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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.

  • 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
  • @Tomás That's a good point but let me explain my reasoning. But first to address you first correction, I made a small typo which I corrected, it now reads $u(x) = u(a) + \int_{x}^{a}\nabla u(t)dt$. Secondly, $u \in W^{1,\infty}(I)$ so we have that there exists a continuous representative(I think it might be unique), such that we can extend $u$ on the a.e domain to that which is Lipschitz continuous everywhere while still considering $u \in W^{1,\infty}(I)$. What do you think? –  Aug 24 '14 at 17:42
  • And how would you do this extension? – Tomás Aug 24 '14 at 18:02
  • @Tomás Sorry that's $u(x) = u(a) + \int_{a}^{x}\nabla u(tx+(1-t)y)\cdot(x-y)dt$ as you stated. –  Aug 24 '14 at 18:02
  • @Tomás Uhm I'm not entirely sure about the details of a proof at the moment but consider the following MSE post and then consider the result in Brezis book 'Functional analysis, Sobolev Spaces and Partial Differential Equations'. Theorem 8.2 page 204 which states: Let $u \in W^{1,p}$ with $1 \leq p \leq \infty$ and $I$ bounded ot unbounded then there exists a function $\bar{u} \in C(\bar{I})$ such that $u = \bar{u}$ a.e. on $I$. Do you think considering $u$ as a Lipschitz representative is a valid conclusion? –  Aug 24 '14 at 18:13
  • @Tomás If you are not in possession of this Brezis text I would be happy to email it to you or send you the extract. –  Aug 24 '14 at 18:22
  • The representative will be Lipschitz, but the question is, how do you prove it? You argument only applies a.e.. I fail to see how would you extend it to all $I$. – Tomás Aug 24 '14 at 18:26
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    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
  • @Tomás Yes I see your point. Well we know it's true, since we know that $u$ is Lipschitz continuous on $I$(from the other post). So I guess I have to prove it then that $u$ is absolutely continuous on a.e. segment. Given that, how exactly would you prove that you can extend the Lipschitz continuity as suggested in the other post? –  Aug 24 '14 at 20:30
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    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
  • @Tomás Yes, I assumed that would be the definition of the extension, I will try to prove the Lipschitz continuity. Do you have any idea on how to prove that if $u \in W^{1,\infty}(I)$ as above, then $u$ is absolutely continuous on a.e line segment so that I can write $u(x) = u(a) = \int_{a}^{x}\nabla u(tx + (1-t)y)\cdot (x-y)dt$? –  Aug 26 '14 at 11:27
  • Look here: http://math.stackexchange.com/questions/908175/is-a-sobolev-function-absolutely-continuous-with-respect-to-a-e-segment-of-line – Tomás Aug 26 '14 at 12:37
  • @Tomás Okay thanks, very helpful. I'm wondering why the result of absolute continuity for Sobolev space functions are commonly presented by stating that it is restricted to almost every line parallel to the coordinate directions in $\mathbb{R}^{n}$, when it actually holds for every line segment. –  Aug 26 '14 at 16:05
  • @Tomás Hi Tomas, your first commented states "FIrst, it is $$u(x) = u(a) + \int_{a}^{x}\nabla u(tx+(1-t)a)(x-a)$$..." upon revision it seems to me that it should be $$u(x) = u(a) + \int_{0}^{1}\nabla u(tx+(1-t)a)\cdot (x-a)dt$$ What do you think? –  Aug 27 '14 at 12:36
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    Yes, you are right. – Tomás Aug 27 '14 at 12:47

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