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Recall the hypercontractive inequality:

Let $rho = sqrtfracp-1q-1$, 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 $g = T_frac1rho(f)$,

$||f||_q leq ||T_frac1rho(f)||_p$

Now if $p=2$ it's clear how to finish since it's clear how $T_frac1rho$(f) affect the fourier coefficients, and it's clear how the fourier coefficients relate to the $l_2$ norm. However for $pneq 2$, it's not clear to me how to show $||T_frac1rho(f)||_p leq frac1rho ^d ||f||_p$, 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 $qgeq 2geq pgeq 1$. Let's suppose the Hypercontractivity Theorem holds as you have stated, i.e. for $qgeq pgeq 1$, suppose for $rho=sqrtfracp-1q-1$, we have
beginequation
|T_rhof|_qleq |f|_p.
endequation

To use this statement to get this revised version (or almost, you forgot to take the reciprocal of $rho$), suppose $qgeq 2$ and first take $p=2$, so $rho^-1=sqrtq-1$. As you seem to have figured out, hypercontractivity gives
beginalign
|f|_q^2&=|T_rhoT_rho^-1f|_q^2\
&leq |T_rho^-1f|_2^2\
&=sum_k=0^d rho^-2kW_k(f)\
&leq rho^-2dsum_k=0^dW_k(f)\
&=rho^-2d|f|_2^2\
&=sqrtq-1^2d |f|_2^2,
endalign
where $W_k(f)$ is the level-k Fourier weight of $f$, and we use Parseval and the degree assumption. Taking squareroots yields $|f|_qleq sqrtq-1^d|f|_2$ in this simple case.

We also have for $1leq pleq 2$
beginalign
|f|_2^2&=langle f,frangle\
&leq |f|_p|f|_p/(p-1)\
&leq |f|_psqrtp/(p-1)-1^d |f|_2\
&=sqrtfrac1p-1^d|f|_p|f|_2,
endalign
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 $qgeq 2geq pgeq 1$, that
beginequation
|f|_qleq sqrtq-1^d |f|_2leq sqrtfracq-1p-1^d |f|_p,
endequation
as claimed.

According to O'Donnell's website, this paper gives the state of the art on this problem (though I haven't looked through it).