Shravas Rao

We attempt to prove that small discrepancy implies small spectral expansion for Cayley hypergraphs over certain types of abelian groups.  In the case of regular graphs, this is known to hold for Cayley graphs over any group by a result of Conlan and Zhao, but does not hold for general graphs~\cite{Conlon:2016}.

We start by giving definitions.  Because these are just my personal notes, I will let the definition of Cayley hypergraphs be the exact same as that in my paper with Jop, except we let $\mathbf{g} = (0, g, g^2, \ldots, g^{t-1})$ and $q = \mathbf{1}$ for now.  In particular, $K = \cay^{(t)}(\Gamma, q, S)$ is a $t$-uniform Cayley hypergraph with edges induced by the set $S \subseteq \Gamma$, and we let $A_K$ be the corresponding adjacency tensor.  Given a Cayley hypergraph $K$, we say that a sub-hypergraph of $K$, $H = \cay^{(t)}(\Gamma, q, S')$ for some $S' \subseteq S$ has spectral expansion $\lambda$ with respect to $K$ if \[\lambda \geq \|A_H-A_K\|_{t, \ldots, t}.\]  We say that $H$ has discrepancy $\epsilon$ with respect to $K$ if for all $T_1, \ldots, T_t \subseteq \Gamma$, 

\[|A_H(1_{T_1}, \ldots, 1_{T_t})-A_K(1_{T_1}, \ldots, 1_{T_t})| \leq \epsilon |\Gamma|.\]  

We can lower bound discrepancy in terms of the $\ell_{\infty}, \ldots, \ell_{\infty}$-norm of $A_H-A_K$ as stated in the following Lemma (the upper bound is trivial).  The proof follows almost exactly that of the same fact for matrices in Lemma 2.1 in~\cite{Conlon:2016}.

If $A$ is an $n \times \cdots \times n$ $t$-linear form, there exist $T_1, \ldots, T_t \subseteq [n]$ such that 

\[\|A\|_{\ell_{\infty}, \ldots, \ell_{\infty}} \leq \pi^t|A(1_{T_1}, \ldots, 1_{T_t})|\]

By a straightforward generalization of the Expander Mixing Lemma, it follows that if a Cayley hypergraph has spectral expansion $\lambda$, then it also has discrepancy $\epsilon$ for some constant $\epsilon$ depending on $\lambda$.  For completeness, we will state a more general form of the Expander Mixing Lemma below.

\begin{lem}

Let $f \in \R^{\Gamma}$, let $A_g$ be the adjacency tensor of the hypergraph $\cay^{(t)}(\Gamma, g)$.

Then for every $\lambda$, there exists an $\eps$ such that if

\[

\left\|\sum_{g \in \Gamma} f(g)A_g\right\|_{t, \ldots, t} \leq \lambda,

\]

then

\[

\left\|\sum_{g \in \Gamma} f(g)A_g\right\|_{\infty, \ldots, \infty} \leq \lambda|\Gamma|.

\]

\end{lem}

\begin{proof}

Let $T_1, \ldots, T_t$ be the sets obtained by applying Lemma~\ref{lem:disctoinfty} to $\sum_{g \in \Gamma} A_g$.

Then,

\begin{align*}

\left\|\sum_{g \in \Gamma} f(g)A_g\right\|_{\infty, \ldots, \infty} &\leq \sum_{g \in \Gamma} f(g)A_g(1_{T_1}, \ldots, 1_{T_t}) \\

&\leq C\left\|\sum_{g \in \Gamma} f(g)A_g\right\|_{t, \ldots, t}\|1_{T_1}\|_t\cdots\|1_{T_t}\}_t \\

&\leq C\lambda |\Gamma|

\end{align*}

as desired.

\end{proof}

We attempt to prove the converse, stated below.

Let $f \in \R^{\Gamma}$, let $A_g$ be the adjacency tensor of the hypergraph $\cay^{(t)}(\Gamma, g)$.

Then for every $\lambda$, there exists an $\eps$ such that if

\[

\left\|\sum_{g \in \Gamma} f(g)A_g\right\|_{\infty, \ldots, \infty} \leq \eps|\Gamma|.

