ACL characterization of functions in $W^{1,\infty}(\Omega)$

Question

I am reading "A First Course in Sobolev Space" by Leoni. Theorem 10.35 states that for $1\leq p<\infty$, a function $u\in L^p(\Omega)$ belongs to $W^{1,p}(\Omega)$ if and only it has a representative $\bar{u}$ that is absolutely continuous on $a.e.$ line segments of $\Omega$ that are parallel to the coordinate axes. The proof in the book is not valid for $p=\infty$. However, according to Wikipedia, this theorem also holds for $p=\infty$. Can anyone provide a proof?

Answer 1

First, what you wrote is not quite correct. $u \in W^{1,p}(\Omega)$ if and only if there is a $\bar{u} \in L^{1}(\Omega)$ such that, for any $i \in \{1,2,\dots,d\}$, for a.e. $s \in \mathbb{R}^{d-1}$, if $\Omega_{i,s} = \{x \in \Omega \, \mid \, (x_{1},\dots,\hat{x}_{i},\dots,x_{d}) = s\}$, then $\bar{u}$ restricts to an absolutely continuous function $\bar{u}_{i,s}$ in $\Omega_{i,s}$ and \begin{equation*} \int_{\mathbb{R}^{d-1}} \left( \|\bar{u}_{i,s}\|_{L^{p}(\Omega_{i,s})}^{p} + \|\bar{u}_{i,s}'\|_{L^{p}(\Omega_{i,s})}^{p} \right) \, ds < \infty. \end{equation*} My point is you need to know that the $W^{1,p}$ norms on the line segments are summable. (One direction should be plausible if you think about Fubini and what $u_{i,s}'$ is liable to be.)

$p = \infty$ means Lipschitz in balls: The rule of thumb is "$\|Du\| \leq M$ in a weak sense" means that $|u(x) - u(y)| \leq M\|x -y\|$ (at least locally). I put "a weak sense" in quotations because this is true both for distributional derivatives and viscosity theory --- quotations because it should "morally" be true whenever you are working with some generalized notion of derivative.

To see this, you only need to use convolution. Suppose that $u \in W^{1,\infty}(\Omega)$ for some $\Omega \subseteq \mathbb{R}^{d}$ open. Fix a ball $B \subseteq \Omega$. I will show that \begin{equation*} |u(x) - u(y)| \leq \|Du\|_{L^{\infty}(\Omega)} \|x - y\| \quad \text{if} \, \, x,y \in B. \end{equation*}
We could replace $B$ by $\Omega$ if $\Omega$ were convex, but it may not be true otherwise. This is about geometry, though, it has nothing much to do with $u$ --- the same way the Mobius strip looks great locally, but isn't orientable. (Also, it would still be an issue if $u$ were smooth.)

If we define the mollification* $u^{\epsilon} = u * \rho^{\epsilon}$ (where $*$ is convolution and $\rho^{\epsilon}$ is defined in the usual way...), then remember $u^{\epsilon}$ is smooth and, for all $\epsilon > 0$ small enough, \begin{equation*} Du^{\epsilon} = Du * \rho^{\epsilon} \quad \text{in} \, \, B. \end{equation*} Notice this gives \begin{equation*} \|Du^{\epsilon}\|_{L^{\infty}(B)} = \|Du\|_{L^{\infty}(\Omega)} \|\rho^{\epsilon}\|_{L^{1}(B)} \leq \|Du\|_{L^{\infty}(\Omega)}. \end{equation*} Things are smooth and $B$ is convex so a $L^{\infty}$ bound on the derivative gives a Lipschitz bound. Indeed, \begin{equation*} |u^{\epsilon}(x) - u^{\epsilon}(y)| = \left| \int_{0}^{1} Du(x + t (y - x)) \cdot (y - x) \, dt \right| \leq \|Du\|_{L^{\infty}(\Omega)} \|y - x\|. \end{equation*} At the same time, since $\|u^{\epsilon}\|_{L^{\infty}(\mathbb{R}^{d})} \leq \|u\|_{L^{\infty}(\Omega)}$ independently of $\epsilon$ and $u^{\epsilon} \to u$ almost everywhere in $\Omega$, we can invoke the Arzela-Ascoli Theorem to get a $\bar{u} : B \to \mathbb{R}$ such that $u^{\epsilon} \to \bar{u}$ uniformly in $B$ and \begin{equation*} |\bar{u}(x) - \bar{u}(y)| \leq \|Du\|_{L^{\infty}(\Omega)} \|x - y\| \quad \text{if} \, \, x, y \in B. \end{equation*} Notice that $u = \bar{u}$ a.e. in $B$.

By covering $\Omega$ with countably many balls, one deduces that, in fact, we can extend $u$ to a continuous function $\bar{u} : \Omega \to \mathbb{R}$ such that $u = \bar{u}$ a.e.

Lipschitz in balls means Lipschitz on line segments: This is where the $p = \infty$ case is relatively easy. Suppose that $B \subseteq \Omega$ is a ball as before. If $A$ is any subset of $B$, then \begin{equation*} |\bar{u}(x) - \bar{u}(y)| \leq \|Du\|_{L^{\infty}(\Omega)} \|x - y\| \quad \text{if} \, \, x,y \in A \end{equation*} since it is true for $x,y \in B$. That is, $\bar{u}$ restricts to a uniformly Lipschitz function in $A$ with Lipschitz constant $\|Du\|_{L^{\infty}(\Omega)}$. This is true, in particular, if $A$ is a line segment contained in $B$.

