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Suppose that $E \subseteq F$ are measurable and bounded sets. Suppose that for every finite sequence $c$ we have $$A \sum_n |c_n|^2 \leq \left\|\sum_n c_n e^{2 \pi i \alpha_n x}\right\|_{L^2(E)}^2 \leq B \sum_n |c_n|^2,$$ where $A,B > 0$ and $\alpha_n$ is a collection of real numbers. Show that we have a similar chain of inequalities (with potentially different $A$ and $B$) when we replace the $L^2(E)$ norm with $L^2(F)$ norm.

My work: For the lower bound we can get for free by $$\left\|\sum_n c_n e^{2 \pi i \alpha_n x}\right\|_{L^2(F)}^2 \geq \left\|\sum_n c_n e^{2 \pi i \alpha_n x}\right\|_{L^2(E)}^2 \geq A \sum_n|c_n|^2.$$ The upper bound is more difficult. That is how can we prove $$\left\|\sum_n c_n e^{2 \pi i \alpha_n x}\right\|_{L^2(F)}^2 \leq M \sum_n |c_n|^2,$$ for $M > 0$. It seems that this needs to use the boundedness assumption but I am not sure. Trying to apply Holders inequality doesn't do the trick ... Any help would we welcomed.

Thanks! :)

Asuka Langley Sohryu
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  • Maybe try to cover $F$ using copies of $E$? The translation of $x$ by $t$ will incur an additional $e^{2\pi i\alpha_n t}$, which will go into the coefficients $c_n$ and does not change $|c_n|$? – user58955 May 01 '22 at 04:52
  • It suffices to consider $F=[a,b],$ as $F$ is bounded. If $E$ contains an interval $[c,d],$ then by remark made by @user58955 $[a,b]$ is covered by finite number of translates of $[c,d].$ But the problem remains, when $E$ does not contain any interval (up to the Lebesgue measure). – Ryszard Szwarc May 01 '22 at 09:13

1 Answers1

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The solution is inspired by the comment of @user58955.

Assume $E\subset F\subset [-a,a].$ Then for $f\ge 0$ we have $$\displaylines{\int\limits_E\,\int\limits_{-2a}^{2a}f(x+y)\,dx\,dy=\int\limits_E\,\int\limits_{-2a+y}^{2a+y}f(u)\,du\,dy=\int\limits_{-3a}^{3a}f(u)\left (\int\limits_{-a}^a {\bf 1}_{E\cap [u-2a,u+2a]}(y)\,dy\right)\,du \\ \ge \int\limits_{-a}^{a}f(u)\left (\int\limits_{-a}^a {\bf 1}_{E\cap [u-2a,u+2a]}(y)\,dy\right)\,du =\int\limits_{-a}^{a}f(u)\left (\int\limits_{-a}^a {\bf 1}_{E}(y)\,dy\right)\,du \\=|E|\, \int\limits_{-a}^{a}f(u)\,du \ge |E|\, \int\limits_{F}f(u)\,du} $$ Let $$f(x)=\left |\sum_n c_ne^{2\pi i\alpha_nx}\right |^2$$ Then $$\displaylines{\int\limits_E\,\int\limits_{-2a}^{2a}f(x+y)\,dx\,dy}= \int\limits_{-2a}^{2a}\,\int\limits_Ef(x+y)\,dy\,dx\le B\int\limits_{-2a}^{2a}\sum_{n}|c_ne^{2\pi i\alpha_nx}|^2\, dx =4Ba \,\sum_{n}|c_n|^2 $$ Summarizing $$\int\limits_{F}\left |\sum_n c_ne^{2\pi i\alpha_nx}\right |^2\,dx= \int\limits_{F}f(x)\,dx\le {4Ba\over |E|}\,\sum_{n}|c_n|^2$$

Ryszard Szwarc
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