{\varepsilon}{|a_{i^*}|}\exp\left(-\frac{|x|}{\sigma_{J_0}}\right)\lesssim_{\eta} \frac{\varepsilon}{\sigma}\exp\left(-\frac{|x|}{\sigma}\right),\end{align*}$$

where in the last step we used that $a_{i^*}^2\geq \eta /2$ (see above). Thus, we again obtain the desired bound for $\Pr [f-\mathbb {E} f\in [x,x+\varepsilon ]]$ .

Proof. Let $\lambda _1,\ldots ,\lambda _n$ be the eigenvalues of F, ordered such that $|\lambda _1|\ge \cdots \ge |\lambda _n|$ . By Theorem 4.13, we have $s=\sum _{j=r+1}^n\lambda _j^2$ .

As in the proof of Theorem 1.6, we write $f(\vec Z)-\mathbb {E}[f(\vec Z)]=\sum _{j=1}^{n}(a_jW_j + \lambda _i(W_j^2-1))$ , where $(W_1,\ldots ,W_n)\sim \mathcal {N}(0,1)^{\otimes n}$ are independent standard Gaussians. From Lemma 5.4, recall that

$$\begin{align*}|\mathbb{E}\exp(i\tau (a_j W_j+\lambda_j(W_j^2-1)))|=|\mathbb{E}\exp(i\tau (a_j W_j+\lambda_jW_j^2))| \le \frac{1}{(1+4\lambda_j^2\tau^2)^{1/4}}\end{align*}$$

for $j=1,\ldots ,n$ . We then deduce

$$ \begin{align*} |\mathbb{E}[\exp(i\tau f(\vec{Z}))]| &= \prod_{j=1}^{n}|\mathbb{E}[\exp(i\tau (a_jW + \lambda_j(W_j^2-1)))]| \le \prod_{j=1}^{n}\frac{1}{(1+4\lambda_j^2\tau^2)^{1/4}}\\ &\le \prod_{j=1}^{r}\left(1+4\tau^2\sum_{t=0}^{\lfloor{(n-j)/r}\rfloor}\lambda_{j + rt}^2\right)^{-1/4}\le \left(1+4\tau^2\sum_{t=0}^{\lfloor{(n-r)/r}\rfloor}\lambda_{r+rt}^2\right)^{-r/4}\\ &\le \left(1+\frac{4\tau^2}r\sum_{j=r+1}^{n}\lambda_{j}^2\right)^{-r/4}\lesssim_{r} \frac{1}{(1+\tau^2s)^{r/4}}. \\[-48pt] \end{align*} $$