<p>I'm not sure that the statement you want is known to hold in full generality (if you look at O'Donnell's website, he remarks that what he stated in that survey is not in fact known to be true in full generality). That said, it does hold for <span class="math-container">$q\geq 2\geq p\geq 1$</span>. Let's suppose the Hypercontractivity Theorem holds as you have stated, i.e. for <span class="math-container">$q\geq p\geq 1$</span>, suppose for <span class="math-container">$\rho=\sqrt{\frac{p-1}{q-1}}$</span>, we have
<span class="math-container">\begin{equation}
\|T_{\rho}f\|_q\leq \|f\|_p.
\end{equation}</span></p>

<p>To use this statement to get this revised version (or almost, you forgot to take the reciprocal of <span class="math-container">$\rho$</span>), suppose <span class="math-container">$q\geq 2$</span> and first take <span class="math-container">$p=2$</span>, so <span class="math-container">$\rho^{-1}=\sqrt{q-1}$</span>. As you seem to have figured out, hypercontractivity gives
<span class="math-container">\begin{align}
\|f\|_q^2&amp;=\|T_{\rho}T_{\rho^{-1}}f\|_q^2\\
&amp;\leq \|T_{\rho^{-1}}f\|_2^2\\
&amp;=\sum_{k=0}^d \rho^{-2k}W_k(f)\\
&amp;\leq \rho^{-2d}\sum_{k=0}^dW_k(f)\\
&amp;=\rho^{-2d}\|f\|_2^2\\
&amp;=\sqrt{q-1}^{2d} \|f\|_2^2,
\end{align}</span>
where <span class="math-container">$W_k(f)$</span> is the level-k Fourier weight of <span class="math-container">$f$</span>, and we use Parseval and the degree assumption. Taking squareroots yields <span class="math-container">$\|f\|_q\leq \sqrt{q-1}^d\|f\|_2$</span> in this simple case.</p>

<p>We also have for <span class="math-container">$1\leq p\leq 2$</span> 
<span class="math-container">\begin{align}
\|f\|_2^2&amp;=\langle f,f\rangle\\
&amp;\leq \|f\|_p\|f\|_{p/(p-1)}\\
&amp;\leq \|f\|_p\sqrt{p/(p-1)-1}^d \|f\|_2\\
&amp;=\sqrt{\frac{1}{p-1}}^{d}\|f\|_p\|f\|_2,
\end{align}</span>
where we use Holder's inequality and then the case we just proved with <span class="math-container">$q=p'=p/(p-1)$</span>. Putting this together with the previous case, we conclude for all <span class="math-container">$q\geq 2\geq p\geq 1$</span>, that
<span class="math-container">\begin{equation}
\|f\|_q\leq \sqrt{q-1}^d \|f\|_2\leq \sqrt{\frac{q-1}{p-1}}^d \|f\|_p,
\end{equation}</span>
as claimed.</p>

<p>According to O'Donnell's website, <a href="https://arxiv.org/pdf/1902.02406.pdf" rel="noreferrer">this paper</a> gives the state of the art on this problem (though I haven't looked through it).</p>

CC BY-SA 4.0
2019-10-14T21:32:57.873Z
Thanks, I didn't look at the website. I liked your 'duality' solution, seems like I can use whenever I prove something for q>2 to deduce about p<2. Interestingly I can't even find a argument that doesn't use this duality (where the q>2 is done directly from the fourier expression), do you have one?
Andy
CC BY-SA 4.0
2019-10-15T16:11:37.937Z
@Andy Thanks! No, this is the only argument that I know, though a quick scan of the paper seems like it does this without duality (i.e. going through $p=2$) with a neat argument (Theorem 1 shows the heat operator from $L^q$ to itself isn't too big a contraction using a nice averaging argument, and then uses Hypercontractivity as you quoted to get the $L^q$ to $L^p$ result in Corollary 2). Also, what I wrote isn't really my solution, it's essentially from Chapter 9 of O'Donnell's book and the exercises there. I'm not sure where the argument originally comes from.
Jason Gaitonde
CC BY-SA 4.0
2019-10-15T20:34:06.207Z
hang on, it's actually silly, there's no need for duality. $|T_\rho f|_2| \geq \rho^d ||f|||_2$. I wonder if there are examples where the duality is really essential.
Andy
CC BY-SA 4.0
2021-11-28T15:49:48.727Z
I have also asked this from Andy but he does not seem to be active. What does duality mean and how is it generally calculated? @J.G
Pegi
CC BY-SA 4.0
2019-10-14T17:44:36.363Z
2019-10-14T17:44:36.363Z
Jason Gaitonde
<p>Recall the hypercontractive inequality:</p>

<p>Let <span class="math-container">$\rho = \sqrt{\frac{p-1}{q-1}}$</span>, then <span class="math-container">$||T_\rho(f)||_q \leq ||f||_p$</span></p>

<p>In <a href="https://www.cs.cmu.edu/~odonnell/papers/analysis-survey.pdf" rel="nofollow noreferrer">https://www.cs.cmu.edu/~odonnell/papers/analysis-survey.pdf</a> it is stated that the dual of the hypercontractive inequality is equivalent to the following:</p>

<p>If <span class="math-container">$f$</span> has degree at most <span class="math-container">$d$</span>, then</p>

<p><span class="math-container">$||f||_q \leq \rho ^ d ||f||_p$</span></p>

<p>I am unable to show this (Specifically, the direction that the original version implies the dual version), and would like to help.</p>

<p>Attempt-</p>

<p>We know for every <span class="math-container">$g$</span>, <span class="math-container">$||T_\rho(g)||_q \leq ||g||_p$</span></p>

<p>Plugging <span class="math-container">$g = T_{\frac{1}{\rho}}(f)$</span>,</p>

<p><span class="math-container">$||f||_q \leq ||T_{\frac{1}{\rho}}(f)||_p$</span></p>

<p>Now if <span class="math-container">$p=2$</span> it's clear how to finish since it's clear how <span class="math-container">$T_{\frac{1}{\rho}}$</span>(f) affect the fourier coefficients, and it's clear how the fourier coefficients relate to the <span class="math-container">$l_2$</span> norm. However for <span class="math-container">$p\neq 2$</span>, it's not clear to me how to show <span class="math-container">$||T_{\frac{1}{\rho}}(f)||_p \leq \frac{1}{\rho ^d} ||f||_p$</span>, or even relate the norms.</p>

<p>I will note the proof that the dual implies us works well here (I can provide more detalis if needed, but I don't think it will help).</p>

CC BY-SA 4.0
2021-11-28T15:44:05.043Z
Excuse me, what is a dual version? Could you explain\send a link to explain?
Pegi
CC BY-SA 4.0
2021-11-28T22:34:51.920Z
@Pegi I don't remember the details. But the idea is that results about $L_q$ spaces for $q>2$ correspond to ones about $L_p$ with $p<2$ using duality; literally the dual of $L_q$ (As a Banach space) is $L_p$ with $1/p+1/q=1$ (This is often stated as Holder inequality). More concretely suppose you know $||f||_q \leq C||f||_2$, then you can deduce $||f||_2 \leq C||f||_p$; $||f||_2^2 = <f,f> \leq ||f_p|| ||f||_{\frac{p}{p-1}} \leq C||f||_p ||f||_2$ and canceling the $||f_2||$ gives the result.
Andy
<co.combinatorics><boolean-functions>
Dual to hypercontractive inequality
Andy
# Dual to hypercontractive inequality

(Asked by: username_0 on 2019-10-14)

Recall the hypercontractive inequality:

Let $\rho = \sqrt{\frac{p-1}{q-1}}$, then $||T\_\rho(f)||\_q \leq ||f||\_p$

In <https://www.cs.cmu.edu/~odonnell/papers/analysis-survey.pdf> it is stated that the dual of the hypercontractive inequality is equivalent to the following:

If $f$ has degree at most $d$, then

$||f||\_q \leq \rho ^ d ||f||\_p$

I am unable to show this (Specifically, the direction that the original version implies the dual version), and would like to help.

Attempt-

We know for every $g$, $||T\_\rho(g)||\_q \leq ||g||\_p$

Plugging $g = T\_{\frac{1}{\rho}}(f)$,

$||f||\_q \leq ||T\_{\frac{1}{\rho}}(f)||\_p$

Now if $p=2$ it's clear how to finish since it's clear how $T\_{\frac{1}{\rho}}$(f) affect the fourier coefficients, and it's clear how the fourier coefficients relate to the $l\_2$ norm. However for $p\neq 2$, it's not clear to me how to show $||T\_{\frac{1}{\rho}}(f)||\_p \leq \frac{1}{\rho ^d} ||f||\_p$, or even relate the norms.

I will note the proof that the dual implies us works well here (I can provide more detalis if needed, but I don't think it will help).

