Proposed proof of analysis 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.

• Omit line "We can write ... every line)." – Marcin Malogrosz Aug 23 '14 at 20:23
• There are some point which I want to discuss. First, it is $$u(x)=u(a)+\int_a^x \nabla u(tx+(1-t)a)(x-a)$$ and this equality is not true for all $x,a\in I$, because you are concluding it, by using the fact that a Sobolev function is absolutely continuous on "a.e." line. So in fact, you have deduced that $$|u(x)-u(y)|\le \alpha |x-y|,\ a.e.\ x,y\in I.$$ How do you conclude from here that the inequality hold for all $x,y\in I$? – Tomás Aug 24 '14 at 17:13
• Look, there is one thing here, which you are using, however, you have not proved it. We know that $u$ is absolutely continuous on a.e. segment of line paralell to the coordinate axis. You are using more thant it, you are saying that $u$ is absolutely continuous on a.e. segment of line. Is this true? If this is true then, the same argument given by 40 votes, in the anither answer will work here, because the set where $u$ will not be Lipschitz, is a set of zero measure. Then you extend it by continuity as he suggested. – Tomás Aug 24 '14 at 19:43
• If you can prove that $u$ is Lipschitz on a set $E$, such that it's complement has zero measure, then you can extend it by continuity. Indeed take $x\in I\setminus E$. If the is no sequence $x_n\in I$ with $x_n\to x$ then, it would exist a set with postive measure in $I\setminus E$, which is a absurd, therefore $E$ is dense in $I$. Now, for $x\in I\setminus E$, let $u(x)=\lim u(x_n)$. You have to prove that this extension is Lipschitz. – Tomás Aug 24 '14 at 21:26
• Yes, you are right. – Tomás Aug 27 '14 at 12:47