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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

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What differentiates complete from incomplete inner product spaces is that if $V$ is complete, then for all sequences $(a_n)$ such that $\sum |a_n|^2 <\infty$, $\sum a_ne_n$ converges to an element of $V$. If $V$ is not complete, then there exists a sequence $(a_n)$ such that $\sum |a_n|^2<\infty$ but $\sum a_n e_n$ does not converge (although the partial sums form a Cauchy sequence). –  Jonas Meyer May 3 '12 at 5:44
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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.

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There's no guarantee that the maximal orthonormal set is countable, though, right? Or are you building that into your definition of inner product space (like how some people only allow separable Hilbert spaces)? –  Willie Wong May 3 '12 at 10:47
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In fact, this Wikipedia entry seems to contradict your assertion that any inner product space, complete or not, has an orthonormal basis. In particular, a maximal orthonormal system may not need to be an orthonormal basis in the absence of completeness and separability (just one of the two is enough). –  Willie Wong May 3 '12 at 10:57
    
Ah ok, so is it true that: An IPS V has a countable dense subset (is separable) iff it has an ONB ? I think so. [-->]: definition of density. [<--] Just take the linear span of the ONB, and this is a countable dense subset of V. Hence you can say something about non-separable IPS. But that is not of interest. –  Adam Rubinson May 3 '12 at 22:19
    
@AdamRubinson: A non-separable inner product space can certainly have an orthonormal basis, i.e. an orthonormal set whose span is dense. Such a set would necessarily be uncountable in this case. Every non-separable Hilbert space does in fact have an ONB, but as Willie points out a non-separable non-complete inner product space can fail to have one. However, it is true that an inner product space is separable iff it has a countable orthonormal basis. (If: take the $\mathbb{Q}$-span of the ONB. Only if: use Gram-Schmidt as I suggest in my answer.) –  Nate Eldredge May 3 '12 at 22:58
    
I meant to write countable ONB in space of ONB, but thanks for making things clearer anyhow. –  Adam Rubinson May 3 '12 at 23:22
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