Suppose that $B_1$ is the open unit ball in $\mathbb R^n$, denote $W^{2,\infty}(B_1)$ be the sobolev spaces and $C^{1,1}(\bar B_1)$ is the Holder spaces. It seems the equality $W^{2,\infty}(B_1)=C^{1,1}(\bar B_1)$ holds at first glance, but how to move from intuition to a strictly argument?
What's more, I don't know why closed $\bar B_1$ rather than $B_1$?
Another thing: is this some kind of problem suitable for research or just a well known results?
For first-order space, see relation between $W^{1,\infty}$ and $C^{0,1}$ and Sobolev embedding for $W^{1,\infty}$？
The result for higher orders follows from $W^{1,\infty}=C^{0,1}$ directly - one no longer needs any consideration of the geometry of domains. That is, if for some domain $U$ we have $W^{1,\infty}(U)=C^{0,1}(U)$, then $W^{k,\infty}(U)=C^{k-1,1}(U)$ for all positive integers $k$.
Sketch: take $u\in W^{2,\infty}$. The weak gradient $Du$ is in $W^{1,\infty}$, hence in $C^{0,1}$. When the weak gradient is given by a continuous function, it is the classical gradient. (Because the antiderivative of a continuous function is differentiable.) Hence, $u\in C^{1,1}$.