I'm trying to find a counter example to the following if there does indeed exist one. Let us consider $u\in W^{2,\infty}(\mathbb{R}^n)$ does it follow that $u \in C^{1,1}(\mathbb{R}^n)$. It is known to me that $W^{1,\infty}(\mathbb{R}^n) = C^{0,1}(\mathbb{R}^n)$ (well after modifying the elements in the Sobolev space on a null set) and that the norms are equivalent. Additionally, if there are no counter examples would it be possible to get a hint for proving why the Sobolev space lies in the Holder
The only progress I do have is that using Banach-Alaoglu one can show that $C^{k,1}(\mathbb{R}^n) \subset W^{k+1,\infty}(\mathbb{R}^n)$, but of course this is the wrong direction.
EDIT I think I have proved that indeed $C^{k,1}(\mathbb{R}^n) = W^{k+1,\infty}(\mathbb{R}^n)$. I'll post my solution in a couple of days for those of you interested.
Here is the proof I have, I'll do it for $k=2$ but arbitrary $k$ follows by induction and the inductive step follows in a very similar way. Fix $x\in \mathbb{R}^n$, let $u\in W^{2,\infty}(\mathbb{R}^n)$ and suppose WLOG that $Du(x)=0$, otherwise, consider $\tilde{u}(y) = u(y) +(x-y)Du(x)$. Then $Du \in W^{1,\infty}(\mathbb{R}^n) = C^{0,1}(\mathbb{R}^n)$ by Morrey's embedding theorem. Then due to the continuity of $Du$ we have that; $$\lim_{r\downarrow0}\|Du\|_{L^\infty(B_r(x))} = 0 $$.
Then by Morrey's Lemma (applied to $u$ which also in $W^{1,\infty}(\mathbb{R}^n)$ we have that $\forall y \in B_r(x)$;
$$|u(x)-u(y)| \leq |x-y| \|Du\|_{L^\infty(B_r(x))}.$$
Thus , $$\lim_{r\downarrow0} \sup_{y \in B_r(x)} \frac{|u(x)-u(y)|}{|x-y|} = 0$$
Hence $u$ is differentiable at $x$. But we can apply this to all $x\in \mathbb{R}^n$ and conclude that $u$ is differentiable everywhere and we already know that $Du$ is continuous and so $u\in C^1(\mathbb{R}^n)$ and hence we are done.
I've skipped a few details, if anyone would like something more specific please let me know!