# Dual to hypercontractive inequality

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).

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.
• @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. – J.G Oct 15 '19 at 16:11
• 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 Oct 15 '19 at 20:34