# Hypercontractive Inequalities for the Second Norm of Highly Concentrated Functions, and Mrs. Gerber’s-Type Inequalities for the Second Rényi Entropy

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

**Proposition**

**1.**

- –
- The functions ${\left\{{\psi}_{2,q}\right\}}_{q}$ are somewhat cumbersome to describe, and hence we relegate their precise definition to Definition 1 below.
- –
- Inequality (4) upper bounds $En{t}_{2}\left(\right)open="("\; close=")">{f}_{\u03f5}$ in terms of $En{t}_{q}\left(f\right)$ for $q>1$, and $\u03f5$. Taking $q=2$ gives an upper bound on $En{t}_{2}\left(\right)open="("\; close=")">{f}_{\u03f5}$ in terms of $En{t}_{2}\left(f\right)$ and $\u03f5$, in analogy to (2).
- –
- Recall that for a point $x\in {\{0,1\}}^{n}$ and $0\le r\le n$, the Hamming sphere of radius r around x is the set $\{y\in {\{0,1\}}^{n}:\phantom{\rule{3.33333pt}{0ex}}|y-x|=r\}$. As will be seen from the proof of Proposition 1, (4) is essentially tight for a certain convex combination of the uniform distribution on ${\{0,1\}}^{n}$ and the characteristic function of a Hamming sphere of an appropriate radius (depending on q, $\u03f5$, and the required value of $En{t}_{q}\left(f\right)$).
- –
- In information theory one typically considers a slightly different notion of Rényi entropies: For a probability distribution P on $\Omega $, the ${q}^{th}$ Renyi entropy of P is given by ${H}_{q}\left(P\right)=-\frac{1}{q-1}{\mathrm{log}}_{2}\left(\right)open="("\; close=")">{\sum}_{\omega \in \Omega}{P}^{q}\left(\omega \right)$. To connect notions, if f is a nonnegative (non-zero) function on ${\{0,1\}}^{n}$ with expectation 1, then $P=\frac{f}{{2}^{n}}$ is a probability distribution, and $En{t}_{q}\left(f\right)=n-{H}_{q}\left(P\right)$. Furthermore, $En{t}_{q}\left(\right)open="("\; close=")">{f}_{\u03f5}$, where X is a random variable on ${\{0,1\}}^{n}$ distributed accordinng to P and Z is an independent noise vector corresponding to a binary symmetric channel with crossover probability $\u03f5$. Hence, (2) can be restated as$$H\left(\right)open="("\; close=")">X\oplus Z,$$$${H}_{2}\left(\right)open="("\; close=")">X\oplus Z$$Here $\phi $ is an explicitly given function on $[0,1]\times \left(\right)open="["\; close="]">0,1/2$, which is increasing and convex in its first argument ($\phi (x,\u03f5)=1-\psi (1-x,\u03f5)$), and similarly for ${\phi}_{2,q}$.Next, we describe our main result, a hypercontractive inequality for the ${\ell}_{2}$ norm of ${f}_{\u03f5}$ which takes into account the ratio between ${\ell}_{q}$ and ${\ell}_{1}$ norms of f, and more specifically $En{t}_{q}\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}$.

