Hi please advise on my proof of the following result:
Assume that $I \subset \mathbb{R}^{n}$ is convex, bounded open set with Lipschitz boundary and let $u_{m},u$ be such that $$u_{m} \rightharpoonup^{*} u \text{ in } W^{1,\infty}(I)$$ and $$\Vert \nabla u_{m} \Vert_{L^{\infty}(I)} \leq \alpha$$
for some fixed $\alpha > 0$.
Show that $u_{m}$ and $u$ are Lipschitz continuous on $I$ with Lipschitz constant $\alpha$.
Proposed Proof:
I will first do the proof for the case of $u$.
Note that since $u_{m} \rightharpoonup^{*} u$ in $W^{1,\infty}(I)$ it follows that $\nabla u_{m} \rightharpoonup^{*} \nabla u$ in $L^{\infty}(I)$, therefore $\Vert \nabla u \Vert_{L^{\infty}(I)} \leq \liminf\limits_{m \rightarrow \infty}\Vert \nabla u_{m} \Vert_{L^{\infty}(I)} \leq \alpha$.
We can write $u(x) = u(a) + \int_{a}^{x}\nabla u(tx+(1-t)y)\cdot(x-y) dt$(Since $u$ is a Sobolev function, it is absolutely continuous on almost every line). We now use the following argument: Consider $$\psi(t) := u(tx + (1-t)y) ~~~ \text{ for } t \in [0,1]$$ Then $$\psi(1) - \psi(0) = \int_{0}^{1}\psi^{'}(t)dt = \int_{0}^{1}\nabla u(tx+(1-t)y)\cdot (x-y)dt$$ therefore $$|u(x)-u(y)| \leq \int_{0}^{1}|\nabla u(tx+(1-t)y)||x-y|dt \leq \Vert \nabla u \Vert_{L^{\infty}}|x-y| \leq \alpha |x-y|$$ This shows $u$ is Lipschitz on $I$ with Lipschitz constant $\alpha$.
The same process can be carried out for the case involving $u_{m}$ by using the assumed condition $\Vert u_{m} \Vert_{L^{\infty}} \leq \alpha$.
$\square$
Let me know if this proof is fine? Thanks for any assistance, also let me know if something is unclear.