In general, if $I \subseteq \Omega$ is a line segment, then we can find an $N \in \mathbb{N}$ and points $x_{0},x_{1},\dots,x_{N} \in N$ such that $I = \{(1 - t) x_{0} + t x_{N} \, \mid \, t \in [0,1]\}$ and the sub-segment $I_{i} = \{(1 - t)x_{i - 1} + t x_{i} \, \mid \, t \in [0,1]\}$ is contained in some ball contained in $\Omega$ for any $i \in \{1,2,\dots,N\}$. Since $I_{i}$ is contained in a ball, what we just proved gives \begin{equation*} |\bar{u}(x) - \bar{u}(y)| \leq \|Du\|_{L^{\infty}(\Omega)} \|x_{i} - x_{i - 1}\| \quad \text{for each} \, \, x \in I_{i}. \end{equation*} Therefore, by the triangle inequality, \begin{equation*} |\bar{u}(x) - \bar{u}(y)| \leq \|Du\|_{L^{\infty}(\Omega)} \|x - y\| \quad \text{if} \, \, x,y \in I. \end{equation*} (Given $x,y \in I$, we can assume $x \in I_{j}$ and $y \in I_{k}$ for some $j < k$ and then \begin{equation*} |\bar{u}(x) - \bar{u}(y)| \leq |\bar{u}(x) - \bar{u}(x_{j})| + \sum_{i = j + 1}^{k - 1} |\bar{u}(x_{i - 1}) - \bar{u}(x_{i})| + |\bar{u}(x_{k}) - \bar{u}(y)| \leq M \|x - y\|.) \end{equation*} Thus, $\bar{u}$ is a Lipschitz function in $I$, hence it is absolutely continuous in $I$, and \begin{equation*} \|\bar{u}_{I}'\|_{L^{\infty}(I)} \leq M. \end{equation*}

Notice we did not need to assume that $I$ was parallel to a coordinate axis, and we get all the line segments at once. But this is to lose sight of the fact that $\bar{u}$ is itself a very nice (i.e. Lipschitz) function away from $\partial \Omega$ so, of course, it is nice on line segments.

Lipschitz on line segments (with a bound) implies Lipschitz): What I will prove is that if $u$ has a representative $\bar{u} \in L^{\infty}(\Omega)$ such that $\bar{u}_{i,s}$ is Lipschitz on $\Omega_{i,s}$ (as defined above) with $\|\bar{u}_{i,s}\|_{L^{\infty}(\Omega_{i,s})} \leq M$ for a.e. $s \in \mathbb{R}$ for some $M > 0$ (independent of $s$), then $u \in W^{1,\infty}(\Omega)$ and $\|Du\|_{L^{\infty}(\Omega)} \leq \sqrt{d} M$.

To see this, observe that if $\varphi \in C^{\infty}_{c}(\Omega)$ and $i \in \{1,2,\dots,d\}$, then, by Fubini's Theorem, \begin{align*} \left| \int_{\Omega} \varphi_{x_{i}}(x) u(x) \, dx \right| &= \left| \int_{\mathbb{R}^{d-1}} \left(\int_{\Omega_{i,s}} \varphi_{i,s}'(r) \bar{u}_{i,s}(r) \, dr \right) \, ds \right| \\ &= \left| \int_{\mathbb{R}^{d-1}} \left( \int_{\Omega_{i,s}} \varphi_{i,s}(r) \bar{u}_{i,s}'(r) \, dr \right) \, ds \right| \\ &\leq M \int_{\mathbb{R}^{d-1}} \int_{\Omega_{i,s}} \varphi_{i,s}(r) \, dr \, ds \\ &= M \int_{\Omega} \varphi(x) \, dx. \end{align*} It follows (by the Riesz Representation Theorem for the pairing $L^{1}(\Omega)$-$L^{\infty}(\Omega)$ that there is a $v_{i} \in L^{\infty}(\Omega)$ such that \begin{equation*} \int_{\Omega} \varphi_{x_{i}}(x) u(x) \, dx = \int_{\Omega} \varphi(x) v(x) \, dx. \end{equation*} Hence $u \in W^{1,\infty}(\Omega)$ and $Du = (v_{1},\dots,v_{d})$. Furthermore, $\|Du\|_{L^{\infty}(\Omega)} \leq \sqrt{d} M$ provided we use the norm \begin{equation*} \|(\psi_{1},\dots,\psi_{d})\|_{L^{\infty}(\Omega)} = \text{ess-sup}_{\Omega} \, \, \left( \sum_{i = 1}^{d} |\psi_{i}|^{2} \right)^{\frac{1}{2}}. \end{equation*}

It is worth noting that this last part works just as well if $p \in (1,\infty)$ instead. (I suppose slightly more work is needed if $p = 1$...)

Last note: Other references on this topic are Adams's book Sobolev Spaces and Evans and Gariepy, Measure Theory and Fine Properties of Functions are two options. The latter answers your question.

*Let me know if you do not know what I meant by mollification. Arguably, it should be learned before Sobolev spaces are introduced.

Answer 2

It is shown in A. Kufner, O. John, S. Fucik, Function Spaces, Springer, 1977. It is stated as Theorem 5.6.5, but the proof is essentially contained in 5.6.2, 5.6.3.

The idea is to show that the directional distributional derivative and the directional a.e. derivative coincide for functions in $L^1_{\rm loc}$ and with any of the above derivatives in $L^1_{\rm loc}$.