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Is anybody aware of, or can provide at least an outline, of a proof that the Hilbert space of Lebesgue functions square-integrable on the closed real interval [a,b], equipped with the $L^2$ norm, is separable?

I've seen an ugly proof involving truncated functions so I'm not desperate, but would really like to use something nice. By the way, if you refer to a particular dense countable subset, could you please explain why it is dense and countable even if you consider it to be a fairly 'high-profile' set?

Thanks

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    polynomials with rational coefficients2011-04-26

3 Answers 3

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The set of functions $\{e_n\colon n\in\mathbb{Z}\}$ given by $e_n(x) = \exp\left(2\pi in\frac{x-a}{b-a}\right)$ is dense in $C[a,b]$ by the Stone-Weierstrass Theorem.* Since $C[a,b]$ is dense in $L^2[a,b]$, it follows that $\{e_n\colon n \in \mathbb{Z}\}$ is dense in $L^2[a,b]$.

*Technically speaking, they're dense in the space of continuous functions normalized so that $f(a) = f(b) = 1$. However, this doesn't really matter as we can always look at $[a-\epsilon, b+\epsilon]$ instead.

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    To complete the argument you need to use the fact that the uniform norm (or the $L^\infty$ norm) is stronger than the $L^2$ norm. More precisely, if a sequence of function on $[a,b]$ converges uniformly it will also converge in $L^2$ to the same limit.2011-04-26
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The sub-$\mathbb Q$-vector space generated by the characteristic functions of intervals with rational end-points is countable and dense.

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    If/when you post$a$new question giving the result of you completing the details of the argument and asking for validation, I and others will probably do just that. :)2011-04-27