**Theorem**

**1.**

- –
- The precise definition of the functions ${\left\{{\kappa}_{2,q}\right\}}_{q}$ will be given in Definition 1 below. At this point let us just observe that since the sequence ${\left\{En{t}_{q}\left(f\right)\right\}}_{q}$ increases with q, we would expect the fact that $En{t}_{q}\left(f\right)$ is large to become less significant as q increases. This is expressed in the properties of the functions ${\left\{{\kappa}_{2,q}\right\}}_{q}$ in the following manner: If $q\ge 2$ then for any $0<\u03f5<\frac{1}{2}$ the function ${\kappa}_{2,q}(x,\u03f5)$ starts as a constant-$\left(\right)$ function up to some $x=x(q,\u03f5)>0$, and becomes strictly decreasing after that. In other words $x(q,\u03f5)$ is the largest possible value of $\frac{En{t}_{q}\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}}{}$ for which Theorem 1 provides no new information compared to (1). For $1<q<2$ there is a value $0<\u03f5\left(q\right)<\frac{1}{2}$, such that for all $\u03f5\le \u03f5\left(q\right)$ the function ${\kappa}_{2,q}(x,\u03f5)$ is strictly decreasing (in which case we say that $x(q,\u03f5)=0$). However, $x(q,\u03f5)>0$ for all $\u03f5>\u03f5\left(q\right)$. The function $\u03f5\left(q\right)$ decreases with q (in particular, $\u03f5\left(q\right)=0$ for $g\ge 2$). The function $x(q,\u03f5)$ increases both in q and in $\u03f5$.
- –
- Notably, taking $q\to 1$ in Theorem 1 gives (see Corollary 1)$$\parallel {f}_{\u03f5}{\parallel}_{2}\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}{\parallel f\parallel}_{\kappa},$$
- –
- As will be seen from the proof of Theorem 1, (5) is essentially tight for a certain convex combination of the uniform distribution on ${\{0,1\}}^{n}$ and characteristic functions of one or two Hamming spheres of appropriate radii (the number of the spheres and their radii depend on q, $\u03f5$, and the required value of $En{t}_{q}\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}$).
- –
- Let f be a non-constant function and let $0<\u03f5<\frac{1}{2}$ be fixed. Consider the function $F\left(q\right)={F}_{f,\u03f5}\left(q\right)={\kappa}_{2,q}\left(\right)open="("\; close=")">\frac{En{t}_{q}\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}}{}n$. It will be seen that there is a unique value $1<q(f,\u03f5)\le 1+{(1-2\u03f5)}^{2}$ of q for which $F\left(q\right)=q$. Furthermore, $q(f,\u03f5)={\mathrm{min}}_{q\ge 1}F\left(q\right)$. Hence it provides the best possible value for $\kappa $ in Theorem 1. With that, determining $q(f,\u03f5)$ might in principle require knowledge of all the Renyi entropies $En{t}_{q}\left(f\right)$, for $1\le q\le 1+{(1-2\u03f5)}^{2}$, while typically we are in possession of one of the “easier” Rényi entropies, such as $Ent\left(f\right)$ or $En{t}_{2}\left(f\right)$.

#### 1.1. Full Statements of Proposition 1 and Theorem 1

**Definition**

**1.**

- If ${\varphi}_{\u03f5}^{\prime}(1-x)<\frac{1}{q}$, let ${\alpha}_{0}={\left(\right)}^{{\varphi}_{\u03f5}^{\prime}}-1$. Define$${\psi}_{2,q}(x,\u03f5)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2\xb7\left(\right)open="\{"\; close>\begin{array}{c}\frac{q-1}{q}\xb7x+\left(\right)open="("\; close=")">{\varphi}_{\u03f5}\left(\right)open="("\; close=")">{\alpha}_{0}+\frac{1-{\alpha}_{0}}{q}\end{array}if& {\varphi}_{\u03f5}^{\prime}(1-x)\frac{1}{q}$$
- Let $y=\frac{q-1}{q}\xb7x+\frac{1}{q}$. Let ${q}_{0}=1+{(1-2\u03f5)}^{2}$. If $y\ge \frac{1}{{q}_{0}}$, let ${\alpha}_{0}$ be determined by $1-{\alpha}_{0}-\frac{{\alpha}_{0}{\varphi}_{\u03f5}\left({\alpha}_{0}\right)}{1-{\alpha}_{0}}=y$. If $x=0$, define ${\kappa}_{2,q}(x,\u03f5)={q}_{0}$. Otherwise, define$${\kappa}_{2,q}(x,\u03f5)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left(\right)open="\{"\; close>\begin{array}{ccc}{q}_{0}& if& y\le \frac{1}{{q}_{0}}\\ -\frac{x}{{\varphi}_{\u03f5}(1-x)}& if& y\frac{1}{{q}_{0}}\phantom{\rule{3.33333pt}{0ex}}and\phantom{\rule{3.33333pt}{0ex}}-\frac{x}{{\varphi}_{\u03f5}(1-x)}\ge \phantom{\rule{3.33333pt}{0ex}}q\\ \frac{{\alpha}_{0}-1}{{\varphi}_{\u03f5}\left({\alpha}_{0}\right)}& if& y\frac{1}{{q}_{0}}\phantom{\rule{3.33333pt}{0ex}}and\phantom{\rule{3.33333pt}{0ex}}-\frac{x}{{\varphi}_{\u03f5}(1-x)}q\end{array}$$

