Plancherel-Polya Type Inequality for non-compactly Fourier-supported Functions?? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:50:17Z http://mathoverflow.net/feeds/question/34493 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34493/plancherel-polya-type-inequality-for-non-compactly-fourier-supported-functions Plancherel-Polya Type Inequality for non-compactly Fourier-supported Functions?? Philipp 2010-08-04T11:06:19Z 2010-09-12T09:39:47Z <p>Hi!</p> <p>The Plancerel-Polya inequality can be stated as follows:</p> <p>Let $0 &lt; p\le \infty$ 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 $$\sum_{k\in\mathbb{Z}} \sup_{x\in [k2^{-\nu},(k+1)2^{-\nu}]} |g(x)|^p \lesssim 2^{\nu} \|g\|_p^p.$$</p> <p><strong>Question</strong>: 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 $p&lt;1$.</p> <p><strong>EDIT</strong>: 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 $$\sum_{k\in \mathbb{Z}} |c_k|^p \lesssim \|\sum_{k\in \mathbb{Z}}c_k\varphi(\cdot - k)\|_p^p ?$$</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.</p> http://mathoverflow.net/questions/34493/plancherel-polya-type-inequality-for-non-compactly-fourier-supported-functions/34506#34506 Answer by Piero D'Ancona for Plancherel-Polya Type Inequality for non-compactly Fourier-supported Functions?? Piero D'Ancona 2010-08-04T14:29:41Z 2010-08-04T14:29:41Z <p>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.</p> <p>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</p> <p>$$\|g\|_{L^\infty}\lesssim 2^{\nu/p} \|g\| _{L^p}.$$ Thus your function must satisfy Bernstein's inequality. </p> <p>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...</p>