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I need an example of a Hilbert space in which the following does not hold for all $x$:

$$ x=\sum_k^{\infty} \langle x,u_k \rangle u_k. $$

That is, there are elements that are not expressible as countable sums. It is obvious that this can only happen in the case of a Hilbert space that admits an uncountable orthonormal system.

Is the completion of quasi-periodic functions an example of such spaces?

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Consider the space of real-valued functions on $\mathbb{R}$ with the inner-product given by $$ \langle f, g\rangle = \lim_{R\to \infty} \int_{-R}^{+R} f(x)g(x)dx $$ and let $H$ denote the Hilbert space of equivalence classes of functions whose induced norm is finite (modulo functions whose norm is zero).

Now note that $\sin(ax)$ and $\sin(bx)$ are orthogonal in this space if $a\neq b$. Hence, $\{\sin(ax) : a\in \mathbb{R}\}$ forms an uncountable orthogonal set, and so $H$ cannot be separable.

An additional interesting example is that of $L^2$ spaces themselves. See this discussion.

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  • $\begingroup$ I don't think this example works; it's just the usual $L^2(\mathbb R)$ as a Hilbert space. The induced norm of $f(x)=\sin(ax)$ would be infinite. $\endgroup$ – Dap Feb 14 '18 at 8:24
  • $\begingroup$ There should be 1/R in your inner product definition. $\endgroup$ – Infinite Aug 6 '18 at 6:42
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If $H$ is a Hilbert space with orthonormal basis $\{u_\alpha\}_{\alpha\in A}$, then for all $h\in H$, the set $B=\{\alpha\in A: \langle h,u_\alpha\rangle \neq 0 \}$ is countable, and $h$ can be written as the countable sum $h=\sum\limits_{\alpha\in B}\langle h,u_\alpha\rangle u_\alpha$.

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  • $\begingroup$ @Anthony: Because $\|h\|^2=\sum|\langle h,u_\alpha\rangle|^2$. If there were uncountably many nonzero terms, there would be an $n$ with infinitely many terms greater than $1/n$, hence the series would not converge. $\endgroup$ – Jonas Meyer Oct 13 '13 at 6:05
  • $\begingroup$ I just realised myself that the very definition of $u_\alpha$ being an orthonormal basis is that every $h$ can be written as a countable linear combination. $\endgroup$ – Anthony Carapetis Oct 13 '13 at 6:07
  • $\begingroup$ @Anthony: That definition is not usual to me. An onb is a maximal orthonormal subset. But either way it is a consequence that each element is a countable "linear combination" of onb elements. $\endgroup$ – Jonas Meyer Oct 13 '13 at 6:09

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