**Corollary**

**1.**

- 1.
- Taking $q\to 1$ in Theorem 1 gives:$$\parallel {f}_{\u03f5}{\parallel}_{2}\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}{\parallel f\parallel}_{\kappa},\phantom{\rule{1.em}{0ex}}with\phantom{\rule{1.em}{0ex}}\kappa =-\frac{Ent\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}}{/}{\varphi}_{\u03f5}\left(\right)open="("\; close=")">1-Ent\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}/n$$
- 2.
- Taking $q=2$ in Theorem 1 gives, for $x=\frac{En{t}_{2}\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}}{}n$ and ${q}_{0}=1+{(1-2\u03f5)}^{2}$$$\parallel {f}_{\u03f5}{\parallel}_{2}\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}{\parallel f\parallel}_{\kappa},\phantom{\rule{1.em}{0ex}}with\phantom{\rule{1.em}{0ex}}\kappa =\left(\right)open="\{"\; close>\begin{array}{ccc}{q}_{0}& if& \frac{x+1}{2}\le \frac{1}{{q}_{0}}\\ \frac{\alpha -1}{{\varphi}_{\u03f5}\left(\alpha \right)}& otherwise\end{array}$$In the second case α is determined by $1-\alpha -\frac{\alpha {\varphi}_{\u03f5}\left(\alpha \right)}{1-\alpha}=\frac{x+1}{2}$.

**Theorem**

**2.**

#### 1.2. Applications

**Lemma**

**1.**

- 1.
- $${\kappa}^{\prime}\left(0\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{4}{\mathrm{ln}2}\xb7\frac{\left(\right)}{2}-1$$
- 2.
- Let $\u03f5\sim 0$ express the fact that ϵ is a sufficiently small absolute constant. Then for $\u03f5\sim 0$ holds $|{\kappa}^{\prime}\left(\u03f5\right)-{\kappa}^{\prime}\left(0\right)|\le O\left(\u03f5\right)$, where the asymptotic notation hides absolute constants which may depend on x.

**Corollary**

**2.**

**Corollary**

**3.**

**Corollary**

**4.**

- Let D be a subset of ${\{0,1\}}^{n}$ of cardinality $\left|D\right|={2}^{H\left(\rho \right)n}$, for some $0\le \rho \le 1$. Then$$\lambda \left(D\right)\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}2\sqrt{\rho (1-\rho )}\xb7n.$$This is almost tight if D is a Hamming ball of exponentially small cardinality.
- For any $0\le \delta \le \frac{1}{2}$ holds$${R}_{LP}\left(\delta \right)\phantom{\rule{3.33333pt}{0ex}}\ge \phantom{\rule{3.33333pt}{0ex}}\frac{1-H\left(\delta \right)+H\left(\right)open="("\; close=")">1/2-\sqrt{\delta (1-\delta )}}{}2$$

- –
- As discussed above, the first of the these claims answers the question of [19] and shows that a certain approach to bound binary codes does not lead to an improvement of the first JPL bound. The second claim shows that the best possible bound obtainable via the linear programming approach of [15] is not better than the arithmetic average of the Gilbert-Varshamov bound and the first JPL bound. Observe that the first claim is a consequence of the logarithmic Sobolev inequality in Corollary 2, and hence of the behavior of the norm ${\kappa}_{2,2}$ in Theorem 1 as $\u03f5\to 0$. The second claim is a consequence of the uncertainty principle in Corollary 3, and hence of the behavior of the norm ${\kappa}_{2,2}$ in Theorem 1 as $\u03f5\sim 0$. We find these connections between notions to be rather intriguing.
- –
- As we have mentioned, the first of the claims recovers a result of [20], where it was also derived from the appropriate logarithmic Sobolev inequality. (Apart from this claim being a simple corollary of Theorem 1, an additional reason for stating it here is that it has only appeared in the unpublished arXiv preprint [20].) The second claim of recovers a result of [17].

