Assume $f\in L^1(\mathbb R)$ and $(1+x^2)f''\in L^1(\mathbb R)$. Can we say that $xf'\in L^1(\mathbb R)$?
It looks like a weighted Gagliardo-Nirenberg interpolation result, but it is not so easy to find information on this estimate online. I would like to see a proof, a reference that discusses this or related estimates, or a counterexample.
Remarks: we know that $f'$ is bounded, continuous and goes to zero at infinity. Moreover, $f'\in L^1(\mathbb R)$ by the usual Gagliardo-Nirenberg estimates.
One can try to estimate the absolute value of $f'(x)$ using $$ |f'(x)|\leq\int_x^{+\infty}|f''(s)|ds\leq\int_x^{+\infty}\frac{1}{1+s^2}[(1+s^2)|f''(s)|]ds\leq \frac{C}{1+x^2}$$ (the last inequality follows only for $x\geq 0$, but the bound holds for all $x$ by symmetry). Thus, $xf'$ is "almost in $L^1(\mathbb R)$". So, the question sounds at least meaningful.