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I was wondering whether the Schwartz functions are also dense in $$\{f \in L^2(\mathbb{R}^n); \int_{\mathbb{R}^n} |x|^2 |f(x)|^2 dx + \int_{\mathbb{R}^n}|\xi|^2 |\hat{f}(\xi)|^2 d \xi < \infty\}$$

where the norm is given by $$||f||_{L^2}^2 = \int (1+|x|^2) |f(x)|^2 dx + \int (1+|\xi|^2) |\hat{f}(\xi)|^2 d \xi.$$

If we would only have one summand, then this would be a $H^1-$ Sobolev summand norm and for this space it is of course true, but what about this case here, where we have an additional term in the norm?

I suspect it is true, but how can I see it?

Edit: I want to note that the argument given in the comments is incomplete.

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    $\begingroup$ Schwartz functions are dense in $H^1$ and convergence in $H^1$ of $f_n \to f$ ($f_n$ Schwartz, $f$ in your space) implies convergence in $H^0$, which implies convergence of $\sqrt{1+|x|^2} f_n \to \sqrt{1+|x|^2} f$ in $H^0$ as well (multiplication by $\sqrt{1+|x|^2}$ is a closed operator because it has a bounded inverse). So $f_n \to f$ also in your norm. $\endgroup$
    – Igor Khavkine
    Jun 3, 2016 at 1:25
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    $\begingroup$ @IgorKhavkine Doesn't $f_n(x)=n^{-1}\exp(-x^2/n^2)$ contradict your argument? Its $L^2$ norm is $C/\sqrt{n}$ while that of $xf_n$ is $C'\sqrt{n}$. $\endgroup$
    – Jean Duchon
    Jun 3, 2016 at 15:07
  • $\begingroup$ @JeanDuchon yes, I agree it is completely unclear how this closedness argument can be applied here. You just don't have that $\sqrt{1+|x|^2 } f_n$ is a convergent sequence. $\endgroup$
    – Leroy
    Jun 4, 2016 at 14:44
  • $\begingroup$ @JeanDuchon, you and Leroy are of course right. For my argument to work, I would need to know that $\sqrt{1+|x|^2} f_n$ converges to something, which a priori may not be true. One could improve the argument with a better choice of the sequence $f_n$, but Christian Remling already seems to have given a good answer below. $\endgroup$
    – Igor Khavkine
    Jun 4, 2016 at 21:33

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We can do this by approximating the derivative of such an $f$ in $L^2$, as follows: Given $\epsilon>0$, take $L>0$ so large that $\int_{|x|>L}(x^2|f|^2+|f'|^2)<\epsilon$.

Now approximate $f'$ in $L^2(-L,L)$ by an $h\in C_0^{\infty}(-L,L)$, and put $$ g(x)=f(-L)+\int_{-L}^x h(t)\, dt . $$ Finally, we modify $g$ on the two intervals $L\le |x|\le L+1$ to obtain a compactly supported smooth function. Notice that this can be done in such a way that $|g|, |g'|\lesssim |f(\pm L)|$ on these intervals.

We're now hoping that both $\|(1+|x|)(g-f)\|_2$ and $\|g'-f'\|_2$ will be small, and that is pretty much clear already, except for the contributions of the type $\int_{L\le x\le L+1} x^2|g|^2\simeq L^2|f(L)|^2$. However, these are harmless also because it can not be the case that for all large $L$, at least one of $|f(L)|, |f(-L)|$ is $\ge\epsilon/L$ (in that case $xf$ would not be in $L^2$), so we just need to choose our $L$'s carefully.

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