**Corollary**

**5.**

#### 1.3. Related Work

## 2. Proof of Proposition 1

**Definition**

**2.**

**Lemma**

**2.**

**Lemma**

**3.**

- 1.
- ${\varphi}_{\u03f5}\left(\alpha \right)$ is strictly concave and increasing from ${\varphi}_{\u03f5}\left(0\right)=-\frac{{\mathrm{log}}_{2}\left(\right)open="("\; close=")">\frac{4}{{q}_{0}}}{}2$ to 0 on $0\le \alpha \le 1$.
- 2.
- ${\varphi}_{\u03f5}^{\prime}\left(0\right)=1$, ${\varphi}_{\u03f5}^{\prime}\left(1\right)=\frac{1}{{q}_{0}}$.
- 3.
- $\frac{\alpha -1}{{\varphi}_{\u03f5}\left(\alpha \right)}$ is strictly increasing in α, going up to ${q}_{0}$, as $\alpha \to 1$.
- 4.
- The function $g\left(\alpha \right)=1-\alpha -\frac{\alpha}{1-\alpha}\xb7{\varphi}_{\u03f5}\left(\alpha \right)$ is strictly decreasing on $[0,1]$. Moreover, $g\left(0\right)=1$ and $g\left(1\right)=\frac{1}{{q}_{0}}$.

**Proof.**

- ${\varphi}_{\u03f5}^{\prime}\left(\right)open="("\; close=")">1-\frac{qN}{q-1}$. In this case h is increasing on I and we get$$\underset{(\alpha ,\nu )\in \Omega}{\mathrm{max}}\left(\right)open="\{"\; close="\}">{\varphi}_{\u03f5}\left(\alpha \right)+\nu \phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}$$$${\varphi}_{\u03f5}\left(\right)open="("\; close=")">1-\frac{qN}{q-1}.$$The last equality follows from the definition of ${\psi}_{2,q}$ in this case.
- ${\varphi}_{\u03f5}^{\prime}\left(\right)open="("\; close=")">1-\frac{qN}{q-1}$. Note that, by Lemma 3, $1={\varphi}_{\u03f5}^{\prime}\left(0\right)>\frac{1}{q}$. Hence, in this case the maximum of h on I is located at the unique zero of its derivative, that is at the point ${\alpha}_{0}$ such that ${\varphi}_{\u03f5}^{\prime}\left(\right)open="("\; close=")">{\alpha}_{0}$. Using the definition of ${\psi}_{2,q}$ in this case, we get$$\underset{(\alpha ,\nu )\in \Omega}{\mathrm{max}}\left(\right)open="\{"\; close="\}">{\varphi}_{\u03f5}\left(\alpha \right)+\nu \phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}N+\left(\right)open="("\; close=")">{\varphi}_{\u03f5}\left(\right)open="("\; close=")">{\alpha}_{0}$$

