Let
- $d\in\left\{2,3\right\}$
- $D\subseteq\mathbb R^d$ be a bounded domain
- $\lambda$ be the Lebesgue measure on $D$
- $H^k(D)$ be the Sobolev space
Can we show that each element of $H^k(D)$ has a continuous representative?
For $d=1$ and $D=(a,b)$ with $0<a<b<\infty$, it's easy to show that $$f(x)=\underbrace{f(a)+\int_{(a,\;x)}f'\;{\rm d}\lambda}_{=:\;g(x)}\;\;\;\text{for }\lambda\text{-almost all }x\in D$$ for all $f\in H^1$ (where $f'$ is the weak derivative of $f$). By Lebesgue's dominated convergence theorem, $g\in C^0(D)$ and since two continuous functions which equal $\lambda$-almost everywhere are identical, we can uniquely identify $f$ with $g$.
However, what about $d=3$? I'm aware of the Sobolev embedding theorem which states that $$H^k(D)\hookrightarrow C^{\ell}(\bar D)\;\;\;\text{for all }\ell\in\left[0,k-\frac d2\right)\;,\tag 1$$ i.e. we need to choose $k\ge 2$.
Now, the statement of $(1)$ for $k=2$ and $\ell=0$ is that there is a bounded, linear and injective function $\iota:H^2(D)\to C^0(\bar D)$ where $H^2(D)$ is equipped with the Sobolev norm and $C^0(\bar D)$ is equipped with the supremum norm. Thus, each $f\in H^2(D)$ can be uniquely identified with $\iota(f)\in C^0(\bar D)$.
The question is: Is $\iota$ the inclusion map? That's the case in the situation with $d=1$ described above. But is it the case in the situation with $d=3$ too?
If not, I assume that the answer to the first highlighted question is "no".