# Span of functions dense in $L^2$

This is an exercise from Werner's Funktionalanalysis. I have to show that the linear span of the functions $f_n(x)=x^ne^{-x^2/2}, n\geq0$ is dense in $L^2(\mathbb{R})$. The book gives the hint to first show that the Fourier transform of $\overline{f(x)}e^{-x^2/2}$ vanishes if $\langle f,f_n\rangle=0$ for all $n$.

Now I don't see how this knowledge about the Fourier transform would help with the original statement. The $f_n$ obviously don't contain some orthogonal set, so even if I could conclude that $f=0$ from this I couldn't use some maximal orthogonal system argument for Hilbert spaces.

Also, I am unable to actually show the hint because for $f \in L^2(\mathbb{R})$ the formula for the Fourier transform does not hold and most of the nice properties of the Fourier transform I know are only valid for Schwartzfunctions.

So any kind of hint as to how this fits together would be helpful.

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Let $M=span\{f_n\}$. One way is to show that $M^{\perp}=\{0\}$ or equivalently, if $f\in L^2$ satisfies $(f,g)=0\ \forall\ g\in M$ then $f=0$. If you take $g=f_n$ in the last equality, you hae that $(f,f_n)=0\ \forall\ n$. Now try to use the hint. – Tomás Dec 7 '12 at 14:06
But I actually have to show $\overline{M}^{\perp}=\{0\}$ because it only says dense. It probably need not be true that $M^{\perp}=\{0\}$. – erlking Dec 7 '12 at 14:26
The Hilbert space $L^2(\mathbb{R})$ splits as $\overline{M} \oplus \overline{M}^{\perp}$. You need to show that $\overline{M}^{\perp} = \{0\}$. As $M^{\perp} = \overline{M}^{\perp}$, it is enough to show that $M^{\perp} = \{0\}$. – levap Dec 7 '12 at 14:31

Let $t\in\Bbb R$. We have for $t\in\Bbb R$ that $$\sum_{n=0}^{+\infty}\int_{\Bbb R}f(x)e^{-x^2/2}\frac{(itx)^n}{n!}=0.$$ As for all integer $n$, $$\left|f(x)e^{-x^2/2}\frac{(itx)^n}{n!}\right|= |f(x)|e^{-t^2/2}\frac{|x|^n|t|^n}{n!},$$ and the RHS is integrable for the product of Lebesgue and counting measure, we can switch the integral and the sum, which gives that for all $t\in\Bbb R$, $$\int_{\Bbb R}f(x)e^{-x^2/2}e^{itx}dx=0.$$ As $x\mapsto f(x)e^{-x^2/2}$ is integrable, by uniqueness of Fourier transform $f(x)e^{-x^2/2}=0$ almost everywhere, hence $f\equiv 0$.
If $f\in L_2(\mathbb{R})$, then $f(x)e^{-x^2/2}$ is Fourier transformable. You can prove this by Schwartz inequality.