\]

then

\[

\left\|\sum_{g \in \Gamma} f(g)A_g\right\|_{t, \ldots, t} \leq \lambda,

\]

As a corollary, we see that small discrepancy implies small spectral expansion.

\begin{cor}

Let $\Gamma = F_p^n$. For $t \leq p$ and every $\lambda$ there exists an $\epsilon$ such that if a Cayley $t$-uniform hypergraph $H = \cay^{(t)}(\Gamma, q, S')$ has discrepancy $\epsilon$ with respect to $K = \cay^{(t)}(\Gamma, q, S)$, then it also has spectral expansion $\lambda$ with respect to $K = \cay^{(t)}(\Gamma, q, S)$.

\end{cor}

\begin{proof}

If for all $T_1, \ldots, T_t \subseteq \Gamma$ it holds that

\[

|A_H(1_{T_1}, \ldots, 1_{T_t})-A_K(1_{T_1}, \ldots, 1_{T_t})| \leq \epsilon |\Gamma|,

\]

it follows by Lemma~\ref{lem:disctoinfty} that

$\|A_H-A_K\|_{\infty, \ldots, \infty} \leq \pi^t \epsilon |\Gamma|$.

Finally, by Lemma~\ref{lem:disctospec} it follows that $\|A_H-A_K\|_{t, \ldots, t} \leq \lambda$.

\end{proof}

To prove Lemma~\ref{lem:disctospec}, we use the generalized von Neumann inequality and the inverse Gowers theorem, which we state below.  We first state the definition of the Gowers norm.

The generalized von Neumann inequality can be thought of as an analogue of H\"{o}lder's inequality, in which the upper bound involves the Gower's norm of some function.

In the proof of Lemma~\ref{lem:disctospec}, we will explain how to choose $f$ and $\{a_{ij}\}$ so that the left-hand side of Equation~\eqref{eq:genvn} is equal to $\sum_{g \in \Gamma} f(g)A_g(x_1, \ldots, x_t)$ for vectors $x_i$.

Let $\{a_{ij}\} \in (\F_p^n)^{t \times t}$ be so that $a_{ij} = 0$ for all $i = j$, and non-zero otherwise.

Let $f: (\F_p^n) \rightarrow \R$ be a function, and  $g_i :(\F_p^n) \rightarrow \R$ for each $1 \leq i \leq t$ such that $|g_i| \leq 1$ for all $i$.

For all distinct $i_1, i_2 \in [t]$,

\begin{multline}\label{eq:genvn}

\left| \mathbb{E}_{z_1, \ldots, z_t \in \F_p^n} \left[f(z_1 + \cdots+z_t) \prod_{i=1}^t g_i(a_{i1} z_1+ \ldots + a_{it} z_t))\right]\right|

\leq \\

\|f\|_{U^{3}}\mathbb{E}_{z \in \F_p^n}[g_{i_1}(z)^2]^{1/2}\mathbb{E}_{z \in \F_p^n}[g_{i_2}(z)^2]^{1/2}

%\mathbb{E}_{z \in \F_p^n}[g_3(z)^4]^{1/4}

\end{multline}

\end{lem}

We note that this version of the generalized von Neumann inequality is slightly tighter than the usual version.

Rather than bounding in terms of $\max_{z \in \F_p^n} g_i(z)$, we bound in terms of the $2$nd moments.

For completeness, we include the proof below.

\begin{proof}[Proof of Lemma~\ref{lem:genvn}]

Without loss of generality, assume that $i_1 = 1$ and $i_2 = 2$.

If this is not the case, one can exchange $g_{1}$ and $g_2$ with $g_{i_1}$ and $g_{i_2}$ and the first and second rows of $a$ with the $i_1$st and $i_2$nd rows of $a$, and the same with the columns of $a$.

Note that by the assumptions of the lemma,

\mathbb{E}_{z_1, \ldots, z_t \in \F_p^n}[g_i(a_{i1} z_1+ \ldots + a_{it} z_t))^2]^{1/2} = \mathbb{E}_{z \in \F_p^n}[g_i(z)^2]^{1/2}

for all $i$.

By the above, the fact that $g_1$ does not depend on $z_1$, and the Cauchy-Schwarz inequality, the right-hand side of Eq.~\eqref{eq:genvn} is bounded above by

\begin{align*}

\mathbb{E}&_{z_2, \ldots, z_t \in \F_p^n} \left[g_1(a_{12} z_2+ \cdots + a_{1t} z_t)) \mathbb{E}_{z_1 \in \F_p^n}\left[f(z_1 + \cdots+z_t) \prod_{i=2}^t g_i(a_{i1} z_1+ \cdots + a_{it} z_t))\right]\right]

\leq \\

&\mathbb{E}_{z_2, \ldots, z_t \in \F_p^n} \left[\mathbb{E}_{z_1 \in \F_p^n}\left[f(z_1 + \cdots+z_t) \prod_{i=2}^t g_i(a_{i1} z_1+ \cdots + a_{it} z_t))\right]^2\right]^{1/2}\mathbb{E}_{z \in \F_p^n} \left[g_1(z)^2\right]^{1/2}.

