# The number of $\langle x, e_k\rangle$ such that $| \langle x, e_k\rangle | > 1/m$ is less than $m^2 ||x||^2.$

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 $$n_m < m^2 ||x||^2.$$

• @Thomas Shellby I tried using Bessel inequality – Invisible Man Oct 18 '19 at 18:32
• @ Thomas Shellby but I dont know how to exactly get $n_m$ < $m^2$ $||x||^2$ – Invisible Man Oct 18 '19 at 18:33
• Use $\sum_{k}|\langle x,e_k\rangle|^2 \le \|x\|^2$ as a starting point. Then reduce the sum to those for which $|\langle x,e_k\rangle| > 1/m$. – DisintegratingByParts Oct 18 '19 at 18:37
• 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? – 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 $$\frac {n_m}{m^2}\lt \sum_{k=1}^{n_m} |\langle x,e_{l_k}\rangle |^2\leq \|x\|^2.$$
• 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 $S_m$. – Robert Shore Oct 18 '19 at 18:54