Consider the question: In an inner product space $V$, when does the Fourier series of $x$, $\sum\limits_{n=1}^k\langle e_n,x\rangle e_n$ converges to $x$ as $k\to\infty$? Well, certainly is converges for all $x$ if $V$ is a Hilbert space. But what if $V$ is not a Hilbert space? Is the completeness property that a Hilbert space possesses necessary to ensure the Fourier series converges in $V$ for all $x\in V$?
Might there be incomplete inner product space $V$, e.g. maybe $c_{00}$ – sequences in $\mathbb{F}$ with finitely many non-zero entries, along with associated inner product $\langle(a_n),(b_n)\rangle = \sum\limits_{n=1}^\infty a_n \overline{b_n}$. Then we choose the obviously countable orthonormal basis. Isn't it true that for each $x$ the Fourier series converges? If this is the case then what can we say about such inner product space whose Fourier series of $x$ converges to any given $x$ in the space, assuming the space is not a Hilbert space, i.e. not complete? Is the point that these spaces are arbitrary and the completeness property will guarantee us Fourier convergence? Perhaps Hilbert spaces are, in turn, easier to deal with in general because they guarantee us nice properties.
Also (more basic question), can we always find an ONB of a well-defined inner product space?
Edit: I just realized Gram-Schmidt does this for us in all Hilbert spaces. I guess the question still remains for incomplete inner product spaces.
Apologies for my lack of LaTeX skills….
So yeah lots of questions. Hopefully someone can tell me if I am on the right lines of thought. Thanks
Best Answer
First of all, I think it's confusing to refer to an arbitrary series of the form $\sum_n \langle e_n, x \rangle e_n$ as a Fourier series. I'd reserve that name for the special case when $e_n$ are trigonometric functions.
To answer your second question first, any separable inner product space, complete or not, has an orthonormal basis (i.e. an orthonormal set whose linear span is dense). Just pick up a countable dense subset and apply the Gram-Schmidt algorithm to it.
(A non-separable inner product space can fail to have an orthonormal basis, as Willie Wong's comment below points out. I got this wrong in my original answer. Thanks, Willie, for the correction.)
For your first question, yes, in any inner product space, complete or not, if $\{e_n\}$ is a (countable) orthonormal basis in the above sense (so that the space is necessarily separable), then $\sum_{n=1}^k \langle e_n, x \rangle e_n \to x$ in norm. Since the span of $\{e_n\}$ is dense, for any $\epsilon$ there exists an integer $k$ and $a_1, \dots, a_k$ such that $\lVert x - \sum_{n=1}^k a_n e_n\rVert^2 < \epsilon$. On the other hand, a simple computation will show that $$\lVert x - \sum_{n=1}^k a_n e_n\rVert^2 - \lVert x - \sum_{n=1}^k \langle x, e_n \rangle e_n\rVert^2 = \sum_{n=1}^k |a_n - \langle x, e_n\rangle|^2 \ge 0$$ hence we have $\lVert x - \sum_{n=1}^k \langle x, e_n \rangle e_n\rVert^2 < \epsilon$. A similar argument will also show that $\lVert x - \sum_{n=1}^k \langle x, e_n \rangle e_n\rVert^2$ is non-increasing with $k$, and so we get the desired convergence.