- If ${\varphi}_{\u03f5}^{\prime}(1-x)<\frac{1}{q}$, then by the discussion above, the point $\left(\right)$ lies on the line $\frac{\alpha}{q}+\nu =N+\frac{1}{q}$, but not on the line $\alpha +\nu =1$. Observe that ${2}^{{\alpha}^{*}n-o\left(n\right)}\le \left|S\right|\le {2}^{{\alpha}^{*}n}$ (the first estimate follows from the Stirling formula, for the second estimate see, e.g., Theorem 1.4.5. in [28]). As the first attempt, let $g={2}^{{\nu}^{*}n}\xb7{1}_{S}$. Then $N-o\left(n\right)\le \frac{{\alpha}^{*}-1}{q}+{\nu}^{*}-o\left(n\right)\le \frac{1}{n}{\mathrm{log}}_{2}{\parallel g\parallel}_{q}\le \frac{{\alpha}^{*}-1}{q}+{\nu}^{*}=N$. That is, $x-{o}_{n}\left(1\right)\le \frac{En{t}_{q}\left(g\right)}{n}\le x$. However, $\mathbb{E}g$ is exponentially small. To correct that, we define f to be $v={2}^{\left(\right)open="("\; close=")">{\nu}^{*}-\delta}$ on S, and $\frac{{2}^{n}-\left|S\right|v}{{2}^{n}-\left|S\right|}$ on the complement of S. Then $\mathbb{E}f=1$. We choose $\delta $ to be as small as possible, while ensuring that $\frac{En{t}_{q}\left(f\right)}{n}\le x$. Since the contribution of the constant-1 function to ${\parallel f\parallel}_{q}$ is exponentially small w.r.t. ${\parallel f\parallel}_{q}$, we can choose $\delta ={o}_{n}\left(1\right)$. We now have $\mathbb{E}f=1$, $\frac{En{t}_{q}\left(f\right)}{n}\le x$, and$$\frac{En{t}_{2}\left({f}_{\u03f5}\right)}{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{n}{\mathrm{log}}_{2}{\parallel {f}_{\u03f5}\parallel}_{2}^{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{n}{\mathrm{log}}_{2}\left(\right)open="\langle "\; close="\rangle ">{f}_{2\u03f5(1-\u03f5)},f$$$$2\xb7\left(\right)open="("\; close=")">{\varphi}_{\u03f5}\left(\right)open="("\; close=")">{\alpha}^{*}-{o}_{n}\left(1\right)\phantom{\rule{3.33333pt}{0ex}}\ge \phantom{\rule{3.33333pt}{0ex}}{\psi}_{2,q}(x,\u03f5)-{o}_{n}\left(1\right).$$Here the second equality follows from the semigroup property of the noise operator: ${T}_{\u03f5}\circ {T}_{\u03f5}={T}_{2\u03f5(1-\u03f5)}$. The first inequality follows from the tightness part of Theorem 2 and the definition of ${\varphi}_{\u03f5}$. The second inequality follows from Lemma 2.The tightness of (4) in this case now follows, taking into account the fact that ${\psi}_{2,q}$ is strictly increasing.
- If ${\varphi}_{\u03f5}^{\prime}(1-x)\ge \frac{1}{q}$, the point $\left(\right)$ lies on the intersection of the lines $\frac{\alpha}{q}+\nu =N+\frac{1}{q}$, and $\alpha +\nu =1$. Hence the function $g={2}^{{\nu}^{*}n}\xb7{1}_{S}$ has both $x-{o}_{n}\left(1\right)\le \frac{En{t}_{q}\left(g\right)}{n}\le x$, and $1-{o}_{n}\left(1\right)\le \mathbb{E}g\le 1$. It is easy to see that g can be corrected as in the preceding case, by decreasing it slightly on S and adding a constant component, to obtain a function f with expectation 1 and $En{t}_{q}\left(f\right)\le x$, and with $\frac{En{t}_{2}\left({f}_{\u03f5}\right)}{n}\ge {\psi}_{2,q}(x,\u03f5)-{o}_{n}\left(1\right)$, proving the tightness of (4) in this case as well. We omit the details.