\end{align*}

Note that

\begin{align*}

\mathbb{E}_{z_1 \in \F_p^n}&\left[f(z_1 + z_2 + z_3) \prod_{i=2}^t g_i(a_{i1} z_1+ \cdots + a_{it} z_t)\right]^2 \nonumber

\\

&=  \mathbb{E}_{z_1 \in \F_p^n}\left[f(z_1 + \cdots+z_t) \prod_{i=2}^t g_i(a_{i1} z_1+ \cdots)\right]

\mathbb{E}_{z_1' \in \F_p^n}\left[f(z_1' + z_2 + \cdots+z_t) \prod_{i=2}^t g_i(a_{i1} z_1'+ \cdots)\right] \\

&= \mathbb{E}_{z_1, h_1 \in \F_p^n}\left[\Delta_{h_1} \left(f(z_1 + \cdots+z_t) \prod_{i=2}^t g_i(a_{i1} z_1+ \cdots + a_{it} z_t)\right)\right].

\end{align*}

Applying Cauchy-Schwarz again, we get an upper bound of

\begin{multline*}

\mathbb{E}_{z_1, h_1, z_3, z_4, \ldots \in \F_p^n} \left[\mathbb{E}_{z_2 \in \F_p^n}\left[\Delta_{h_1} \left(f(z_1 + \cdots+z_t) \prod_{i=3}^t g_i(a_{i1} z_1+ \cdots + a_{it} z_t)\right)

\right]^2\right]^{1/2} \\

\mathbb{E}_{z_1, h, z_3, z_4, \ldots \in \F_p^n}\left[g_2(a_{21} z_1+  a_{23} z_3+a_{24}z_a+\cdots)^2g_2(a_{21} (z_1+h_1)+ a_{23} z_3+a_{24}z_a+\cdots)^2\right]^{1/2}.

\end{multline*}

Note that

\begin{align*}

\mathbb{E}&_{z_1, h, z_3, z_4, \ldots\in \F_p^n}\left[g_2(a_{21} z_1+  a_{23} z_3+a_{24}z_a+\cdots)^2g_2(a_{21} (z_1+h_1)+ a_{23} z_3+a_{24}z_a+\cdots)^2\right] = \\

&= \mathbb{E}_{z_3, z_4, \ldots\in \F_p^n}\left[\mathbb{E}_{z_1}\left[g_2(a_{21} z_1+ a_{23} z_3+a_{24}z_a+\cdots)^2\right]\mathbb{E}_{z_1'}\left[g_2(a_{21} z_1'+  a_{23} z_3+a_{24}z_a+\cdots)^2\right]\right] \\

&= \mathbb{E}_{z}\left[g_2(z)^2\right]^2.

\end{align*}

The rest of the proof follows by the usual generalized von Neumann inequality.

%Then,

%\begin{multline*}

%\mathbb{E}_{z_2 \in \F_p^n}\left[\Delta_{h_1} f(z_1 + z_2+z_3)  g_3(a_{31} z_1+ a_{32} z_2)g_3(a_{31} (z_1+h_1)+ a_{32} z_2)\right]^2 = \\

%\mathbb{E}_{z_2, h_2 \in \F_p^n}\left[\Delta_{h_1, h_2} f(z_1 + \cdots + z_t) \prod_{s \in \{0, 1\}^2} g_3(a_{31} (z_1+s_1h_1)+ a_{32}(z_2+s_2h_2))\right]^2

%\end{multline*}

%One more application of Cauchy-Schwarz gives

%\begin{multline*}

%\mathbb{E}_{z_1, h_1, z_2, h_2 \in \F_p^n}\left[\mathbb{E}_{z_3}\left[\Delta_{h_1, h_2} f(z_1 + \cdots + z_t)\right]^2\right]^{1/2}

%\mathbb{E}_{z_1, h_1, z_2, h_2 \in \F_p^n}\left[\prod_{s \in \{0, 1\}^2} g_3(a_{31} (z_1+s_1h_1)+ a_{32}(z_2+s_2h_2))^2\right]^{1/2}.

