TITLE: ACL characterization of functions in $W^{1,\infty}(\Omega)$
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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.