$\dot{H}^1\subset L^6 $

Question

I am reading about Lebesgue space and Sobolev spaces that are homogeneous. An exercise that my professor gave me is: How can I use the Sobovev's inject to demonstrate that

$$\dot{H}^1(\mathbb{R}^3)\subset L^6(\mathbb{R}^3) ?$$

I was thinking in to use the respective norms, but I cannot solve it.

Answer 1

For $\Omega \subset \mathbb{R}^3$, a bounded open subset with regular boundary, if $u \in H^1(\Omega) = W^{1,2}(\Omega)$, using Evans-2010, Theorem 6, page 284 we have $u \in L^{q}(\Omega)$, where $1/q = 1/2 - 1/3$. Note that $q = 6$. The case $\Omega = \mathbb{R}^{3}$ can be found in Kesavan S.Topics in functional analysis and applications.(2003), in the page 79, Theorem 2.4.5.