$\nabla \sqrt{\rho} \in L^2(\mathbb{R}^3) \implies \rho \in L^3(\mathbb{R}^3)$

Question

I found this in the

INTERNATIONAL JOURNAL OF QUANTUM CHEMISTRY, VOL XXIV, 250 (1983)

inside the paper of Elliot H. Lieb with the title Density Functionals for Coulomb Systems and I really don't get it. If somebody could tell me why this holds or give me a hint i would be very thankful.

EDIT:

$\rho : \mathbb{R}^3 \to \mathbb{R}_{\ge 0}$ and $\sqrt{\rho} \in H^1(\mathbb{R}^3) = W^{1,2}(\mathbb{R}^3)$.

Answer 1

By Sobolev embeeding, from $\sqrt\rho\in H^1(\mathbb R^3)$ it follows that $$ \sqrt \rho \in L^6(\mathbb R^3), $$ which implies $$ \rho \in L^3(\mathbb R^3). $$

Answer 2

A short answer, if $\sqrt{\rho} \in H^1(\mathbb{R})$, then by the Sobolev embedding ( https://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem with $1/p^* = 1/6 = 1/2 -1/3 = 1/p - 1/n$), we have $\sqrt{\rho} \in L^6(\mathbb{R}^3)$, which then results in $\rho \in L^3(\mathbb{R}^3)$. You can even get estimates on the norm this way.

edit: 36 seconds too late...