How about the inclusion relation between $H^1$ and $C^2$? If $u$ satisfies a Poisson's equation in a compact support domain $U$
$$
- \nabla \cdot ( \nabla u) = f
$$
then $u$ must be a $C^2$ function? Otherwise $\Delta u$ may be not well-defined.
The weak form of the Poisson's reads for any test function $v \in H^1(U)$ such that
$$
\int_U \nabla u \nabla v ~dx = \int_U f v ~dx
$$
This, on the other hand, requires $u$ an $H^1$ function?
The $u$ satisfies the Poisson's must also satisfies its weak form, but the converse is not true.
Claim:  $C^2 \subset H^1$ but $H^1 \not\subset C^2$.

*

*Is this claim true? For the compact support domain $U$? For any open bounded domain $V$? For $\mathbb{R}^n$?

*Can this be extended such as $C^{k+1} \subset H^k$ but $H^k \not\subset C^{k+1}$? This result conflicts with the Sobolev embedding theorem?

 A: I will use Evans' definitions of the $C^k$ spaces, that is
Definition.
Let $\newcommand{\R}{\mathbb R}n,m,k\in\mathbb N$ or $k=\infty$, let $U\subset\mathbb R^n$ be an open set and let $V\subset\mathbb R^m$ or $V\subset\mathbb C^m$. We will use the notations ($V$ may be omitted; also, "$0$-times continuously differentiable" means "continuous")
\begin{align*}
 C^k(U;V):=\{u:U\to V:u\text{ is $k$-times continuously differentiable}\}, \\
 C^k(\bar U;V):=\{u:U\to V:D^{\mathbf k} u\text{ is uniformly continuous on all bounded subsets of $U$,} \\\text{for all $\mathbf k\in\mathbb N_0^n$ with $\lvert\mathbf k\rvert\le k$}\}.
 \end{align*}

Then we see: If $U$ is bounded, then $C^k(U)\subset H^k(U)$ is false (because $C^k(U)\subset L^2(U)$ is false) but $C^k(\bar U)\subset H^k(\bar U)$ is true. In particular, we have $$C^{k+1}(\bar U)\subset C^k(\bar U)\subset H^k(\bar U).$$
Also, for $U$ non-empty, $H^k(\bar U)\subset C^k(\bar U)$ is false. But Sobolev embedding (Winfried Kaballo. Aufbaukurs Funktionalanalysis und Operatortheorie. Springer Spektrum, 2013, Theorem 4.11) does give $$H^{k+1}(\mathbb R)\subset C^{k}(\bar{\mathbb R})$$ and there are analogous results for other domains than $\mathbb R$. There are two abuses of notation here:

*

*$\subset$ actually means "can be embedded by changing each function only on a set of Lebesgue measure zero"

*$\bar{\mathbb R}$ means "the $C^{k}$-norm is finite".


Exercise. Prove that we always have (under the right conditions, but I am tired now) $$C^{k+1}(\bar U)\subset W^{k+1,\infty}(\bar U)\subset W^{k+1,\infty}(\bar U).$$
