Show that $2\nabla \sqrt f\,+\,x \sqrt f=0$ (a.e.). $\implies$ $\sqrt f\in \mathcal C_0$. (Derivatives are in weak sense)

Question

Show that $2\nabla \sqrt f\,+\,x \sqrt f=0$ (a.e.). $\implies$ $\sqrt f\in \mathcal C_0$. (Derivatives are in weak sense)

Given that $f\in L^1(\mathbb R^d),f\geq 0,\int_{\mathbb R^d}f=1, \int_{\mathbb R^d}f(x)|x|^2\leq d$

As $f\in L^1(\mathbb R^d)$ so $\lim_{|x|\rightarrow \infty}\sqrt f(x)=0$ only thing remaining to show is continuity.

Author feels that it has to be shown by using Sobolev inequality and bootstrap arguments.

Regards.

Harish

Answer 1

Well, I don't see why you can claim that $$ f\in L^1\text{ so }\lim_{|x|\rightarrow \infty} \sqrt{f}=0. $$ In general that is true \emph{ if there exists a limit}. What may happen is that you have step functions whose maximum tends to infinity in such a way the integral (actually the series) is finite. For instance, we can consider the function $$ g(x)=n\textbf{1}_{[n,n+1/n^3)}, $$ so $$ \int_0^\infty g(x)dx=\sum_{n}\frac{1}{n^2}<\infty. $$

Said that, notice that, using the pde, we have $$ \nabla\sqrt{f}=-0.5x\sqrt{f}\in L^2. $$ This, together with $f\in L^1$, means that $\sqrt{f}\in H^1$. If we want to use bootstrap arguments, I think that we need something on $x\nabla\sqrt{f}$. Right? Hopefully this helps.