I am self studying Kreyszig book of functional analysis and couldn't solve following problem which is 8 th of section 3.4 Problem is

Show that an element $x$ of an inner product space $X$ cannot have too many Fourier coefficients $\langle x, e_k\rangle$ which are big, $\langle e_k\rangle$ is orthonormal sequence. Precisely, show that number $ n_m $ of $\langle x, e_k\rangle$ such that $| \langle x, e_k\rangle | > 1/m$ must satisfy $ n_m < m^2 ||x||^2.$

I tried using Bessel's inequality but I don't know how to exactly get $ n_m < m^2 ||x||^2.$

  • $\begingroup$ @Thomas Shellby I tried using Bessel inequality $\endgroup$ – Invisible Man Oct 18 '19 at 18:32
  • $\begingroup$ @ Thomas Shellby but I dont know how to exactly get $ n_m $ < $ m^2 $ $ ||x||^2 $ $\endgroup$ – Invisible Man Oct 18 '19 at 18:33
  • $\begingroup$ Use $\sum_{k}|\langle x,e_k\rangle|^2 \le \|x\|^2$ as a starting point. Then reduce the sum to those for which $|\langle x,e_k\rangle| > 1/m$. $\endgroup$ – DisintegratingByParts Oct 18 '19 at 18:37
  • $\begingroup$ So, you know that $\sum_{k=1}^\infty|\langle x,e_k\rangle|^2\le\|x\|^2$. Now, assume that, e.g., the first $K$ summands are greater than $\frac 1{m^2}$. What follows from there? $\endgroup$ – amsmath Oct 18 '19 at 18:38

Let $S_m=\{e_k\mid\frac1m\lt |\langle x,e_k\rangle |\} $. Assume we have $n_m$ distinct elements in $S_m $, say $e_{l_1},\ldots,e_{l_{n_m}} $. Then $$\frac {n_m}{m^2}\lt \sum_{k=1}^{n_m} |\langle x,e_{l_k}\rangle |^2\leq \|x\|^2.$$

  • $\begingroup$ @ Thomas Shellby Thanks $\endgroup$ – Invisible Man Oct 18 '19 at 18:51
  • $\begingroup$ The logic is correct but I think the indexing within the sum is not. The sum as written picks up all indices $\leq n_m$, whether or not the index is within $S_m$. $\endgroup$ – Robert Shore Oct 18 '19 at 18:54
  • 1
    $\begingroup$ @RobertShore Thank you. I have changed the indexing. $\endgroup$ – Thomas Shelby Oct 18 '19 at 18:59

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