TITLE: $x f'$ bounded by $x^2f $ and $f''$?
QUESTION [11 upvotes]: Consider the Hilbert space of functions $f \in L^2(\mathbb R)$ such that $x^2f \in L^2(\mathbb R) $ and $ f'' \in L^2(\mathbb R).$
I am wondering whether it is true that $xf'\in L^2(\mathbb R)$ as well? 
It seems natural and perhaps some sort of interpolation should yield the claim but I fail to see how to show this.

REPLY [15 votes]: By integration by parts,
$$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f.$$
$$-\int 2xf'f = \int f^2=\|f\|^2.$$
By the Cauchy-Schwarz inequality
$$\bigg|\int x^2 f'' f\bigg|\le \|x^2f\|\|f''\|.$$
The conclusion is proved.