Is there any problem with the following, please advise:
Take $I \subset \mathbb{R}^{n}$ convex, closed and bounded.
I want to show that if I have $u_{m} \rightharpoonup^{*} u$ in $W^{1,\infty}(I)$ and $\Vert \nabla u_{m} \Vert_{L^{\infty}(I)} \leq \sigma$ where $\sigma > 0$. It then follows that $u$ is Lipshitz continuous and has $\sigma$ as a Lipschitz constant.
We have that $\Vert \nabla u \Vert_{L^{\infty}} \leq \liminf\limits_{m\rightarrow \infty} \Vert \nabla u_{m} \Vert \leq \sigma$ since the norm is lower semi-continuous.
We first note that since $u \in W^{1,\infty}(I)$ it follows that $u$ is locally Lipschitz (since $I$ is compact it follows that $u$ is Lipschitz on $I$). So $u$ is differentiable almost everywhere in the classic sense of differentiation.
We now use the following:
Consider $\phi(t):= u(tx + (1-t)y)$ $\text{ }$ for $\text{ }$ $t \in [0,1]$
Then $\phi(1) – \phi(0) = \int_{0}^{1}{\phi}^{'}(t)dt = \int_{0}^{1}\nabla u(tx + (1-t)y)\cdot(x-y)dt$
$\therefore |u(x) – u(y)| \leq \int_{0}^{1}|u(tx+(1-t)y)||x-y|dt \leq \Vert \nabla u \Vert_{L^{\infty}}|x-y| \leq \sigma|x-y|$
This shows that $u$ is locally Lipschitz with Lipschitz constant $\sigma$.
Thanks for any assistance.
Best Answer
The proof is reasonable. When working with Sobolev spaces, one should keep in mind that the elements are equivalence classes of functions (agreeing except on a null set). However, in $W^{k,p}$ with $kp>n$ we have a canonical representative of each equivalence class - namely, a continuous function. So it is understood, often without saying, that this representative is being considered.
In general, every representative of a Sobolev function is absolutely continuous only on almost every line (ACL property, mentioned by Giuseppe Negro). This is something that must be considered even when we have continuity, i.e., when $kp>n$. A standard approach is to use the fundamental theorem of calculus on the line segments where we have absolute continuity, and then use the fact that such segments are dense. So, if you don't know already that $W^{1,\infty}\implies $ Lipschitz, this is the way to go. If this implication is known, then we have absolute continuity on every line automatically.
Related: