is $H^2\subset L^{\infty}$

Question

Is the space of function $H^2$ subset of the space $L^{\infty}$ on any set (such as $\mathbb{R}^3$ or $\mathbb{R}^3\times\mathbb{R}_+$)?

Is there a reference for the proof?

Thank you in advance.

Answer 1

Since the question is a bit vague, I can only give a rather vague answer. In general, in order to determine if a Sobelev function is in $L^{\infty}(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is a not necesarrily bounded subset, is usually done via the Morrey estimate $$ ||u||_{C^{0,\gamma}(\Omega)}\leq C||u||_{W^{1,p}(\Omega)} $$ for $p>n$. In order to get the $p$ "high enough", you can use the Gagliardo–Nirenberg–Sobolev inequality inequality to show that $$ ||u||_{W^{1,p}(\Omega)}\leq C ||u||_{H^2(\Omega)} $$ for some suitable value of $p>2$. Depending on $n$, you can check for which dimensions your function is Hölder and then you have to consider many cases. For example, if $\Omega$ is a bounded domain, then your Hölder function is in $C(\overline{\Omega)})$ and therefore bounded (i.e. in $L^{\infty}(\Omega))$. For unbounded domains, like the whole $\mathbb{R}^n$, you have to work a bit harder. If I remember correctly, square integrable Hölder functions tend to $0$ as $|x| \to \infty$, but I dont have a reference at hand.