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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.

Svetoslav
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Lucio D
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    The last part, in which you pass from $\lVert \nabla u \rVert_{L^\infty} \le \sigma$ to the Lipschitz continuity of $u$ with constant $\sigma$, is totally standard and OK (Modulo one fact. You should specify that the restriction of $u$ to any line is absolutely continuous, and so it is the integral of its own derivative. Saying only "u is differentiable almost everywhere" is not enough. A step function is differentiable almost everywhere and yet it is not the integral of its derivative). – Giuseppe Negro Mar 14 '14 at 17:22
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    I think the comment o @GiuseppeNegro resumed what I was trying to explain you Lucio. – Tomás Mar 14 '14 at 17:26
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    On the other hand, I am not convinced by your proof of the inequality $$\lVert \nabla u\rVert_{L^\infty} \le \liminf_{m\to \infty} \lVert \nabla u_m\rVert_{L^{\infty}}.$$ This is surely true but IMHO it is not justified enough – Giuseppe Negro Mar 14 '14 at 17:26
  • Firstly, I agree that I didn't justify $\Vert \nabla u \Vert_{L^{\infty}} \leq \liminf\limits_{m \rightarrow \infty} \Vert \nabla u_{m} \Vert_{L^{\infty}}$, I was trying to keep the proof concise, so I just stated that. Secondly, I thought that being differentiable almost everywhere was sufficient. Is the fact that $u$ to any line is absolutely continuous implied or do I have to add that as an extra condition? – Lucio D Mar 14 '14 at 17:40
  • Since Lipschitz continuity implies absolute continuity I would assume that is is implied? – Lucio D Mar 14 '14 at 17:57
  • @Tomas Okay, I thought you were saying that it is only possible to form the integral if you have a function that is differentiable everywhere. – Lucio D Mar 14 '14 at 18:03
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    @LucioD: Actually, I think that functions belonging to any Sobolev class satisfy this property. It is called APL-property of Sobolev spaces or something like that, I am no expert. Anyway, since here we are dealing with $W^{1, \infty}$, which contains Lipschitz continuous functions, things are easier as you rightfully point out. Therefore your proof is OK. – Giuseppe Negro Mar 14 '14 at 19:56
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    (P.S.: I was not notified of your previous comment. Next time, consider adding the @...: construct to your comments so that posters are notified of them) – Giuseppe Negro Mar 14 '14 at 19:57
  • @GiuseppeNegro Don't you think that the line $|u(x) - u(y)| \leq \int_{0}^{1}|u(tx+(1-t)y)||x-y|dt \leq \Vert \nabla u \Vert_{L^{\infty}}|x-y| \leq \sigma|x-y|$ from the OP should be actually $|u(x) - u(y)| \leq \int_{0}^{1}|\nabla u(tx+(1-t)y)|{l^2}|x-y|{l^2}dt \leq \sup\limits_{t\in[0,1]}{\Vert \nabla u (tx+(1-t)y) \Vert_{l^2}}|x-y|{l^2} \leq \sqrt{n}\sigma |x-y|{l^2}$ ? Thanks. – Svetoslav Mar 15 '16 at 17:13
  • @Svetoslav: Yes, surely. – Giuseppe Negro Mar 16 '16 at 10:18

1 Answers1

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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.

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user127096
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  • Thanks for your response. I just want to confirm that for $u \in W^{1,\infty}(I)$ I can use the fact that I can write $u$ as the integral of it's derivative since it is absolutely continuous everywhere(because it is Lipschitz continuous everywhere) but for an arbitrary $u \in W^{k,p}(I)$ this is not necessarily possible possible since it is only true for almost every line segment. – Lucio D Mar 16 '14 at 17:16
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    @LucioD That is correct. I'll add that some $W^{k,p}$ embed into $W^{1,\infty}$, namely those with $(k-1)p>n$. For those spaces you also have a Lipschitz representative. – user127096 Mar 16 '14 at 17:26