I'm using Muscalu and Schlag's textbook to study harmonic analysis and I encountered the following claim:

Given some function $f \in \mathcal{S}(\mathbb{R}^{d})$, where $\mathcal{S}(\mathbb{R}^{d})$ denotes the Schwartz space of functions. Let $\hat{f}$ denote the Fourier transform of $f$. Assume that there exists some measurable set $E$, such that $\text{supp}(\hat{f}) \subset E \subset \mathbb{R}^d$. Then for any $1 \leq p \leq q \leq \infty$, we have the following inequality: ($|E|$ below denotes the Lebesgue measure of $E$) $$||f||_{L^q} \leq |E|^{\frac{1}{p}-\frac{1}{q}}||f||_{L^p}$$ I have managed to show the special case when $q=+\infty$ and $p=2$ by using Young's inequality and Plancherel identity. However, the hint says that we still need to use duality and interpolation to deduce the general conclusion. Any ideas on this?

Moreover, how might this estimate be related to the probability version of Bernstein inequality? Thanks in advance!

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    $\begingroup$ I see how to prove your inequality under the assumption that $p \leq q \leq \infty$ and $0<p\leq 2$. It suffices to verify the case $q=\infty$ (By Holder right?). $\|f\|_{\infty} \leq \|\mathcal{F} f\|_{1} \leq |E|^{1/2} \|\mathcal{F} f\|_{2} = |E|^{1/2} \|f\|_{2}$. So $\|f\|_{2} \leq \|f\|_{p}^{p/2}\|f\|_{\infty}^{1-p/2} \leq |E|^{1/2 - p/4}\|f\|_{p}^{p/2} \|f\|_{2}^{1-p/2}$. Hence $\|f\|_{2} \leq |E|^{1/p -1/2} \|f\|_{p}$. Combining these we get $\|f\|_{\infty} \leq |E|^{1/p}\|f\|_{p}$. $\endgroup$ Apr 20, 2021 at 16:55

1 Answer 1


As I have explained in the comment the inequality holds under the assumption $p\leq q\leq \infty$ and $0<p\leq 2$. The inequality does not hold if $p>2$. Indeed, let $q=\infty$. Let me construct a counterexample on the unit circle. Euclidean case can be adapted in a similar way. Consider $f=\frac{1}{\sqrt{N}}\sum_{1\leq k \leq N} e^{2^{k}xi}$. Then $\|f\|_{\infty}=N^{1/2}$. Also $\|f\|_p=C(p)$ as for large $N$ our function $f$ behaves as complex gaussian random variable. So $\|f\|_{\infty} \leq |E|^{1/p} \|f\|_{p}$ implies $N^{1/2}\lesssim N^{1/p}$ which is impossible for $p>2$. That being said perhaps there is a typo in this claim, and the assumption $0<p\leq 2$ is missing.

If $E$ is a ball or some other "nice set" then it is possible to remove the restriction $p\leq 2$.

The name “Bernstein” has nothing to do with the inequality in probability. It has to do with estimating Lp norms $f'$ in therms of Lp norms of $f$ for polynomials.

  • $\begingroup$ Nice counterexample. Are there counterexamples if you allow only $q < \infty$? $\endgroup$ Apr 21, 2021 at 18:33
  • $\begingroup$ @WillieWong, Theorem 2 here link.springer.com/article/10.1007%2FBF01137055 gives an example of a function for $p>2$ and $q<\infty$ showing that the tight constant should be at least $|E|^{\frac{p}{2}(\frac{1}{p} -\frac{1}{q})}$. $\endgroup$ Apr 22, 2021 at 2:30
  • $\begingroup$ thanks for the reference. $\endgroup$ Apr 22, 2021 at 13:45
  • $\begingroup$ Hmmm, in the paper you linked to, equation (10) relies on the fact that on the circle you have $\|u\|_2 \lesssim \|u\|_p$; it is not clear to me that the bound can be expected to hold on $\mathbb{R}$. $\endgroup$ Apr 22, 2021 at 16:24
  • $\begingroup$ I think it should be possible to transfer these examples on the torus to Euclidean space using "transference principles" (see section "transference theorems" extremal010101.wordpress.com/2020/05/06/…) $\endgroup$ Apr 23, 2021 at 14:07

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