The Plancerel-Polya inequality can be stated as follows:

Let $0 < p\le \infty$ and $ \nu \in \mathbb{Z}$. Suppose that $g$ is a (smooth) function satisfying $\mbox{supp }\hat g \subset \lbrace\xi: |\xi| \le 2^{\nu + 1}\rbrace$. Then $$ \sum_{k\in\mathbb{Z}} \sup_{x\in [k2^{-\nu},(k+1)2^{-\nu}]} |g(x)|^p \lesssim 2^{\nu} \|g\|_p^p. $$

Question: Does an analogous inequality hold if the support condition on $\hat g$ is relaxed? Say, if we assume that $g_\nu = 2^{-\nu}\varphi\left(2^{\nu}\cdot\right)$, where $\varphi$ is some smooth function? I am mainly interested in the case $p<1$.

EDIT: I will try a concrete example which is in the same spirit: Assume that $\varphi$ is some nice function (but not with compact frequency support). When does an inequality of the form $$ \sum_{k\in \mathbb{Z}} |c_k|^p \lesssim \|\sum_{k\in \mathbb{Z}}c_k\varphi(\cdot - k)\|_p^p ? $$

hold? Certainly if $\hat \varphi$ has compact frequency support, this follows from the PP inequality. But also if e.g. $\varphi$ is a B-spline, so compact frequency support of $\varphi$ is not necessary.


1 Answer 1


The following argument is particularly easy since $p\le1$, but it should not difficult to prove the same for all $p$, and the answer to your question is essentially negative.

For a generic function $g$ write $g_k=g\chi_k$ where $\chi_k$ is the characteristic function of the annulus $$|x|\in[k2^{-\nu},(k+1)2^{-\nu}].$$ Then for $p\le1$ you can write $$|g(x)|^p\le\sum|g_k(x)|^p.$$ Now assume $g$ satisfies Plancherel-Polya, then the above implies

$$\|g\|_{L^\infty}\lesssim 2^{\nu/p} \|g\| _{L^p}.$$ Thus your function must satisfy Bernstein's inequality.

So you are looking for a class of functions which satisfy in particular Bernstein's inequality with a constant $\sim 2^{\nu/p}$. Take a function $g$ in your class such that $\hat g$ has compact support (you will not want to exclude them I hope) and rescale it, $g_t(x)=g(tx)$. Since the two sides of Bernstein have different scaling properties, you see that the functions $g_t$ drop out of your class for $t$ too large, i.e., when the support of $\hat g_t$ becomes too large. In other words, you need some control on how much mass is spread on large frequencies. A way to control this would be to use weighted norms with weights in Fourier space, but this is nothing else than Sobolev embedding...

  • $\begingroup$ I just edited my original question to note that the function $g$ should be scaled properly in order that e.g. $\|g_\nu\|_p = 1$. $\endgroup$
    – Philipp
    Aug 4, 2010 at 15:20
  • $\begingroup$ So you are dividing both sides of the inequality by $\|g\|_{L^p}$. What does this change? $\endgroup$ Aug 4, 2010 at 15:48
  • $\begingroup$ Ok, so the formula also needs the right scaling. Scaling is not the issue here. I am fairly sure that the inequality also holds in the non-compactly supported frequency case. $\endgroup$
    – Philipp
    Aug 4, 2010 at 16:06
  • 2
    $\begingroup$ I think you are missing the point here, although I might be wrong. The essential information of the PP (or Bernstein) inequality is contained precisely in the constant $2^\nu$, and in its relation with the size of the support of $\hat g$. Anyway, I just wanted to help, certainly not to convince you :) $\endgroup$ Aug 4, 2010 at 16:28
  • $\begingroup$ Yes I misunderstood you. You show that such a constant cannot exist uniformly for all functions. This is clear anyway. I have added another question which should illustrate better what I am after. But thanks a lot, anyway! $\endgroup$
    – Philipp
    Aug 4, 2010 at 20:11

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