<p>The hypercontractive theorem (or Bonami Beckner inequality) says (Ryan O'Donnell):
<a href="https://i.sstatic.net/JvfQG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JvfQG.png" alt="enter image description here"></a></p>

<p>This of course also directly gives the 'intermediate' inequality $\|T_{\rho\sigma}f\|_q  \le \|T_\sigma f\|_{1+(q-1)\rho^2}$.</p>

<p>The theorem also has a 'two function' version, which can be proven, for example using Holders inequality and then the standard theorem:</p>

<p><a href="https://i.sstatic.net/s3SXL.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/s3SXL.png" alt="enter image description here"></a></p>

<p>I'm wondering if anyone has studied 'intermediate' inequalities of this two-function version of the inequality? That would mean something like</p>

<p>$$\langle T_{\rho\sigma} f,g \rangle \le \langle T_{\sigma} f,g \rangle^q$$</p>

<p>for some value of $q$ depending on $\rho$ and $\sigma$.
It doesn't appear to follow easily from any of the individual theorems.</p>

<p>One thing we might try is to look at the normalized case, where we write: </p>

<p>$\langle T_{\rho\sigma} f,g \rangle
= \sum_{S\subseteq[n]}\rho^{|S|}\hat f(S)\hat g(S)$ and divide both sides by $\|f\|\|g\|$. Then we can use the means inequality to show</p>

<p>$$
\left(\frac{\langle T_{\rho} f,g \rangle}{\|f\|\|g\|}\right)^{1/\log\rho}
=
\left(\sum_{S\subseteq[n]}e^{|S|\log\rho}\frac{\hat f(S)\hat g(S)}{\|f\|\|g\|}\right)^{1/\log\rho}
\le
\left(\sum_{S\subseteq[n]}e^{|S|\log\sigma}\frac{\hat f(S)\hat g(S)}{\|f\|\|g\|}\right)^{1/\log\sigma}
=
\left(\frac{\langle T_{\sigma} f,g \rangle}{\|f\|\|g\|}\right)^{1/\log\sigma}
$$</p>

<p>whenever $\log\rho\le\log\sigma$.
However to be valid, this requires the values
$p_k = \sum_{|S|=k}\hat f(S)\hat g(S)$ to be non-negative for all $k$, which is certainly not the case for general functions $f,g:\{-1,1\}^n\to\mathbb R$.</p>

<p>It also doesn't appear to be very sharp, since it becomes equality when all the fourier mass is on level 1, which the hypercontractive inequalities usually tell us isn't possible for many (smaller) functions.</p>

CC BY-SA 3.0
2018-04-19T22:14:03.193Z
Your proposed inequality isn't homogeneous (multiplying $f$ or $g$ by a constant doesn't preserve it), which should allow you to disprove it.
Yuval Filmus
CC BY-SA 3.0
2018-04-19T23:11:17.083Z
@Yuval Great point! I guess that means the "normalized" version is really the right way to go. Or do you see anything that may end up disproving that one as well?
Thomas Ahle
<reference-request><boolean-functions><fourier-analysis>
Relative Two-Function Hypercontractive Inequality $\langle T_{\rho\sigma} f,g \rangle \le \langle T_{\sigma} f,g \rangle^q$
Thomas Ahle
# Relative Two-Function Hypercontractive Inequality $\langle T_{\rho\sigma} f,g \rangle \le \langle T_{\sigma} f,g \rangle^q$

(Asked by: username_0 on 2018-04-18)

The hypercontractive theorem (or Bonami Beckner inequality) says (Ryan O'Donnell):
[![enter image description here](https://i.sstatic.net/JvfQG.png)](https://i.sstatic.net/JvfQG.png)

This of course also directly gives the 'intermediate' inequality $\|T\_{\rho\sigma}f\|\_q \le \|T\_\sigma f\|\_{1+(q-1)\rho^2}$.

The theorem also has a 'two function' version, which can be proven, for example using Holders inequality and then the standard theorem:

[![enter image description here](https://i.sstatic.net/s3SXL.png)](https://i.sstatic.net/s3SXL.png)

I'm wondering if anyone has studied 'intermediate' inequalities of this two-function version of the inequality? That would mean something like

$$\langle T\_{\rho\sigma} f,g \rangle \le \langle T\_{\sigma} f,g \rangle^q$$

for some value of $q$ depending on $\rho$ and $\sigma$.
It doesn't appear to follow easily from any of the individual theorems.

One thing we might try is to look at the normalized case, where we write:

$\langle T\_{\rho\sigma} f,g \rangle
= \sum\_{S\subseteq[n]}\rho^{|S|}\hat f(S)\hat g(S)$ and divide both sides by $\|f\|\|g\|$. Then we can use the means inequality to show

$$
\left(\frac{\langle T\_{\rho} f,g \rangle}{\|f\|\|g\|}\right)^{1/\log\rho}
=
\left(\sum\_{S\subseteq[n]}e^{|S|\log\rho}\frac{\hat f(S)\hat g(S)}{\|f\|\|g\|}\right)^{1/\log\rho}
\le
\left(\sum\_{S\subseteq[n]}e^{|S|\log\sigma}\frac{\hat f(S)\hat g(S)}{\|f\|\|g\|}\right)^{1/\log\sigma}
=
\left(\frac{\langle T\_{\sigma} f,g \rangle}{\|f\|\|g\|}\right)^{1/\log\sigma}
$$

whenever $\log\rho\le\log\sigma$.
However to be valid, this requires the values
$p\_k = \sum\_{|S|=k}\hat f(S)\hat g(S)$ to be non-negative for all $k$, which is certainly not the case for general functions $f,g:\{-1,1\}^n\to\mathbb R$.

It also doesn't appear to be very sharp, since it becomes equality when all the fourier mass is on level 1, which the hypercontractive inequalities usually tell us isn't possible for many (smaller) functions.

**Tags:** reference-request, boolean-functions, fourier-analysis

> **Comments:**
> **username_1**: Your proposed inequality isn't homogeneous (multiplying $f$ or $g$ by a constant doesn't preserve it), which should allow you to disprove it.  
> **username_0**: @Yuval Great point! I guess that means the "normalized" version is really the right way to go. Or do you see anything that may end up disproving that one as well?  

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## 0 answers

