Does $\|f\|_{L^2}\,\le\,C(\|f'\|_{L^2} + \|xf\|_{L^2})$ for $f\in H^1(-1,1)$?

Question

I would like to prove that there is a constant $C>0$ such that for all $f\in H^1(-1,1)$ (the usual Sobolev space) we have $$ \|f\|_{L^2}\,\le\,C(\|f'\|_{L^2} + \|xf\|_{L^2}). $$ Here, $xf$ is just the function $x\mapsto xf(x)$. Does anyone have an idea how to prove this?

Answer 1

This is basically the uncertainty principle in disguise. I'm going to leave a few details to sort out to you, but the basic idea is as follows. From AM-GM

$$2\sqrt{\|f'\|\|xf\|} \le \|f'\| + \|xf\|.$$

But then from Cauchy-Schwarz

$$|\langle f', xf\rangle| \le \|f'\|\|xf\|.$$

Moving the derivative to the other term,

$$\langle f',xf\rangle = -\langle f,(xf)'\rangle = -\langle f,f\rangle - \langle f, xf'\rangle.$$

The last term looks eerily similar to the leftmost term, and indeed $\langle f, xf'\rangle = \overline{\langle f', xf\rangle}$. Moving it to the other side,

$$\langle f', xf\rangle + \overline{\langle f', xf\rangle} = -\langle f,f\rangle.$$

Note that this does not quite match our second inequality, but I'll let you work out the detail there.