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I have a Sobolev space related question. In the book 'Measure theory and fine properties of functions' by Lawrence Evans and Ronald Gariepy. I know the result that states that for $f: \Omega \rightarrow \mathbb{R}$. $f$ is locally Lipschitz in $\Omega$ if and only if $f \in W^{1,\infty}_{loc}(\Omega)$.

I recently saw that it was stated in a book 'Nonlinear partial differential equations with applications' by Roubicek, that $W^{1,\infty}(\Omega) = C^{0,1}(\Omega)$.

Can anyone confirm this stronger result and maybe recommend a book or text which provides a proof? Thanks a lot for any help.

Mike_MasterinMath
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  • It's not stronger, see http://en.wikipedia.org/wiki/H%C3%B6lder_condition. You may not be used to the notation. – Jeb Jul 22 '14 at 17:39
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    @Jeb It is stronger; see my answer. –  Jul 22 '14 at 18:44

2 Answers2

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"Locally" is ambiguous here

$f$ is locally Lipschitz in $\Omega$ if and only if $f \in W^{1,\infty}_{loc}(\Omega)$

The validity of this claim depends on interpretation of "locally Lipschitz". Does it mean

  1. every point of $\Omega$ has a neighborhood in which $f$ is Lipschitz, or
  2. there is $L$ such that every point of $\Omega$ has a neighborhood in which $f$ is $L$-Lipschitz (i.e., satisfies $|f(x)-f(y)|\le L|x-y|$).

With interpretation 1) the above would be false, because $f(x)=1/x$ is locally Lipschitz on the interval $(0,1)$. The authors meant interpretation 2, but you should be aware that it may be less common. Saying "locally $L$-Lipschitz" would be more precise.

Counterexample and quasiconvexity

$W^{1,\infty}(\Omega) = C^{0,1}(\Omega)$

This is not always true. For example, let $\Omega$ be the plane with the slit along negative $x$-axis. Using polar coordinates $r,\theta$, with $-\pi<\theta<\pi$, define $u(r,\theta) = r\theta$. You can check that this is a $W^{1,\infty}$ function (it is locally $10$-Lipschitz, say), but it is not in $C^{0,1}(\Omega)$ because the values of $f$ just above the slit and just below it are far apart.

The counterexample is taken from here, where you can also find a result in the positive direction; for sufficiently nice $\Omega$ the equality holds. Also, see the answer and references in relation between $W^{1,\infty}$ and $C^{0,1}$.

Vanishing on the boundary

The definition of $W^{1,\infty}_0(\Omega)$ is a bit tricky since the usual approach (complete the space of smooth compactly supported functions with respect to the Sobolev norm) does not apply. Instead one can define $W_0^{1,\infty}(\Omega)$ as follows. Every element of $W^{1,\infty}(\Omega)$ has a continuous representative $u$. If $u$ satisfies $\lim_{x\to a}u(x)= 0$ for every $a\in \partial \Omega$, then we say that $u\in W_0^{1,\infty}(\Omega)$. This is a closed subspace, because convergence in $W^{1,\infty}$ norm implies uniform convergence.

As an aside: one can give a unified definition of $W^{1,p}_0(\Omega)$ that works for all $1\le p\le \infty$: a function is in $W^{1,p}_0(\Omega)$ if its zero extension to $\mathbb R^n$ is in $W^{1,p}(\mathbb R^n)$.

For every domain, $W_0^{1,\infty}(\Omega)$ is the same as the set of Lipschitz functions on $\Omega$ that tend to $0$ at the boundary. Indeed, we can extend by zero to the rest of $\mathbb R^n$ and use the fact that $\mathbb R^n$ is convex.

