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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

2 Answers 2

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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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    Can someone verify (or argue against) my comment after the asterisk? It's late where I am, and I can never be 100% sure of myself at this hour...2011-04-26
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    It's not the set of functions but the vector space generated by these functions, and after the asterisk the condition is just $f(a)=f(b)$.2011-04-26
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    Thanks Jesse. Where did you find these functions?2011-04-26
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    @Josef K.: http://en.wikipedia.org/wiki/Fourier_series2011-04-26
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    One needs to be slightly careful. Stone-Weierstrass Theorem gives that the trig polynomials are dense in $C[a,b]$ in the topology given by uniform convergence of functions, while $C[a,b]$ is dense in $L^2$ in the Hilbert space topology of $L^2$. So just this is not enough to guarantee that the trignometric polynomials are dense in $L^2$. This is the distinction [I wrote about in my answer](http://math.stackexchange.com/questions/33645/) to another question.2011-04-26
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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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    Very good answer: simple, elegant and almost trivial after observing this fact in the first place. :-) (the upvote wasn't enough.)2011-04-26
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    Thank you. Could you please explain a bit about the endpoints? In particular, if $a$ is irrational then any simple function $\phi$ will vanish at $a+\delta$, leaving a set of positive measure on which $|f-\phi|=|f|$. Now $\delta$ tends to zero, but why doesn't $f$ escape fast enough for convergence in $L^2$ to be avoided?2011-04-26
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    @Josef: I'm not sure I understand your question. The subspace Mariano describes approximates the space of all step functions since $\mathbb{Q}$ is dense in $\mathbb{R}$, and the step functions are dense in $L^2$ by the standard argument.2011-04-26
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    @Qiaochu: I could rephrase my question as follows: given that $\mathbb Q$ is dense in $\mathbb R$, how do you conclude that the step (simple) functions with endpoints in $\mathbb Q$ are dense in the step functions with endpoints in $\mathbb R$. But I can see now that it may be so because - unlike my arbitrary function $f$ above, a step function in $\mathbb R$ cannot escape because it has finitely many values. (I hope this makes sense to someone other than me, but it makes me happy anyway ;) )2011-04-26
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    The $L^2$ distance between the characteristic function of $[a,b]$ and $[p,q]$, where $p$ and $q$ are rational approximations for $a$ and $b$, is $\sqrt{|a-p|+|b-q|}$. Clearly you can make this arbitrarily small.2011-04-26
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    I was talking about step functions though.2011-04-26
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    Step functions are just (finite) linear combinations of characteristic functions of intervals...2011-04-26
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    @R. Thank you. I am not sure whether you aim to agree or disagree with the comment I addressed to Qiaochu above where I refer to step functions. I don't feel the need to decompose the argument beyond this to indicator functions unless there is something wrong in my conclusion.2011-04-26
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    I don't aim to disagree with your comment, just to show a simple line of reasoning why it follows from the density of $\mathbb{Q}$ in $\mathbb{R}$ that the step functions with rational endpoints will be dense in $L^2$.2011-04-26
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    @josef, but to go from one of those facts to the next in order to get a detailed proof should be easy for anyone familiar enough with measure theory to understand your question!2011-04-27
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    Sorry, I fail to see what the argument is that either of you have in your minds. What comes across on the page is just mathematical facts (eg denseness of $\mathbb Q$ in $\mathbb R$, that the difference betweeen two indicator functions can be made arbitrarily small), which I do not disagree with but whose consequences to the present context are not made evident in your text.2011-04-27
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    Well: a corollary of my last comment is that you should think about this for a while and find yourself a way of actually filling in the details. That I, or others, do it for you will help significantly less than that.2011-04-27
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    I have done a first course in Measure Theory and in Functional Analysis. I think the follow-on thinking from the above is easy, but I am taking my first steps in a complex field and feel the need for validation of even simple conclusions in order to grow in confidence. (By the way, the last comment wasn't in relation to you proof, Mariano.)2011-04-27
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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