Timeline for $\ell^2 \to L^\infty$-inequality for almost periodic functions
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 22 at 14:40 | comment | added | Aleksei Kulikov | @JasonZhao it is easy to construct an example of equidistributed $z_k$ such that $\sum z_k \frac{1}{k} cos(kx)\to +\infty$ for $x = 0$ say, while for the second one it is obviously bounded by Cauchy--Schwarz. | |
| Sep 22 at 5:22 | answer | added | Onur Oktay | timeline score: 3 | |
| Sep 21 at 23:26 | comment | added | Jason Zhao | @ChristianRemling perhaps a more concrete question is, can you cook up $\{z_k\}_k \subseteq \mathbb S^1$ which are equidistributed, such that $u_n(x) := \sum_{k = 1}^n z_k \tfrac1k \cos(kx)$ blows up in $L^\infty$? More generally, given coefficients $c_k \in \ell^2$, is the map $c_k \mapsto \sum_{k = 1}^n z_k |c_k| \cos(kx)$ bounded from $\ell^2 \to L^\infty$? If not, a counter-example would be greatly appreciated. | |
| Sep 21 at 22:36 | comment | added | Christian Remling | I think this is hopeless. Even for a periodic function, $\|\widehat{u}\|_{\ell^2}=\|u\|_{L^2}$, but of course the $L^2$ does not control the $L^{\infty}$ norm at all. | |
| Sep 21 at 21:59 | history | asked | Jason Zhao | CC BY-SA 4.0 |