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.



$x f'$ bounded by $x^2f $ and $f''$?
By a cutoff function argument, it suffices to assume $f$ is compactly supported, so we can integrate by parts without picking up boundary terms.


Thus
$$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f$$
Hence using Cauchy-Schwarz,
$$\|xf'\|\_2^2 \le \int |2xf f'| + \int |x^2 f f''| \le 2\|xf\|\_2 \|f'\|\_2 + \|x^2f\|\_2 \|f''\|\_2.$$
The second term is finite by our assumption. For the first term, note that
$$\|xf\|\_2^2 = \int x^2 f^2 = \int |x^2f| |f| \le \|x^2f\|\_2 \|f\|\_2$$
and 
$$\|f'\|^2 = \int (f')^2 = -\int f f'' \le \int |f f''| \le \|f\|\_2 \|f''\|\_2.$$


Putting everything together, we have
$$\|xf'\|\_2^2 \le 2 \|f\|\_2 \sqrt{\|f''\|\_2 \|x^2f\|\_2} + \|x^2f\|\_2 \|f''\|\_2.$$


