Fourier coefficients $\langle x,e_k\rangle$ are at countable

Question

Say we have an Inner Product space $(X, \langle\cdot ,\cdot \rangle)$. Let $(e_k)$ be an orthonormal sequence, where $k \in I$. Then we know by Bessel's inequality that $$ \sum_1^\infty |\langle x,e_k\rangle|^2 \le \|x\|^2$$ The $\langle x,e_k\rangle$ are called Fourier coefficients of x (with respect to $e_k$). How do we then show that any $x \in X$ can have at most countably many non-zero Fourier coefficients?

First we suppose that the orthonormal family $(e_k)$ is uncountable, otherwise it's trivial. So my first question is, how can there exist uncountably many $e_k$? Is there an example?

What I thought could help is the following: form the coefficients $\langle x,e_k\rangle$. We know that for every $n=1,2,3,\ldots$ $$ \sum_1^n |\langle x,e_k\rangle|^2 \le \|x\|^2$$

But I couldn't make any progress.

Answer 1

Let $(e_i)_{i\in I}$ be an arbitrary orthonormal set, possibly uncountable. I claim that for any $\delta>0$ there are at most finitely many $i\in I$ with $|\langle x,e_i\rangle|\ge\delta$.

If not then there would be a countable sequence within the $e_i$ with $|\langle x,e_i\rangle|\ge\delta$ within that sequence. That would contradict Bessel's inequality as applied to that sequence.

So for $n\in\Bbb N$ there are finitely many $i$ with $|\langle x,e_i\rangle|\ge 1/N$. Therefore there are only countably many $i$ with $|\langle x,e_i\rangle|>0$.

Answer 2

If $I$ is incountable and $x : I \to \mathbb R_{>0}$ is any application, then $\sum_{i \in I} x_i$ is divergent. Indeed, consider $ A_n = \{i \in I : \frac{1}{n}> x_i > \frac{1}{n + 1} \}$. As $I = \bigcup A_n$, there is an uncountable $A_n$, and $\sum_i x_i > \sum_{i \in A_n} x_i > + \infty$.