Why $W^{2,n}_\text{loc}(\Omega)\hookrightarrow W^{1,q}_\text{loc}$ for all $q>0$?

Question

I have saw that $W^{2,n}_\text{loc}(\Omega)\hookrightarrow W^{1,q}_\text{loc}$ for all $q>0$ where $\Omega \subset \mathbb{R}^n$. I know that by Sobolev inequalities $W^{1,p}(\Omega)\subset C^{0,1-n/p}$ if $p>n$(Suppose $\Omega$ smooth if you need). Then $u \in C^{0,\alpha}$ for all $\alpha \in (0,1)$ if $u$ is $W^{1,p}$ for all $p>n.$

Answer 1

There is a typo either in your question or in your source. It is true that $W^{2,n}$ embeds into $W^{1,q}$ (not $W^{2,q}$) for all $q<\infty$, on smooth domains. This is just the Sobolev embedding applied to the first derivatives. It follows that $W^{2,n}_{\rm loc}(\Omega) \subseteq W^{1,q}_{\rm loc}(\Omega)$, by applying the above to balls compactly contained in $\Omega$.

Your conclusion about $C^{0,\alpha}$ is correct.