Twice differentiable sobolev function Suppose $u \in W^{2,\infty}(0,1)$. Therefore, $u$ has a bounded Lipschitz representative $g$, with $u'$ being represented by the classical $g'$. Analogously, $u'$ is represented by the bounded Lipschitz $ h$, with $u''$ being represented by $h'$. Can one conclude that $g \in C^{1,1}$ and $g' = h$? How to do so?
We can say that $g'=h$ almost everywhere for sure, but one here needs to show then that $g'$ is continuous. One way would be to define $g'=h$ outside of where the Rademacher theorem gives existence of $g'$. And now one has to prove that $h$ is the everywhere derivative of $g$.
What can one do here?
Also, how to extend this to more dimensions?
 A: We start with a general observation:
Let $f \in W^{1, \infty}(0,1)$. Then we know that there exists $f' \in L^\infty(0, 1)$ such that
$$\langle f, \varphi' \rangle = -\langle f', \varphi \rangle \qquad (\varphi \in C_c^\infty(0, 1)). $$
In particular, $f' \in L^1(0, 1)$. Hence, $g : (0, 1) \to \mathbb C, \, g(x) = \int_0^x f'(t) \, \mathrm dt$ is well defined.
But we can say even more, namely that $g \in C(0, 1)$. To see this, let $x \in (0, 1)$ and $(x_n)_{n \in \mathbb N}$ be a sequence in $(0, 1)$ such that $x_n \to x$ as $n \to \infty$. Then
$$g(x_n) - g(x) = \int_x^{x_n} f'(t) \, \mathrm dt = \int_{(0, 1)} \mathbf 1_{(x, x_n)}(t) f'(t) \, \mathrm dt \longrightarrow 0$$
by the dominated convergence theorem since $\mathbf 1_{(x, x_n)}f' \in L^1(0, 1)$ and  $\mathbf 1_{(x, x_n)}f' \to 0$ pointwise. Thus, $ g \in C(0, 1)$. Moreover,
\begin{align*}
\langle g, \varphi' \rangle &= \int_0^1 g(t) \varphi'(t) \, \mathrm d t = \int_0^1 \int_0^t f'(s) \varphi'(t) \, \mathrm d s \, \mathrm d t \\
&= \int_0^1 \int_s^1 f'(s) \varphi'(t) \, \mathrm d t \, \mathrm d s = \int_0^1 f'(s) \Bigl[ \varphi(t) \Bigr]^{t = 1}_{t = s} \, \mathrm d s \\
&= - \int_0^1 f'(s) \varphi(s) \, \mathrm d s = -\langle f', \varphi \rangle = \langle f, \varphi' \rangle
\end{align*}
for all $\varphi \in C_c^\infty(0, 1)$ by Fubini's theorem. Thus $g = f$ almost everywhere.
Now back to your question:
Let $u \in W^{2, \infty}(0,1)$. Then $u'' \in L^\infty(0, 1)$ and if we apply the observation above twice, we observe $u' \in C(0,1)$ and $u \in C(0,1)$. By the fundamental theorem of calculus, we get $u \in C^1(0,1)$ since $u(x) = \int_0^x u'(t) \, \mathrm d t$. Finally, $u \in C^{1,1}(0,1)$ since
$$\lvert u(x) - u(y) \rvert \leq \sup_{x \in (0, 1)} \lvert u'(x) \rvert \cdot \lvert x - y \rvert \leq \lVert u' \rVert_\infty \cdot \lvert x - y \rvert \qquad (x, y \in (0, 1))$$
by the mean value theorem.
The situation in more dimensions is very subtle. If $\Omega \subseteq \mathbb R^n$ is open, then the question about the Sobolev embeddings into the (Hölder-)continuous functions relies on the regularity of $\partial \Omega$ and there are many different results in that direction (see, e.g. Sobolev inequality).