## 3. Proof of Theorem 1

**Proposition**

**2.**

**Proof**

**of**

**Proposition**

**2.**

**Lemma**

**4.**

**Proposition**

**3.**

- 1.
- $$M\left(\right)open="("\; close=")">1-\frac{qN}{q-1},\frac{qN}{q-1}.$$
- 2.
- If $\frac{-\frac{qN}{q-1}}{{\varphi}_{\u03f5}\left(\right)open="("\; close=")">1-\frac{qN}{q-1}}$, then for any choice of $\left(\right)$ holds$$M\left(\right)open="("\; close=")">{\alpha}_{1},{\nu}_{1}.$$
- 3.
- If $\frac{-\frac{qN}{q-1}}{{\varphi}_{\u03f5}\left(\right)open="("\; close=")">1-\frac{qN}{q-1}}$, then for any choice of $\left(\right)$ holds$$M\left(\right)open="("\; close=")">1-\frac{qN}{q-1},\frac{qN}{q-1}\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}M\left(\right)open="("\; close=")">0,N+\frac{1}{q}$$$$\mathrm{where}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{\alpha}_{0}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{is}\phantom{\rule{3.33333pt}{0ex}}\mathrm{determined}\phantom{\rule{3.33333pt}{0ex}}\mathrm{by}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}1-{\alpha}_{0}-\frac{{\alpha}_{0}{\varphi}_{\u03f5}\left(\right)open="("\; close=")">{\alpha}_{0}}{}1-{\alpha}_{0}$$

#### 3.1. Proof of Lemma 4

**Lemma**

**5.**

- 1.
- ${\alpha}_{1}\le 1-\frac{qN}{q-1}$ and ${\nu}_{1}\ge \frac{qN}{q-1}$.
- 2.
- $\frac{1-{\alpha}_{1}}{{\nu}_{1}}<{q}_{0}$.

**Proof.**

**Corollary**

**6.**

- 1.
- For any $(\alpha ,\nu )\in \Omega $ holds ${\varphi}_{\u03f5}\left(\alpha \right)+\nu =\mathrm{max}\left(\right)open="\{"\; close="\}">\frac{{\alpha}_{1}-1}{f(\alpha ,\nu )}+{\nu}_{1},\frac{\alpha -1}{f(\alpha ,\nu )}+\nu $.
- 2.
- $0<f\le M\left(\right)open="("\; close=")">{\alpha}_{1},{\nu}_{1}$ on Ω.
- 3.
- For any $(\alpha ,\nu )\in \Omega $ holds $f(\alpha ,\nu )\le \frac{\alpha -1}{{\varphi}_{\u03f5}\left(\alpha \right)}$. (If $\alpha =1$ we replace the RHS of this inequality with ${q}_{0}$.)

**Proof.**

#### 3.2. Proof of Proposition 3

**Lemma**

**6.**

- $\frac{{\alpha}_{1}-1}{f\left(\right)open="("\; close=")">{\alpha}^{*},{\nu}^{*}}+{\nu}^{*}$.
- ${\alpha}^{*}\le 1-\frac{qN}{q-1}$.

**Lemma**

**7.**

- $\frac{-\frac{qN}{q-1}}{{\varphi}_{\u03f5}\left(\right)open="("\; close=")">1-\frac{qN}{q-1}}$. Let $\phantom{\rule{0.166667em}{0ex}}\left(\right)open="("\; close=")">{\alpha}^{*},{\nu}^{*}$ be a maximum point of $\phantom{\rule{0.166667em}{0ex}}f$ of the second type in this case. Then ${\alpha}^{*}\le 1-\frac{qN}{q-1}$.
- $\frac{-\frac{qN}{q-1}}{{\varphi}_{\u03f5}\left(\right)open="("\; close=")">1-\frac{qN}{q-1}}$. In this case f has a unique maximum point $\phantom{\rule{0.166667em}{0ex}}\left(\right)open="("\; close=")">{\alpha}^{*},{\nu}^{*}$. This point is of the second type. Furthermore, ${\alpha}^{*}>1-\frac{qN}{q-1}$, and it is uniquely determined by the following identity:$$\frac{{\alpha}^{*}-1}{{\varphi}_{\u03f5}\left({\alpha}^{*}\right)}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{{\alpha}^{*}-{\alpha}_{1}}{{\alpha}^{*}-\left(\right)open="("\; close=")">1-{\nu}_{1}}$$

**Lemma**

**8.**

#### 3.3. Proofs of the Remaining Lemmas

**Proof**

**of**

**Lemma**

**6.**

**Proof**

**of**

**Lemma**

**7.**

- $\frac{{\alpha}_{1}-1}{f\left(\right)open="("\; close=")">{\alpha}^{*},{\nu}^{*}}+{\nu}^{*}={\varphi}_{\u03f5}\left(\right)open="("\; close=")">{\alpha}^{*}$.
- $1-\frac{qN}{q-1}<{\alpha}^{*}\le 1$ and ${\alpha}^{*}+{\nu}^{*}=1$.

**Proof**

**of**

**Lemma**

**8.**

- $q\ge {q}_{0}$. In this case, by the third claim of Lemma 3, $-\frac{x}{{\varphi}_{\u03f5}(1-x)}$ is never larger than q, and hence$${\kappa}_{2,q}(x,\u03f5)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left(\right)open="\{"\; close>\begin{array}{ccc}{q}_{0}& if& y\le \frac{1}{{q}_{0}}\\ \frac{{\alpha}_{0}-1}{{\varphi}_{\u03f5}\left({\alpha}_{0}\right)}& if& y\ge \frac{1}{{q}_{0}}\end{array}$$Here $y=\frac{q-1}{q}\xb7x+\frac{1}{q}$, ${q}_{0}=1+{(1-2\u03f5)}^{2}$, and ${\alpha}_{0}$ is determined by $1-{\alpha}_{0}-\frac{{\alpha}_{0}{\varphi}_{\u03f5}\left({\alpha}_{0}\right)}{1-{\alpha}_{0}}=y$. Note that ${\alpha}_{0}$ is well-defined, by the fourth claim of Lemma 3. The fact that ${\kappa}_{2,q}$ is decreasing in x follows from combining the third and the fourth claims of Lemma 3. In fact, ${\kappa}_{2,q}$ is a constant-$\left(\right)$ function for $0\le x\le \frac{q-{q}_{0}}{(q-1){q}_{0}}$, and it is strictly decreasing for larger x.
- $q<{q}_{0}$. In this case y is always greater than $\frac{1}{{q}_{0}}$ and we have that$${\kappa}_{2,q}(x,\u03f5)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left(\right)open="\{"\; close>\begin{array}{ccc}-\frac{x}{{\varphi}_{\u03f5}(1-x)}& if& -\frac{x}{{\varphi}_{\u03f5}(1-x)}\ge \phantom{\rule{3.33333pt}{0ex}}q\\ \frac{{\alpha}_{0}-1}{{\varphi}_{\u03f5}\left({\alpha}_{0}\right)}& if& -\frac{x}{{\varphi}_{\u03f5}(1-x)}\le q\end{array}$$It suffices to show that ${\kappa}_{2,q}$ is decreasing on both relevant subintervals of $[0,1]$, and this again follows from the third and the fourth claims of Lemma 3. In this case ${\kappa}_{2,q}$ is strictly decreasing on $[0,1]$.

## 4. Remaining Proofs

**Proof**

**of**

**Lemma**

**3.**

**Proof**

**of**

**Lemma**

**1.**

**Lemma**

**9.**

- 1.
- $$\varphi (t,0)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{t-1}{2}\phantom{\rule{1.em}{0ex}}andfor\u03f5\sim 0holds\phantom{\rule{1.em}{0ex}}|\varphi (t,\u03f5)-\frac{t-1}{2}|\le O\left(\u03f5\right).$$
- 2.
- $$\frac{\partial \varphi}{\partial \u03f5}\left(\right)open="("\; close=")">t,0\phantom{\rule{1.em}{0ex}}andfor\u03f5\sim 0holds\phantom{\rule{1.em}{0ex}}$$$$|\frac{\partial \varphi}{\partial \u03f5}\left(\right)open="("\; close=")">t,\u03f5|\le O\left(\u03f5\right).$$
- 3.
- $$\frac{\partial \varphi}{\partial t}\left(\right)open="("\; close=")">t,0-\frac{1}{2}|\le O\left(\u03f5\right).$$

