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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.

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  • $\begingroup$ It still troubles me that I don't know what the sum $\sum_{k \in I}$ means if $I$ is an uncountable set. Maybe we should just abandon Bessel's inequality and try a differernt approach? $\endgroup$
    – Kenny Wong
    May 10, 2017 at 15:28
  • $\begingroup$ This should help: math.stackexchange.com/questions/20661/… $\endgroup$ May 10, 2017 at 15:33
  • $\begingroup$ It troubles me too, but not the sum. Bessel's inequality talks about orthonormal sequences. And that's why, after supposing the $e_k$ is uncountable, I wrote the finite sum-form of the inequality, which we can use just fine, using finitely many of the uncountable $e_k$ (naming them $e_1 , e_2 , ..$ for convinience). The question is a Lemma I found in Kreyszig's introduction to functional analysis, the proof left for the reader. $\endgroup$
    – Arbiter
    May 10, 2017 at 15:34
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    $\begingroup$ Let $X$ be the functions $f: [0,1]\rightarrow\mathbb{C}$ that are non-zero on at most a countable number of points, such that $\|f\|^2=\sum_x |f(x)|^2 < \infty$, which is the same as $\|f\|^2 = \int |f(x)|^2d\mu(x)$, where $\mu$ is counting measure on $[0,1]$. Let $\delta_y$ be the function that is $1$ at $y$ and is $0$ otherwise. $\{ \delta_y \}_{y\in [0,1]}$ is an uncountable orthonormal basis of $L^2_{\mu}[0,1]$. You can construct any cardinality of basis you want. $\endgroup$ May 10, 2017 at 15:48
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    $\begingroup$ @KennyWong : if $I$ is a set, $$\sum_{k \in I} x_i = \sup_{I' \subset I, I' \text{ is finite} } \{ \sum_{i \in I'} x_i \}$$ $\endgroup$
    – user171326
    May 10, 2017 at 15:53

2 Answers 2

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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 $|\langle x,e_i\rangle|>0$.

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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 $\sum_i x_i > \sum_{i \in A_n} x_i > + \infty$.

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  • $\begingroup$ I understand your reasoning, however, unless I have it wrong, Bessel's inequality doesn't tell us that $ \sum_{i \in I} $ is finite, because it talks about sequences, i.e. $I$ countable, which is not the case we want. Right? $\endgroup$
    – Arbiter
    May 10, 2017 at 15:48
  • $\begingroup$ Bessel tells you that the sum is convergent so that there are only countably many $i$ such that $x_i$ is not zero. $\endgroup$
    – user171326
    May 10, 2017 at 15:49

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