6
$\begingroup$

The classical Hausdorff-Young inequality states that

$$ \Vert \widehat{f} \Vert_{p'} \leq \Vert f \Vert_p \text{ for } 1 \leq p \leq 2. $$

For $p=2$, we even have equality due to Plancherel.

If we additionally assume that $f \geq 0$, we also get

$$ \Vert \widehat{f} \Vert_\infty = \widehat{f}(0) = \int f(x) \, dx = \Vert f \Vert_1, $$

i.e. we get equality in the Hausdorff-Young inequality for $p=1$ also.

My question is, wether this generalizes to $1 \leq p \leq 2$ (at least asymptotically), i.e. do we have

$$ \Vert \widehat{f} \Vert_{p'} \asymp \Vert f \Vert_p \text{ for } 1 \leq p \leq 2 \text{ and } f \geq 0 \text{?} $$

We can not use interpolation here (at least I do not see it), because the estimate on the "boundary" points (at least at $p=1$) is only valid for $f \geq 0$ (and the whole inequality can also only be valid for those $f$), so that the usual "splitting" (for real interpolation) can not be applied. Similarly, complex interpolation does not seem to work.

But also the classical method for constructing a counterexample does not work, i.e. one can not take something like

$$ f = \sum_{m=1}^n M_{\xi_m} g, $$

where $M_\xi g (y) = e^{2\pi i \xi y} g(y)$ denotes modulation, because this will violate the non-negativity.

Taking

$$ f = \sum_{m=1}^n T_{x_n} g $$

does not violate this assumption and I can asymptotically calculate $\Vert f \Vert_p$ in this case (for $\min_{n \neq m} |x_n - x_m| \to \infty$), but I am unable to calculate the integral

$$ \Vert \widehat{f} \Vert_{p'} = \bigg( \int \big| \sum_{m=1}^n e^{\pm 2\pi i x_n \xi }\big|^{p'} \cdot |\widehat{g}(\xi)|^{p'} \, d\xi \bigg)^{1/p'}. $$

precisely enough.

Any ideas would be appreciated.

Improve this question
$\endgroup$
3
  • $\begingroup$ If the spatial domain is the integers (so the frequency domain is the unit circle), then one can take $f$ to be the indicator function of a lacunary sequence such as $1, 2, \dots, 2^n$ and use Rudin's inequality (see e.g. Lemma 4.33 of my book with Van Vu) to contradict the inequality; the case $p = 4/3$ can be worked out by hand for instance. One can then transfer to Euclidean spaces by standard methods (e.g. blurring each integer by an approximation to the identity to pass from ${\bf Z}$ to ${\bf R}$). $\endgroup$
    – Terry Tao
    Commented Oct 7, 2014 at 23:30
  • 4
    $\begingroup$ In fact, it is very rare for the Hausdorff-Young inequality to be close to sharp; the function $f$ basically has to look like the indicator function of the sum of a generalised arithmetic progression and an ellipsoid. A precise formulation of this fact (which comes from additive combinatorics) can be found in Proposition 6.4 of this paper of Christ: arxiv.org/pdf/1406.1210.pdf $\endgroup$
    – Terry Tao
    Commented Oct 8, 2014 at 0:02
  • $\begingroup$ Dear Professor Tao, thank you very much for your comments. I will definitely try to work through the paper by Christ, because I was (more or less exactly) looking for functions which make Hausdorff Young sharp and was just hoping (perhaps naively) that non-negative functions might already work. $\endgroup$
    – PhoemueX
    Commented Oct 9, 2014 at 20:57

2 Answers 2

Reset to default
2
$\begingroup$

If $1<p<2$, then it is not possible to have the inequality $$ \|f\|_p \lesssim \|\widehat{f}\|_{p'} \quad\quad\quad\quad\quad (1) $$ for all $f\ge 0$. This follows from the existence of (positive) purely singular measures $\mu$ with $\widehat{\mu}\in L^{p'}$ (in fact, $\widehat{\mu}$ can have power decay). (I used this fact also in my answer to this question; here, however, decay of $\widehat{\mu}$ in an average sense is enough, which is much easier to obtain.)