%\end{multline*}

%We have

%\[

%\mathbb{E}_{z_1, h_1, z_2, h_2 \in \F_p^n}\left[\mathbb{E}_{z_3}\left[\Delta_{h_1, h_2} f(z_1 + \cdots + z_t)\right]^2\right] = \|f\|_{U^{3}}^{8},

%\]

%and

%\begin{align*}

%\mathbb{E}&_{z_1, h_1, z_2, h_2 \in \F_p^n}\left[\prod_{s \in \{0, 1\}^2} g_3(a_{31} (z_1+s_1h_1)+ a_{32}(z_2+s_2h_2))^2\right]\\

%&= \mathbb{E}_{z_1, z_1'}\left[\mathbb{E}_{z_2}\left[g_3(a_{31} z_1 + a_{32}z_2)^2g_i(a_{31} z_1' + a_{32}z_2)^2\right]\mathbb{E}_{z_2'}\left[g_3(a_{31} z_1 + a_{32}z_2')^2g_i(a_{31} z_1' + a_{32}z_2')^2\right]\right] \\

%&= \mathbb{E}_{z_1, z_1'}\left[\mathbb{E}_{z_2}\left[g_3(a_{31} z_1 + a_{32}z_2)^2g_3(a_{31} z_1' + a_{32}z_2)^2\right]^2\right] \\

%&\leq \mathbb{E}_{z_2}\left[\mathbb{E}_{z_1}\left[g_3(a_{31} z_1 + a_{32}z_2)^4\right]\mathbb{E}_{z_1'}\left[g_3(a_{31} z_1' + a_{32}z_2)^4\right]\right] \\

%&= \mathbb{E}_{z}[g_3(z)^4]^2,

%\end{align*}

%as desired, where the inequality follows from Cauchy-Schwarz.

%\end{proof}

%

%We also prove the following claim that will be used to further upper bound the bound in Lemma~\ref{lem:genvn}.

%

%\begin{claim}\label{claim:expbound}

%Let $X \in \R$ be a random variable taking on positive values.

%Then,

%\[

%\mathbb{E}[X^2]\mathbb{E}[X^4]^{1/4} \leq \mathbb{E}[X^3].

%\]

%\end{claim}

%\begin{proof}

%Note that for a random variable $Z$ taking on positive values,

%\[

%\frac{\mathbb{E}[Z^x]}{\mathbb{E}[Z]^x}

%\]

%increases as $x$ increases, when $x$ is positive.

%This follows from normalizing so that $\mathbb{E}[Z] = 1$, and using Jensen's inequality.

%Then if $Z = X^4$,

%\[

%\frac{\mathbb{E}[Z^{1/2}]}{\mathbb{E}[Z]^{1/2}} \leq \frac{\mathbb{E}[Z^{3/4}]}{\mathbb{E}[Z]^{3/4}},

%\]

%which implies that

%\[

%\mathbb{E}[Z^{1/2}]\mathbb{E}[Z]^{1/4} \leq \mathbb{E}[Z^{3/4}],

%\]

%as desired.

\end{proof}

For certain groups $\Gamma$, a lower bound on the Gowers norm of a function $f$ implies certain structural properties of the function.  

Before stating the exact result, we define the notion of a non-classical polynomial.

Let $P: \mathbb{F}_p^n \rightarrow \mathbb{T}$ be a non-classical polynomial of degree $d < p$.  Then there exist non-classical polynomials $P_0, \ldots, P_{d}$ such that

\sum_{j=0}^d P_i(x+jy) = P(y)

Decompose $P$ into a sum of linear forms as in Lemma~\ref{lem:classtonon}.  For each $T_i$, we show that there exist $\alpha_0, \ldots, \alpha_{d}$ such that

\sum_{j=0}^d \alpha_j T_i(x+jy, \ldots, x+jy) = T_i(y, \ldots, y).\label{eq:need}

This is enough to construct the $P_i$ desired.

\sum_{j=0}^d \sum_{s \in \{0, 1\}^i} \alpha_j T_i((1-s_1)x+js_1y, \ldots, (1-s_i)x+js_iy) = 

\sum_{j=0}^d \sum_{s \in \{0, 1\}^i} \alpha_j j^{|s|}T_i((1-s_1)x+s_1y, \ldots, (1-s_i)x+s_iy)

\sum_{j=0}^d \alpha_j j^{k} = 0

\sum_{j=0}^d \alpha_j j^{i} = 1.