Community
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  • From what I read it seems that the equality holds for a quasi-convex domain. Would I be right in saying that for quasi-convex domain $\Omega$, $f \in W^{1,\infty}{0}(\Omega)$ is a Lipschitz function with the property that the unique extension, say $\overline{f}$, to $\overline{\Omega}$ results in $\overline{f}|{\partial \Omega} = 0$ ? –  Jul 23 '14 at 19:27
  • @JohnJack That's true, but with subscript zero you don't need quasiconvexity. See the edit. –  Jul 23 '14 at 19:37
  • By "$W^{1,\infty}_{0}(\Omega)$ is the same as the set of Lipschitz functions on $\Omega$ that tend to 0"...are you referring to the unique extension of uniformly continuous functions to the boundary? –  Jul 23 '14 at 19:52
  • @JohnJack Basically yes, though I do not always think in these terms. I say that $f$ tends to $0$ on the boundary if for every boundary point $b$ we have $f(x)\to 0$ as $x\to b$ (with $x$ in the domain). This is equivalent to saying that if $f$ is defined to be $0$ on $\partial \Omega$, the extended function is continuous. –  Jul 23 '14 at 21:31
  • Your idea seems like a stronger idea than the extension of uniformly continuous functions, since there the idea is that for any boundary point $b$ there is a convergent sequence $x_{n} \rightarrow b$ which implies $f(x_{n}) \rightarrow f(b)$ by uniform continuity. What do you think? –  Jul 23 '14 at 21:50
  • Look at how the extension is described here rexel28.wordpress.com/2010/11/03/extending-uniformly-continuous-functions/ and compare with what you stated. I'm saying that your statement is stronger, that's it. –  Jul 23 '14 at 22:00
  • @Thisismuchhealthier. See MSE post "Characterization of Sobolev space" by Moses. The discussion seems to disagree with your definition of $W^{1,\infty}{0}$ as being simply the closure of $C^{\infty}{0}$ in $W^{1,\infty}$. –  Jul 24 '14 at 20:52
  • @JohnDoe Thanks for catching that. Edited. –  Jul 25 '14 at 03:40
  • @900sit-upsaday There is a Theorem which states that there exists a continuous representation of $u \in W^{1,p}(\Omega)$ for $1 \leq p \leq \infty$ which is some function $\bar{u} \in C(\overline{\Omega})$ for bounded or unbounded $\Omega$. So you could simply state that $u \in W^{1,\infty}{0}(\Omega)$ if $\bar{u}|{\partial \Omega} = 0$. I'm not following your reasoning regarding why $W^{1,\infty}_{0}$ is closed, could you expand on that? –  Jul 25 '14 at 13:52
  • @JohnDoe That "theorem" is false. Having a continuous representative requires $p>n$. For continuity up to the boundary some geometric assumptions on the domain are also needed. As for my reasoning: uniform convergence commutes with taking limits. –  Jul 25 '14 at 14:06
  • @900sit-upsaday Yes actually that would be consistent with the proof I am referring to(Brezis functional analysis book). He assumes that $\Omega$ is some open interval of $\mathbb{R}$. The theorem is proved for $n=1$ only. Do you have a reference for a proof of the existence of a continuous representative for Sobolev spaces with $p>n$? –  Jul 25 '14 at 21:41
  • @JohnDoe Sobolev-Morrey inequality. It's in Chapter 9 of Brezis, theorem 9.12. –  Jul 25 '14 at 21:43
  • @900sit-upsaday Okay is the following consistent with what you are saying: Take $f \in \overline{W^{1,\infty}{0}(\Omega)}$(consider $f$ as the continuous respresentative) then there exists a sequence ${f{j}}{j} \subset W^{1,\infty}{0}(\Omega)$ such that $f_{j} \rightarrow f$ in $W^{1,\infty}$ norm which implies $$f_{j} \rightarrow f \text{ } \text{ uniformly}$$ By assumption, for any $a \in \partial \Omega$ we have that $\lim\limits_{x \rightarrow a}f_{j}(x) \rightarrow 0$ for all $j$. –  Jul 27 '14 at 13:20
  • @900sit-upsaday So by commutativity of uniform convergence with limits: $$0 = \lim_{j \rightarrow \infty}(\lim_{x \rightarrow b}f_{j}(x)) = \lim_{x \rightarrow b}(\lim_{j \rightarrow \infty}f_{j}(x)) = \lim\limits_{x \rightarrow b}f(x)$$ $\therefore f \in W^{1,\infty}_{0}(\Omega)$. –  Jul 27 '14 at 13:21
  • In the previous comments I am trying to prove the result that $W^{1,\infty}_{0}(\Omega)$ is closed in $W^{1,\infty}(\Omega)$. –  Jul 27 '14 at 14:18
  • @900sit-upsaday I made a small typo. My $a \in \partial \Omega$ becomes a $b$ in my previous comments. –  Jul 31 '14 at 20:02
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    I don't think that interpretation 1) of "locally" causes a problem for "local Lipschitz implies $W^{1,\infty}0(\Omega)$. I think your counterexample $f(x)=1/x$ is contained in $W^{1,\infty}{loc}((0,1))$ - any $x\in (0,1)$ has a neighbourhood where $1/x$ and $-1/x^2$ are bounded. – 1Rock Jul 01 '21 at 01:33
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For my future reference!

