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It is generally known that polynomials are dense in the Hilbert space. How do i show this?

I know, it suffices to show completeness and i would like to use this defination: The vectors $\{e_n\}_{n=0}^{\infty}$ are complete in the Hilbert space $L^2(w)$ if $\int_{-\infty}^{\infty}f(x)e_nw(x)dx=0$ for all $n\ge0$ and $f\in L^2(w)$ implies that $f=0$ almost everywhere.

But again, i know that $f=0$ almost everywhere if its fourier transform is zero from the following: If the function f is integrable on $(-\infty,\infty)$ and $$\hat{f}(x)=\int_{-\infty}^{\infty}f(t)e^{ixt}dt\equiv0,$$ then $f=0$ almost everywhere.

So, how do i show that the fourier transform is identically zero? Lets take the weight to be $w(x)=1/2\cosh\frac{\pi}{2}x$ but any other weight can do since the polynomials should be dense regardless of the weight chosen.

Thanking you in advance.

$Proof.$ I have tried to work it out using the defination that $\{e_n\}_{n=0}^{\infty}$ is complete in $L^2(w)$ if $\langle f,e_n\rangle=0$ implies that $f=0$ a.e.

Let $f\in L^2(w)$ and suppose that $\int_{-\infty}^{\infty}f(x)x^nw(x)dx=0$ for all $n\ge0$, then \begin{align} \int_{-\infty}^{\infty}f(x)e^{itx}w(x)dx=\sum_{k=0}^{\infty}\frac{(it)^n}{n!}\int_{-\infty}^{\infty}f(x)x^nw(x)dx=0 \end{align} Thus, since the fourier transform is zero, $f(x)=0$ a.e. and so $\{x^n\}_{n=0}^{\infty}$ is complete in $L^2(w)$ by definition.

Could this be ok and how can i be sure that this fourier transform exists or is analytic so that even the interchange of sum and integral can be valid by beppo-levi lemma?

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With the weight function $w$ you give, polynomials are not even in $L^2(w)$. –  Nate Eldredge Mar 31 '12 at 5:10
    
Thank you Nate. I know that $w(x)=1/2\cosh\frac{\pi}{2}x$ is even but i would like to know how that will affect the completeness on polynomials. –  Davie Roberts Mar 31 '12 at 7:15
    
Wait, I may have misread it. I understood you to mean $w(x) = (1/2) \cosh(\frac{\pi}{2} x)$, and in that case, since $w(x) \to \infty$ as $x \to \pm \infty$, polynomials will not be square-integrable with respect to this weight. But if you mean $w(x) = 1/(2 \cosh (\frac{\pi}{2} x))$, then everything is okay and I can post more details. –  Nate Eldredge Mar 31 '12 at 14:46
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Sure Nate, the weight is $w(x)=\frac1{2\cosh\frac{\pi}{2}x}$, thank you. –  Davie Roberts Mar 31 '12 at 18:12
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1 Answer 1

You could invoke Weierstrass approximation theorem for continuous functions on a compact interval under the sup-norm.

Then use the density of continuous functions of compact support in $L_{2}(\mathbb{R},\omega)$ to show the polynomials are dense in $L_{2}(\mathbb{R},\omega)$

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The Weierstrass approximation theorem lets you approximate functions uniformly on a compact interval, which is quite different from approximating in $L^2$ norm on all of $\mathbb{R}$. So some extra work would be needed there. –  Nate Eldredge Mar 31 '12 at 5:15
    
I would like to use the defination that $\{e_n\}_{n=0}^{\infty}$ is complete in $L^2(w)$ if $\langle f,e_n\rangle=0$ implies that $f=0$ a.e. –  Davie Roberts Mar 31 '12 at 6:17
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