This is a system of linear equations whose coefficients come from the first $i+1$ rows of a $(d+1) \times (d+1)$ Vandermonde matrix.  Because $d < p$, the determinant of such a matrix is non-zero and thus the matrix is invertible and $\alpha_j$ exist as desired.

Finally, we will use the following result of Carlen, Loss, and Lieb~\cite{Carlen:2006}.

\begin{thm}[Multi-linear Riesz--Thorin Interpolation Theorem]\label{thm:mlrt}

Let~$T$ be a~$k$-linear form on~$\R^n$.

Let~$C: [0,1]^k\to \R_+$ be the function defined by

\[

C(\tfrac{1}{p_1},\dots,\tfrac{1}{p_k}) = \|T\|_{p_1, \ldots, p_k},

\]

for any~$p_i\in [1,\infty]$.

Then, $\ln C$ is a convex function on~$[0,1]^k$.

\end{thm}

We are now ready to prove Lemma~\ref{lem:disctospec}.

We prove the converse, and thus start with the assumption that 

\[

\left\|\sum_{g \in \Gamma} f(g)A_g(x_1, \ldots, x_t) \right\|_{t, \ldots, t} > \lambda.

\]

Next we relate the value on the left-hand side to the Gowers norm of the vector $|\Gamma|^2f$.

%where $1_{H, K} = \frac{|\Gamma|^2}{|S'|}1_{S'}-\frac{|\Gamma|^2}{|S|}1_{S}$.

Let $x_1, \ldots, x_t$ be any set of vectors such that $\|x_i\|_{\infty} \leq 1$ for all $i$.

Then

\sum_{g \in \Gamma} f(g)A_g(x_1, \ldots, x_t) = \mathbb{E}_{u, g}[x_1(u)x_2(u+g)\cdots x_t(u+(t-1)g)(|\Gamma|^2f)]. \label{eq:unravel}

%For $z \in (\F_p^n)^m$, let $f(z) = 1_{H, K}(z)$.

%Additionally, l

Let $a_{ij} = i-j$ for all $i$ and $j$, and $g_i = x_i$ for all $i$.

Then the right-hand side of Eq.~\eqref{eq:unravel} can be rewritten as

\[

\mathbb{E}_{z_1, \ldots, z_t} \left[(|\Gamma|^2f(z_1+\cdots+z_t)) \prod_{i=1}^t g_i(a_{i1} z_1 + \cdots + a_{it} z_t)\right].

\]

%From now on, let $t = 3$.

By Lemma~\ref{lem:genvn}, the right-hand side of Eq.~\eqref{eq:unravel} is bounded above by 

\[

\|(|\Gamma|^2f)\|_{U^t}\mathbb{E}[g_{i_1}^2]^{1/2}\mathbb{E}[g_{i_2}^2]^{1/2} = \left\|(|\Gamma|f)\right\|_{U^t}\|x_{i_1}\|_2\|x_{i_2}\|_2

\]

for all distinct $i_1, i_2 \in [t]$, and thus

\[

\left\|\sum_{g \in \Gamma} f(g)A_g(x_1, \ldots, x_t)\right\|_{\mathbf{p}} \leq \left\|(|\Gamma|f)\right\|_{U^t}

\]

where $\mathbf{p}$ is any vector whose coordinates are all $\infty$ except $\mathbf{p}_{i_1} = \mathbf{p}_{i_2} = 2$.

By Theorem~\ref{thm:mlrt}, it follows that

\[

\left\|\sum_{g \in \Gamma} f(g)A_g(x_1, \ldots, x_t)\right\|_{t, \ldots, t} \leq \left\|(|\Gamma|f)\right\|_{U^t}.

\]

Note that by our assumption, there exist $x_1, \ldots, x_t$ such that $\|x_i\|_{t} = 1$ for all $i$ and such that

\[

\sum_{g \in \Gamma} f(g)A_g(x_1, \ldots, x_t) \geq \lambda,

\] and thus $\lambda$ is also a lower bound on $\left\|(|\Gamma|f)\right\|_{U^t}$.