I second 1Rock's comment to the accepted answer. Interpretation 1 of locally Lip seems to be ok and the right one. What does interpretation 2) mean anyway? What does L come from as the function is only locally in $W^{1,\infty}$ and the bound changes from open subdomain to another.

Here is the subtle thing per my research.

Theorem: A function $u\in L^1_{loc}(\mathbb{R}^n)$ has a representative that is bounded and Lipschitz continuous if and only if $u \in W^{1,\infty}(\mathbb{R}^n)$.

(Theorem 2.23 in notes by Juha Kinnunen, Aalto University)

Note the global nature of the claims.

Now what if $U$ is a proper open subset of $\mathbb{R}^n$? As the example in the other answers show, even if $u \in W^{1,\infty}(U)$ we only can get that any point has a (bounded) neighborhood on which $u$ has a representative that is Lipschitz. Let me sketch the idea of proof of this direction as it is important to my discussion.

If one follows the proofs, for example in case of whole $\mathbb{R}^n$, the idea behind showing Lipschitz is to bound $|u_\epsilon(x)-u_\epsilon(y)|$ by the integral of the derivative of the mollifications $u_\epsilon$ of $u$ along a curve from $x$ to $y$ and use the universal bounds on the derivative. (We will ultimately let $\epsilon \to 0$.)

In case of convex domain we simply take the line segment, whose length happens to be ideally equal to $|x-y|$.

In case of quasiconvex domain we are still good since we can find curves with length less than a universal multiple of $|x-y|$. The counter example works because quasiconvexity fails. There is no special thing about the counterexample, just take any domain where intrinsically far away parts come very close in the ambient space, and then have $u$ attain very different values at these portions.

Now, it must be clear that within every ball contained in $U$ the argument will go through and $u$ will have a Lip representative. We will get the same conclusion for $W^{1,\infty}_{loc}(U)$ as for $W^{1,\infty}(U)$ because we only need the L-infinity bound on derivative on a slightly larger ball.

So, this wraps up the proof that $W^{1,\infty}_{loc}(U)$ has Lip representatives on small neighborhoods.

For the reverse implication, observe that if $u$ has a Lipschitz representative on a bounded open subdomain $V$, then $u$ is bounded and hence in $W^{1,\infty}(V)$.

So, Evans and Garipey's statement is correct after all (with interpretation 1 in the other answer.) To be super clear:

Theorem: Let $U$ be open in $\mathbb{R}^n$, and $f \in L^1_{loc}(U)$. Then the following are equivalent:

  1. Every point in $U$ has a bounded neighborhood on which $f$ has a representative that is bounded and Lipschitz,
  2. On every subdomain compactly contained in $U$, $f$ has a representative that is bounded and Lipschitz,
  3. For every subdomain $V$ compactly contained in $U$, $f \in W^{1,\infty}(V)$,
  4. $f \in W^{1,\infty}_{loc}(U)$.

In 1 we can just say a ball centerd at that point. Equivalence of 1 and 2 follows from topology! Items 3 and 4 are equivalent just by definition.

There is one question remaining: We saw that $u \in W^{1,\infty}(U)$ is not better that $u \in W^{1,\infty}_{loc}(U)$ as both only guarantee local Lipschitz (in sense of theorem above). But how about the reverse direction: what if $u$ has a representative that is bounded (necessary for unbounded domains) and Lipschitz on (all of) $U$ -- does it follows that $u \in W^{1,\infty}(U)$? Or do we only get $u \in W^{1,\infty}_{loc}(U)$?

Behnam Esmayli
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