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I have been trying to prove Lemma 3.5 in "Introduction to Functional Analysis" by Kreyszig, which says that:

Any $x$ in an inner product space $X$ can have at most countably many nonzero Fourier coefficients $\langle x,e_k\rangle$ with respect to an orthonormal family $(e_k), k\in I,$ in X.

This is my attempt at proof using the hint provided in the book:

Let $n_m$ be the number of Fourier coefficients which are greater than $1/m$. By Bessel inequality: $$\|x\|^2\geq\sum_{k=1}^{\infty}\big|\langle x,e_k\rangle\big|^2> n_m\frac{1}{m^2}.$$ $$\implies n_m< m^2\|x\|^2. $$ This last inequality is clearly true for any $m=1,2,....$. Which implies that maximum number of nonzero Fourier coefficients $n_{\frac{1}{\infty}}<\infty.$

But I am stuck here. It is clear that number of nonzero coefficients might be infinite, but how does it imply $n_m$ is still countable.

The same question has been asked at: Fourier coefficients $\langle x,e_k\rangle$ are at countable. But I don't understand the last sentence of the answer: Therefore there are only countably many i with $|\langle x,e_i\rangle|>0.$

Abhinav Gupta
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1 Answers1

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$\{k:\langle x,e_k \rangle \neq 0\}$ is the union of the sets $\{k:|\langle x,e_k \rangle| > 1/n\}$ over all positive integers n. Each of the sets in the union is finite (as you have shown) and countable union of countable sets is countable.

Kavi Rama Murthy
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