TITLE: Uncertainty principle density argument
QUESTION [5 upvotes]: I proved the Heisenberg Uncertainty Principle for $f$ in the Schwartz space $ S(\mathbf R)$:
$$
\int_{\mathbf R} |\xi \hat{f}(\xi)|^2 \int_{\mathbf R} |xf(x)|^2 dx \geq \frac{1}{(4\pi)^2} |f|_{L^2(\mathbf R)}^4.
$$
Now I am having a lot of trouble extending it for $f\in L^2(\mathbf R)$. Of course we have to use density of $S(\mathbf R)$ in $L^2(\mathbf R)$ and use that $xf(x)$, $\xi\hat{f}(\xi) \in L^2(\mathbf R)$ (otherwise the inequality is trivial). But I can't think of a way of approximating everything at the same time.
Any hints or suggestions?
Thanks!

REPLY [3 votes]: Let's just take $f_n=\varphi_n *f$, with $\varphi_n(x)=n\varphi(nx)$, and $\varphi\in C_0^{\infty}$, $\int\varphi =1$. We're assuming that $f,xf,\xi\widehat{f}\in L^2$.
Then $\int_{-L}^L |f_n-f|^2\to 0$ by a standard approximation argument, and if we choose $L>0$ large enough, then both $\int_{|x|>L}x^2|f|^2$ and $\int_{|x|>L}x^2|f_n|^2$ are small. So $\|xf_n-xf\|\to 0$.
Similarly, $\xi\widehat{f}=c\widehat{f'}$, and $f'_n=\varphi_n *f'\to f'$ in $L^2$; alternatively, $\xi \widehat{f_n}-\xi\widehat{f}=\xi\widehat{f}(\widehat{\varphi}(\xi/n)-1)$ goes to $0$ in $L^2$ norm by dominated convergence.
(Essentially, what we do here is make use of the fact that smooth functions are dense in Sobolev spaces.)