By the inverse Gowers theorem, there exists a polynomial $P$ of degree $t-1$ such that 

\[\mathbb{E}_{h}[e(P(h)) (|\Gamma|f(h))] \geq \epsilon\] for some $\epsilon$.  By Lemma~\ref{lem:decomp}, there exist $P_0, \ldots, P_{t-1}$ such that \[e(P_0(x))e(P_1(x+y))\cdots e(P_{t-1}(x+(t-1)y)) = e(P(y)),\] and therefore 

\begin{multline*}

\mathbb{E}_{z_1, \ldots, z_t} \left[|\Gamma|f(z_1+\cdots+z_t) \prod_{i=1}^t e\left(P_{i-1}\left(\sum_{j=1}^t (i-j)z_j\right)\right)\right] = \\

\mathbb{E} _{z_1, \ldots, z_t}\left[|\Gamma|f(z_1+\cdots+z_t) e(P(z_1+\cdots+z_t))\right] \geq \epsilon,

\end{multline*}

\sum_{g \in \Gamma} f(g)A_g(e(P_0), \ldots, e(P_{t-1})) \geq \eps|\Gamma|.

Because the coordinates of $e(P_i)$ are bounded above in absolute value by $1$ for all $i$, 

this implies that $\left\|\sum_{g \in \Gamma} f(g)A_g\right\|_{\infty, \ldots, \infty} \geq \eps\|\Gamma\|$ as desired.

%by Lemma~\ref{lem:disctoinfty}, this implies that $H$ has discrepancy at least $\epsilon/\pi^t$ with respect to $K$.

We can use Lemma~\ref{lem:disctospec} along with Theorem 1.1 of~\cite{BG17} to an improvement over Theorem 4.1 in~\cite{BR16}, for certain types of random tensors.

We first state the following tail bound that's implicit in~\cite{BR16}.

\begin{lem}\label{lem:exptotail}

Let $(X, \|\cdot\|)$ be a Banach space and let $A_1, \ldots, A_k \in X$.

Then there exist constants $C$ and $c$ depending on the Banach space such that

\[

\Pr\left[\left\|\eps_1A_1+\cdots+\eps_kA_k\right\| \geq u\sqrt{k}\right] \leq  C\exp(-cu^2/\sigma^2)

\]

where $\eps_1, \ldots, \eps_k$ are random Rademacher random variables, and $\mathbb{E}\left[\left\|\eps_1A_1+\cdots+\eps_kA_k\right\|\right] \leq C\sigma\sqrt{k}$.

\end{lem}

\begin{cor}

Let $\Gamma = \mathbb{F}_p^n$, $A_1, \ldots, A_k$ be independent and identically distributed random $t$-tensors so that each $A_i = \cay^{(t)}(\mathbb{F}_p^n, q, s)$ for each $s \in \mathbb{F}_p^n$.

Then there exists a constant $C_{p, t}$ that depends on $p$ and $t$ such that,

\[

\Pr\left[\left\|\eps_1A_1+\cdots+\eps_kA_k\right\|_{t, \ldots, t} \geq \frac{u \lambda \sqrt{k}}{\eps|\Gamma|}\right]

\leq

C_{p, t}\exp(-cu^2/\sigma^2),

\]

where $\sigma = {|\Gamma|t}\sqrt{|\Gamma|^{1-\frac{1}{\lceil t/2 \rceil}}\log(|\Gamma|)}$

\end{cor}

\begin{proof}

Using the usual symmetrization trick (see Lemma 4.1 in~\cite{BR16}), it is enough to give an upper bound on

\[

\mathbb{E}\left[\left\|\eps_1A_1+\cdots+\eps_kA_k\right\|_{t, \ldots, t}\right],

\]

where $\eps_1, \ldots, \eps_k$ are random Rademacher random variables.

By Theorem 1.1 from~\cite{BG17}, we have that

\begin{equation}\label{eq:gausswidth}

\mathbb{E}\left[\left\|\eps_1A_1+\cdots+\eps_kA_k\right\|_{\infty, \ldots, \infty}\right]

\leq

C{|\Gamma|t}\sqrt{k|\Gamma|^{1-\frac{1}{\lceil t/2 \rceil}}\log(|\Gamma|)}.

\end{equation}

Choose an aribtrary $\lambda$, and let $\eps$ be the corresponding constant from Lemma~\ref{lem:disctospec}.

The statement of the Corollary follows from Eq.~\eqref{eq:gausswidth}, Lemma~\ref{lem:disctospec} and Lemma~\ref{lem:exptotail}.

\end{proof}