**Proof**

**of**

**Lemma**

**9.**

**Proof**

**of**

**Corollary**

**2.**

**Proof**

**of**

**Corollary**

**3.**

**Proof**

**of**

**Corollary**

**4.**

**The first claim of the corollary**

**The second claim of the corollary**

- f is symmetric, that is $f\left(x\right)$ depends only on $\left|x\right|$.
- $f\left(x\right)\le 0$ for $\left|x\right|\ge d$.
- $\widehat{f}\ge 0$ and $\widehat{f}\left(0\right)=1$.
- $f\left(0\right)\le {2}^{{R}_{LP}\left(\delta \right)\xb7n+o\left(n\right)}$.

**Proof**

**of**

**Corollary**

**5.**

**Proof**

**of**

**Corollary**

**1.**

#### Proofs of Comments to Theorem 1

**Lemma**

**10.**

- If $q\ge 2$ then for any $0<\u03f5<\frac{1}{2}$ the function ${\kappa}_{2,q}(x,\u03f5)$ starts as a constant-$\left(\right)$ function up to some $x=x(q,\u03f5)>0$, and becomes strictly decreasing after that. For $1<q<2$ there is a value $0<\u03f5\left(q\right)<\frac{1}{2}$, such that for all $\u03f5\le \u03f5\left(q\right)$ the function ${\kappa}_{2,q}(x,\u03f5)$ is strictly decreasing (in which case we say that $x(q,\u03f5)=0$). However, $x(q,\u03f5)>0$ for all $\u03f5>\u03f5\left(q\right)$. The function $\u03f5\left(q\right)$ decreases with q (in particular, $\u03f5\left(q\right)=0$ for $g\ge 2$). The function $x(q,\u03f5)$ increases both in q and in ϵ.
- The function ${\kappa}_{2,1}(x,\u03f5)=-\frac{x}{{\varphi}_{\u03f5}(1-x)}$ is strictly decreasing in its first argument for any $0<\u03f5<\frac{1}{2}$. It satisfies ${\kappa}_{2,1}(0,\u03f5)={\mathrm{lim}}_{x\to 0}{\kappa}_{2,1}(x,\u03f5)=1+{(1-2\u03f5)}^{2}$, for all $0\le \u03f5\le \frac{1}{2}$.
- Let f be a non-constant function on ${\{0,1\}}^{n}$. Let $0<\u03f5<\frac{1}{2}$. Let $F\left(q\right)={F}_{f,\u03f5}\left(q\right)={\kappa}_{2,q}\left(\right)open="("\; close=")">En{t}_{q}\left(\right)open="("\; close=")">\frac{f}{{\parallel f\parallel}_{1}}$. There is a unique value $1<q(f,\u03f5)\le 1+{(1-2\u03f5)}^{2}$ of q for which $F\left(q\right)=q$. Moreover, $q(f,\u03f5)={\mathrm{min}}_{q\ge 1}F\left(q\right)$. Furthermore, ${\mathrm{lim}}_{\u03f5\to 0}q(f,\u03f5)=2$ for any f.

**Proof.**

**Lemma**

**11.**

**Proof**

**of**

**Lemma**

**11.**

## Author Contributions

## Funding

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

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Levhari, N.; Samorodnitsky, A.
Hypercontractive Inequalities for the Second Norm of Highly Concentrated Functions, and Mrs. Gerber’s-Type Inequalities for the Second Rényi Entropy. *Entropy* **2022**, *24*, 1376.
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Levhari N, Samorodnitsky A.
Hypercontractive Inequalities for the Second Norm of Highly Concentrated Functions, and Mrs. Gerber’s-Type Inequalities for the Second Rényi Entropy. *Entropy*. 2022; 24(10):1376.
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Levhari, Niv, and Alex Samorodnitsky.
2022. "Hypercontractive Inequalities for the Second Norm of Highly Concentrated Functions, and Mrs. Gerber’s-Type Inequalities for the Second Rényi Entropy" *Entropy* 24, no. 10: 1376.
https://doi.org/10.3390/e24101376