**Tags:** co.combinatorics, boolean-functions
> **Comments:**
> **username_1**: Excuse me, what is a dual version? Could you explain\send a link to explain?  
> **username_0**: @Pegi I don't remember the details. But the idea is that results about $L_q$ spaces for $q>2$ correspond to ones about $L_p$ with $p<2$ using duality; literally the dual of $L_q$ (As a Banach space) is $L_p$ with $1/p+1/q=1$ (This is often stated as Holder inequality). More concretely suppose you know $||f||_q \leq C||f||_2$, then you can deduce $||f||_2 \leq C||f||_p$; $||f||_2^2 = <f,f> \leq ||f_p|| ||f||_{\frac{p}{p-1}} \leq C||f||_p ||f||_2$ and canceling the $||f_2||$ gives the result.  

---

## 1 answers

### Answer by username_2 on 2019-10-14. Score: 5 | Accepted answer

I'm not sure that the statement you want is known to hold in full generality (if you look at O'Donnell's website, he remarks that what he stated in that survey is not in fact known to be true in full generality). That said, it does hold for $q\geq 2\geq p\geq 1$. Let's suppose the Hypercontractivity Theorem holds as you have stated, i.e. for $q\geq p\geq 1$, suppose for $\rho=\sqrt{\frac{p-1}{q-1}}$, we have
\begin{equation}
\|T\_{\rho}f\|\_q\leq \|f\|\_p.
\end{equation}

To use this statement to get this revised version (or almost, you forgot to take the reciprocal of $\rho$), suppose $q\geq 2$ and first take $p=2$, so $\rho^{-1}=\sqrt{q-1}$. As you seem to have figured out, hypercontractivity gives
\begin{align}
\|f\|\_q^2&=\|T\_{\rho}T\_{\rho^{-1}}f\|\_q^2\\
&\leq \|T\_{\rho^{-1}}f\|\_2^2\\
&=\sum\_{k=0}^d \rho^{-2k}W\_k(f)\\
&\leq \rho^{-2d}\sum\_{k=0}^dW\_k(f)\\
&=\rho^{-2d}\|f\|\_2^2\\
&=\sqrt{q-1}^{2d} \|f\|\_2^2,
\end{align}
where $W\_k(f)$ is the level-k Fourier weight of $f$, and we use Parseval and the degree assumption. Taking squareroots yields $\|f\|\_q\leq \sqrt{q-1}^d\|f\|\_2$ in this simple case.

We also have for $1\leq p\leq 2$
\begin{align}
\|f\|\_2^2&=\langle f,f\rangle\\
&\leq \|f\|\_p\|f\|\_{p/(p-1)}\\
&\leq \|f\|\_p\sqrt{p/(p-1)-1}^d \|f\|\_2\\
&=\sqrt{\frac{1}{p-1}}^{d}\|f\|\_p\|f\|\_2,
\end{align}
where we use Holder's inequality and then the case we just proved with $q=p'=p/(p-1)$. Putting this together with the previous case, we conclude for all $q\geq 2\geq p\geq 1$, that
\begin{equation}
\|f\|\_q\leq \sqrt{q-1}^d \|f\|\_2\leq \sqrt{\frac{q-1}{p-1}}^d \|f\|\_p,
\end{equation}
as claimed.

According to O'Donnell's website, [this paper](https://arxiv.org/pdf/1902.02406.pdf) gives the state of the art on this problem (though I haven't looked through it).

> **Comments:**
> **username_0**: Thanks, I didn't look at the website. I liked your 'duality' solution, seems like I can use whenever I prove something for q>2 to deduce about p<2. Interestingly I can't even find a argument that doesn't use this duality (where the q>2 is done directly from the fourier expression), do you have one?  
> **username_2**: @Andy Thanks! No, this is the only argument that I know, though a quick scan of the paper seems like it does this without duality (i.e. going through $p=2$) with a neat argument (Theorem 1 shows the heat operator from $L^q$ to itself isn't too big a contraction using a nice averaging argument, and then uses Hypercontractivity as you quoted to get the $L^q$ to $L^p$ result in Corollary 2). Also, what I wrote isn't really my solution, it's essentially from Chapter 9 of O'Donnell's book and the exercises there. I'm not sure where the argument originally comes from.  
> **username_0**: hang on, it's actually silly, there's no need for duality. $|T_\rho f|_2| \geq \rho^d ||f|||_2$. I wonder if there are examples where the duality is really essential.  
> **username_1**: I have also asked this from Andy but he does not seem to be active. What does duality mean and how is it generally calculated? @J.G  

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