I think it is intuitively clear that such a $\mu$ refutes (1), but to elaborate some more on this, consider $f_t=\varphi_t*\mu$, with $\varphi_t(x)=(1/t)\varphi(x/t)$ and $\varphi\ge 0$, $\varphi\in C_0^{\infty}$. Then, if we had (1), it would follow that $f_t$ is a Cauchy sequence in $L^p$; however, this sequence converges to $\mu$ in the sense of distributions, and as $\mu$ is singular, clearly $\mu\notin L^p$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Thanks for the answer. One technical remark: I think that it does not follow (easily) that $f_t$ is Cauchy in $L^p$; for showing that using (1), we would need to know $f_t - f_s \geq 0$ (or $\leq 0$) for $t \geq s$, which I do not see. But somewhat differently, we can conclude using (1) that $(f_t)_t$ is bounded in $L^p$, so that (by reflexivity) there is a subsequence $f_{t_n}$ with $t_n \to 0$ converging weakly in $L^p$. Using that $\mu$ can in fact be chosen to have compact support, one can then derive that the weak limit is a density of $\mu$, which yields a contradiction. $\endgroup$
    – PhoemueX
    Commented Oct 9, 2014 at 21:02
  • $\begingroup$ @PhoemueX: You are right, I was sloppy (I mindlessly used (1) for the difference). Your argument should replace this part. $\endgroup$
    – Christian Remling
    Commented Oct 9, 2014 at 21:34
5
$\begingroup$

It is an elementary exercise in Banach space theory to show that there is NO operator from $L_p$ into $L_{q}$ that satisfies the inequality when $1\le p \not= 2$ with $p<q<\infty$. First, for such $p$ and $q$ every operator from $\ell_p$ into $L_q$ is strictly singular (because $\ell_p$ is not isomorphic to a subspace of $L_q$). Take any (bounded, linear) $T:\ell_p \to L_q$. If $\|Te_n\|_q $ does not converge to zero, assume by passing to a subsequence and making a small perturbation that $Te_n$ is a block basis of the Haar basis and hence is unconditional. By the strict singularity of $T$ there is a normalized block basis $y_k = \sum_{j=n_k}^{n_{k+1}-1} a_j e_j$ s.t. $\|Ty_k\|_q\to 0$. Since $Te_n$ is unconditional, setting $z_k = \sum_{j=n_k}^{n_{k+1}-1} |a_j| e_j$, we also have $\|Tz_k\|_q \to 0$, and of course $\|z_k\|_p = \|y_k\|_p =1$. This proves the result, because $e_n$$ \mapsto \mu(A_n)^{-1/p} 1_{A_n}$ extends to an isometry from $\ell_p$ into $L_p$ if the $A_n$ are disjoint sets of positive measure.

So your question has nothing to do with the Fourier transform.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Do you consider $L^{p}$ with Lebesgue measure? If not then Nelson's hypercontractivity gives the counterexample. Hermite semigroup $e^{-tH}$ maps $L^{p}(d\gamma)$ into $L^{q}(d\gamma)$ for $1 <p<q<\infty$ where $d\gamma$ is the Gaussian measure. $\endgroup$
    – Paata Ivanishvili
    Commented Dec 28, 2016 at 18:22
  • $\begingroup$ The argument works for every infinite dimensional $L_p$ space since every separable $L_p$ space is isometrically isomorphic as a Banach lattice to $X \oplus_p \ell_p^n$ for some $n=0,1,2,\dots, \infty$, where either $X=\{0\}$ or $X=L_p(0,1)$. $\endgroup$
    – Bill Johnson
    Commented Dec 30, 2016 at 6:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .