\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb,mathrsfs} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 29, pp. 1--19.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/29\hfil Variational characterisation of $\infty$-harmonic maps] {A characterisation of $\infty$-harmonic and $p$-harmonic maps via affine variations in $L^\infty$} \author[N. Katzourakis \hfil EJDE-2017/29\hfilneg] {Nikos Katzourakis} \address{Nikos Katzourakis \newline Department of Mathematics and Statistics, University of Reading, Whiteknights, PO Box 220, Reading RG6 6AX, Berkshire, UK} \email{n.katzourakis@reading.ac.uk} \dedicatory{Communicated by Peter Bates} \thanks{Submitted August 10, 2016. Published January 26, 2017} \subjclass[2010]{35D99, 35D40, 35J47, 35J47, 35J92, 35J70, 35J99} \keywords{$\infty$-Laplacian; $p$-Laplacian; generalised solutions; viscosity solutions; \hfill\break\indent Calculus of Variations in $L^\infty$; Young measures; fully nonlinear systems} \begin{abstract} Let $u: \Omega \subseteq \mathbb{R}^n \to \mathbb{R}^N$ be a smooth map and $n,N \in \mathbb{N}$. The $\infty$-Laplacian is the PDE system \[ \Delta_\infty u :=\Big(Du \otimes Du + |Du|^2[Du]^\bot \otimes I\Big) :D^2u = 0, \] where $[Du]^\bot := \operatorname{Proj}_{R(Du)^\bot}$. This system constitutes the fundamental equation of vectorial Calculus of Variations in $L^\infty$, associated with the model functional \[ E_\infty (u,\Omega')= \big\| |Du|^2\big\|_{L^\infty(\Omega')} ,\quad \Omega' \Subset \Omega. \] We show that generalised solutions to the system can be characterised in terms of the functional via a set of designated affine variations. For the scalar case $N=1$, we utilise the theory of viscosity solutions by Crandall-Ishii-Lions. For the vectorial case $N\geq 2$, we utilise the recently proposed by the author theory of $\mathcal{D}$-solutions. Moreover, we extend the result described above to the $p$-Laplacian, $1
0$, $u$ satisfies \[ \|Du\|_{L^\infty(\Omega')} \leq \big\|Du+\lambda DA\big\|_{L^\infty(\Omega')}. \] Then, we have \[ \max_{z\in \overline{\Omega'}} \big\{Du(z) : DA \big\} \geq 0. \] (b) Given $x\in \Omega$ and $0<\varepsilon <\operatorname{dist}(x,\partial\Omega)$, the set \[ \Omega_\varepsilon (x):= \big\{ y\in \Omega \big| |Du(y)|<|Du(x)|\big\} \cap \mathbb{B}_\varepsilon(x) \] is open and compactly contained in $\Omega$ and also $x\in \big(\Omega_\varepsilon(x)\big)(u)$, that is \[ |Du(x)|= \|Du\|_{L^\infty( \Omega_\varepsilon (x) )}. \] \end{lemma} \begin{proof} (a) By assumption we have \[ \|Du\|^2_{L^\infty(\Omega')} \leq \|Du+\lambda DA\|^2_{L^\infty(\Omega')} \] and hence \begin{align*} \operatorname{ess\,sup}_{\Omega'} |Du|^2 &\leq \operatorname{ess\,sup}_{\Omega'} \big\{|Du|^2 + 2\lambda Du : DA + \lambda^2|DA|^2\big\}\\ &\leq \operatorname{ess\,sup}_{\Omega'} |Du|^2 + 2\lambda \operatorname{ess\,sup}_{\Omega'} \big\{ Du : DA\big\} + \lambda^2|DA|^2. \end{align*} Consequently, \[ \operatorname{ess\,sup}_{\Omega'} \big\{ Du : DA\big\} + \frac{\lambda}{2}|DA|^2 \geq 0 \] and by letting $\lambda \to 0^+$, we obtain the desired inequality. (b) is immediate from the definitions. \end{proof} Lemma \ref{lemma1} is in general true for locally Lipschitz maps, once we replace $|Du|$ by the \emph{local $L^\infty$ norm} \[ \|Du\|_\infty(x):= \lim_{\varepsilon\to0} \|Du\|_{L^\infty(\mathbb{B}_\varepsilon(x))} \] which has enough upper semi-continuity properties. \begin{lemma} \label{lemma2} Let $\Omega\subseteq \mathbb{R}^n$ be open and $u\in C^1(\Omega,\mathbb{R}^N)$. Given $\Omega'\Subset \Omega$, let $\Omega'(u)$ be as in Lemma \ref{lemma1}. Let further $A :\mathbb{R}^n \to \mathbb{R}^N$ be an affine map. We set \[ h(t):= \big\|Du+t DA\big\|^2_{L^\infty(\Omega')} - \|Du\|^2_{L^\infty(\Omega')},\quad t\geq0. \] Then, $h$ is convex, $h(0)=0$ and also the lower right Dini derivative of $h$ at zero satisfies \[ \underline{D} h(0^+):= \underset{t\to 0^+}{\lim\inf} \frac{h(t)-h(0)}{t} \geq \max_{y\in {\Omega'(u)}} \big\{ 2 Du(y):DA \big\}. \] \end{lemma} \begin{proof} Effectively, this is an application of Danskin's theorem \cite{D}, but we may also prove it directly. By setting \[ H(t,y):= \big|Du(y)+t DA\big|^2 \] we have \[ h(t)= \max_{y\in \overline{\Omega'}} H(t,y) - \max_{y\in \overline{\Omega'}} H(0,y)\,. \] Also for any $t\geq0$ the maximum $\max_{y\in \overline{\Omega'}} H(t,y)$ is realised at (at least one) point $y^t \in \overline{\Omega'}$. Hence \begin{align*} \frac{1}{t}\big( h(t)-h(0)\big) &= \frac{1}{t}\big[\max_{y\in \overline{\Omega'}} H(t,y) - \max_{y\in \overline{\Omega'}} H(0,y)\big] \\ &= \frac{1}{t}\big[ H(t,y^t) - H(0,y^0) \Big] \\ &= \frac{1}{t}\big[ \big( H(t,y^t) - H(t,y^0) \big) + \big( H(t,y^0)- H(0,y^0)\big)\big] \\ &\geq \frac{1}{t} \big( H(t,y^0) - H(0,y^0) \big), \end{align*} where $y^0\in \overline{\Omega'}$ is any point such that \[ |Du(y^0)|= H(0,y^0)= \max_{\overline{\Omega'}}H(0,\cdot)= \|Du\|_{L^\infty(\Omega')}. \] Hence, by the definition of the set $\Omega'(u)$ in Lemma \ref{lemma1}, we have \begin{align*} \underline{D} h(0^+) &=\ \underset{t\to 0^+}{\lim\inf} \frac{1}{t}\big( h(t)-h(0)\big) \\ & \geq \max_{y\in {\Omega'(u)}} \big\{ \underset{t\to 0^+}{\lim\inf} \frac{1}{t} \big( H(t,y)- H(0,y) \big) \big\}\\ & = \max_{y\in {\Omega'(u)}} \big\{ \underset{t\to 0^+}{\lim\inf} \frac{1}{t} \big( \big|Du(y)+t DA\big|^2- |Du(y)|^2 \big) \big\}\\ &= \max_{y\in {\Omega'(u)}} \big\{ 2 Du(y):DA \big\}. \end{align*} The lemma follows. \end{proof} Let us also record for later use the elementary inequality \[ h(t) - h(0) \geq \underline{D} h(0^+) t,\quad t\geq0, \] which is an immediate consequence of the definitions of convexity and of the lower right Dini derivative. \section{Scalar case $N=1$} The following is the first main result of this section, for $C^1$ $\infty$-harmonic functions. \begin{theorem} \label{theorem8} Let $\Omega\subseteq \mathbb{R}^n$ be open and $u\in C^1(\Omega)$. Given $\Omega'\Subset \Omega$, let $\Omega'(u)$ be as in Lemma \ref{lemma1} and consider the sets of affine functions \begin{align*} \mathcal{A}^{\pm,\infty}_{\Omega'}(u):=& \Big\{ A : \mathbb{R}^n \to \mathbb{R} : D^2A \equiv 0 \text{ and there exist } \xi\in \mathbb{R}^{\pm}, \\ &\quad x\in \Omega'(u) \text{ and } \mathbf{X}_x \in D^{2,\pm}u(x) \text{ s. t. } DA \equiv \xi \mathbf{X}_x Du(x)\Big\}\\ &\cup \mathbb{R}. \end{align*} Then, we have the equivalences \[ \left. \begin{array}{l} Du \otimes Du :D^2u \geq 0 \text{ on } \Omega, \\[3pt] \text{in the Viscosity sense} \end{array} \right\} \; \Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \mathcal{A}^{+,\infty}_{\Omega'}(u), \\[3pt] \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}, \end{array} \right. \] and \[ \left. \begin{array}{l} Du \otimes Du :D^2u \leq 0 \text{ on }\Omega, \\[3pt] \text{in the Viscosity sense} \end{array} \right\} \;\Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \mathcal{A}^{-,\infty}_{\Omega'}(u), \\[3pt] \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \end{array} \right. \] \end{theorem} We note that by the $C^1$ regularity results for $\infty$-harmonic functions of Savin and Evans-Savin \cite{S,ES}, if $n=2$ the hypothesis that $u$ is a $C^1(\Omega)$ viscosity solution is superfluous. Obviously, for certain subdomains it may happen that $\mathcal{A}^{\pm,\infty}_{\Omega'}(u)$ contain only the trivial (i.e.\ constant) functions if $J^{2,\pm}u(x)=\emptyset$ for all points $x\in \Omega'(u)$. Hence, the minimality property above with respect to affine functions is an effective restatement of the definition of viscosity sub/super solutions. In the event that the solution is smooth, Theorem \ref{theorem8} above simplifies to the following statement for classical solutions of the $\infty$-Laplacian, i.e.\ for $C^2$ $\infty$-Harmonic functions. \begin{corollary} \label{corollary9} Suppose that $\Omega\subseteq \mathbb{R}^n$ is open and $u\in C^2(\Omega)$. Then, we have the equivalence \begin{align*} Du \otimes Du :D^2u = 0 \text{ on }\Omega &\; \Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \big(\mathcal{A}^{+,\infty}_{\Omega'} \cup \mathcal{A}^{-,\infty}_{\Omega'}\big) (u), \\[3pt] \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')} \end{array} \right. \\ &\; \Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \mathcal{A}^{\infty}_{\Omega'}(u), \\ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \end{array} \right. \end{align*} Here $\mathcal{A}^{\infty}_{\Omega'}(u)$ is the set of affine functions \begin{align*} \mathcal{A}^{\infty}_{\Omega'}(u) = \Big\{&A : \mathbb{R}^n \to \mathbb{R} : D^2A \equiv 0 \text{ and there exist } \xi\in \mathbb{R},\, x\in \Omega'(u) \\ & \text{such that $A$ is parallel to the tangent of $\xi|Du|^2$ at }x \Big\}. \end{align*} \end{corollary} \begin{proof}[Proof of Theorem \ref{theorem8}] Suppose that for any $\Omega'\Subset \Omega$ and any affine function in $\mathcal{A}^{+,\infty}_{\Omega'}(u)$, we have \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \] Fix any $x\in \Omega$ such that $(Du(x),\mathbf{X}_x)\in J^{2,+}u(x)$, whence $\mathbf{X}_x\in D^{2,+}u(x)$. Consider the affine function \[ A(z):= \xi \mathbf{X}_x : Du(x) \otimes (z-x),\quad z\in \mathbb{R}^n, \] where $\xi \geq0$. Fix also $\varepsilon>0$ and let $\Omega_\varepsilon(x)$ be as in Lemma \ref{lemma1}. Then, for any $\lambda>0$, the affine function $\lambda A$ is contained in $\mathcal{A}^{+,\infty}_{\Omega_\varepsilon(x)}(u)$. Hence, \[ \|Du\|_{L^\infty(\Omega_\varepsilon(x))} \leq \|Du+\lambda DA\|_{L^\infty(\Omega_\varepsilon(x))}. \] By applying Lemma \ref{lemma1} to $u$ and $A$, we have \begin{align*} 0 &\leq \max_{z\in \overline{\Omega_\varepsilon(x)}} \big\{ Du(z) \cdot DA \big\} \\ &= \max_{z\in \overline{\Omega_\varepsilon(x)}} \big\{ Du(z) \cdot\big(\xi \mathbf{X}_x :Du(x)) \big\} \\ &= \max_{z\in \overline{\Omega_\varepsilon(x)}} \big\{ \xi \big(\mathbf{X}_x :Du(x) \otimes Du(z) \big)\big\} \\ & \to \xi \big(\mathbf{X}_x :Du(x) \otimes Du(x) \big), \end{align*} as $\varepsilon \to 0$. Hence, $Du \otimes Du :D^2u \geq 0$ on $\Omega$ in the viscosity sense. Conversely, fix any $\Omega'\Subset \Omega$ and $x\in \Omega'(u)$. If it happens $J^{2,+}u(x)\neq \emptyset$, then any $A\in \mathcal{A}^{+,\infty}_{\Omega'}(u)$ can be written as \[ A(z)= a + \xi \mathbf{X}_x : Du(x) \otimes z ,\ \ \ z\in \mathbb{R}^n, \] for some $a\in\mathbb{R}$, $\xi \geq0$ and $\mathbf{X}_x \in D^{2,+}u(x)$. Let $h$ be the function of Lemma \ref{lemma2} for such an $A$. By applying Lemma \ref{lemma2} to this setting, we have \begin{align*} \underline{D}h(0^+) &\geq \max_{y\in {\Omega'(u)}} \big\{ 2 Du(y) \cdot DA \big\} \\ & \geq 2 Du(x) \cdot DA \\ &= 2 Du(x) \cdot\big(\xi \mathbf{X}_x :Du(x)) \big\} \\ & = 2 \xi \big(\mathbf{X}_x :Du(x) \otimes Du(x) \big) \geq 0, \end{align*} since by assumption $Du \otimes Du :D^2u \geq 0$ on $\Omega$ in the viscosity sense. Since $h(0)=0$ and $h$ is convex, it follows that \[ h(t) \geq h(0) + \underline{D}h(0^+) t \geq 0, \quad t\geq0, \] and hence, by the definition of $h$ we obtain \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')} \] for any $\Omega'\Subset \Omega$ and any $A\in \mathcal{A}^{+,\infty}_{\Omega'}(u)$. The case of supersolutions follows similarly and hence the theorem has been established. \end{proof} \begin{proof}[Proof of Corollary \ref{corollary9}] The first equivalence of the statement is immediate. Since by assumption $u\in C^2(\Omega)$, we have \[ J^{2,+}u(x) \cap J^{2,-}u(x)= \big\{ \big(Du(x),D^2u(x)\big) \big\} \] and hence $D^{2,+}u(x) \cap D^{2,-}u(x) = \{D^2u(x)\}$. The second equivalence of the statement follows by making the choice $\mathbf{X}_x \in D^{2,\pm}u(x)$ in the proof of Theorem \ref{theorem8} above and repeating all the steps. Then, by noting that \[ \mathbf{X}_x Du(x)= D\big( \frac{1}{2}|Du|^2\big)(x) \] it follows that for any $\Omega'\Subset \Omega$ the set $\mathcal{A}^{\infty}_{\Omega'}(u)$ contains only affine functions of the form \[ A(z)= a + \xi D\big(|Du|^2\big)(x) \cdot (z-x), \ \ \ z\in \mathbb{R}^n, \] for $a,\xi \in \mathbb{R}$ and $x\in \Omega'(u)$. The corollary ensues. \end{proof} Theorem \ref{theorem8} extends relatively easily to the case of the $p$-Laplacian for $1
0$ and let $\Omega_\varepsilon(x)$ be as in Lemma \ref{lemma1} and note that for any $\lambda>0$, $\lambda A \in \mathcal{A}^{+,p}_{\Omega_\varepsilon(x)}(u)$. Hence, by arguing as in Theorem \ref{theorem8} we have that \begin{align*} 0 &\leq Du(x) \cdot DA\\ &= Du(x) \cdot \Big( (p-2)\mathbf{X}_xDu(x) + (I: \mathbf{X}_x ) Du(x)\Big) \\ &= \Big( (p-2) Du(x) \otimes Du(x) + |Du(x)|^2I \Big): \mathbf{X}_x. \end{align*} Hence, $u$ is a feeble viscosity solution on $\Omega$. Conversely, fix any $\Omega'\Subset \Omega$ and $x\in \Omega'(u)$. If $J_0^{2,+}u(x)\neq \emptyset$, then any $A\in \mathcal{A}^{+,p}_{\Omega'}(u)$ can be written as \[ A(z)= a + \xi \big( (p-2)\mathbf{X}_x + (I: \mathbf{X}_x ) I\big): Du(x) \otimes z ,\quad z\in \mathbb{R}^n, \] for some $a\in\mathbb{R}$, $\xi \geq 0$ and some $(Du(x),\mathbf{X}_x)\in J_0^{2,+}u(x)$. Let $h$ be the function of Lemma \ref{lemma2} for such an $A$. By applying Lemma \ref{lemma2}, we have \begin{align*} \underline{D}h(0^+) & \geq 2 Du(x) \cdot DA \\ & = 2 \xi \big( (p-2)Du(x) \otimes Du(x) : \mathbf{X}_x + |Du(x)|^2 I: \mathbf{X}_x \big) \geq 0, \end{align*} since by assumption $u$ is a subsolution on $\Omega$ in the feeble viscosity sense. By using that $h(0)=0$ and that $h$ is convex, we deduce as in Theorem \ref{theorem8} that $h(t)\geq0$ for $ t\geq0$ and hence \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')} \] for any $A\in \mathcal{A}^{+,p}_{\Omega'}(u)$ and any $\Omega'\Subset \Omega$. Thus, $(b) \Leftrightarrow (c)$. The case of supersolutions follows analogously and hence the theorem ensues. \end{proof} \section{Vectorial case $N\geq 2$} In this section we extend the results of the previous section to the full case of the $\infty$-Laplace system. We begin by noting that \eqref{1.1} actually consists of two independent systems, the second of which is identically trivial in the scalar case. Namely, if $u:\Omega\subseteq \mathbb{R}^n \to \mathbb{R}^N$ is smooth, then \[ \Delta_\infty u =0 \; \Longleftrightarrow\; \left\{ \begin{aligned} &Du \otimes Du :D^2u = 0,\\ &|Du|^2[Du]^\bot \Delta u = 0. \end{aligned} \right. \] This is an immediate consequence of the mutual perpendicularity of the vector fields $Du \otimes Du :D^2u$ and $|Du|^2[Du]^\bot \Delta u$; indeed, it suffices to recall that $[Du]^\bot$ is the projection on the orthogonal complement of $R(Du)$ and to note the identity \[ 2 Du \otimes Du :D^2u = Du D\big(|Du|^2 \big). \] Our last main result is the following resutl for $C^1$ $\infty$-Harmonic mappings. \begin{theorem} \label{theorem11} Let $\Omega\subseteq \mathbb{R}^n$ be open and $u\in C^1(\Omega,\mathbb{R}^N)$. Given a set $\Omega'\Subset \Omega$, let $\Omega'(u)$ be as in Lemma \ref{lemma1}. Consider first the set of affine maps \begin{align*} &\mathcal{A}^{\top,\infty}_{\Omega'}(u)\\ &:= \Big\{A : \mathbb{R}^n \to \mathbb{R}^N : D^2A \equiv 0 \text{ and there exist } \xi\in \mathbb{R}^N, x\in \Omega'(u) \\ &\quad \mathcal{D}^2u \in \mathscr{Y}\big(\Omega, \overline{\mathbb{R}}^{Nn^2}_s \big) , \mathbf{X}_x \in \operatorname{supp}_*\big(\mathcal{D}^2u(x)\big) \text{ s. t. } DA \equiv \xi \otimes \big(\mathbf{X}_x : Du(x)\big) \Big\}\\ &\cup \mathbb{R}^N. \end{align*} Then, we have the equivalence \[ \left. \begin{array}{l} Du \otimes Du :D^2u = 0 \\[3pt] \text{on } \Omega, \text{ in the $\mathcal{D}$-sense} \end{array} \right\} \; \Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \mathcal{A}^{\top,\infty}_{\Omega'}(u) , \\[3pt] \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \end{array} \right. \] Further, consider the set of affine maps \begin{align*} \mathcal{A}^{\bot,\infty}_{\Omega'}(u) &:= \Big\{A : \mathbb{R}^n \to \mathbb{R}^N : D^2A \equiv 0 \text{ there exist } x\in \Omega'(u), \mathcal{D}^2u \in \mathscr{Y}\big(\Omega, \overline{\mathbb{R}}^{Nn^2}_s \big), \\ &\quad \mathbf{X}_x \in \operatorname{supp}_*\big(\mathcal{D}^2u(x)\big) \text{ s. t. } A(x) \in R\big(Du(x)\big)^\bot, DA \in \mathscr{L}^{\mathbf{X}_x}\big(A(x)\big) \Big\}\\ & \cup \mathbb{R}^N \end{align*} where for any $a\in \mathbb{R}^N$, $\mathscr{L}^{\mathbf{X}_x}(a)$ is an affine matrix space defined as \[ \mathscr{L}^{\mathbf{X}_x}(a):= \begin{cases} \big\{ X\in \mathbb{R}^{Nn} : Du(x):X = -(a\otimes I):\mathbf{X}_x\big\}, &\text{if }Du(x)\neq0 \\ \{0\}, &\text{if }Du(x)=0. \end{cases} \] Then, we have the equivalence \[ \left. \begin{array}{l} |Du|^2[Du]^\bot \Delta u = 0 \\ \text{on } \Omega, \text{ in the $\mathcal{D}$-sense} \end{array} \right\} \; \Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \mathcal{A}^{\bot,\infty}_{\Omega'}(u), \\ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \end{array} \right. \] \end{theorem} In view of Theorem \ref{theorem11}, a mapping is $\infty$-Harmonic in the $\mathcal{D}$-sense if and only if it minimises with respect to the union of the sets of affine variations of the tangential and the normal component: \[ \left. \begin{array}{l} \Delta_\infty u = 0 \text{ on } \Omega, \\[3pt] \text{in the $\mathcal{D}$-sense} \end{array} \right\} \; \Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \big(\mathcal{A}^{\top,\infty}_{\Omega'} \cup \mathcal{A}^{\bot,\infty}_{\Omega'} \big) (u) , \\[3pt] \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \end{array} \right. \] In the event that $u\in C^2(\Omega,\mathbb{R}^N)$, Theorem \ref{theorem11} simplifies to the following statement for classical solutions of the $\infty$-Laplace system, i.e. for $C^2$ $\infty$-Harmonic mappings. \begin{corollary} \label{corollary12} Suppose that $\Omega\subseteq \mathbb{R}^n$ is open and $u\in C^2(\Omega,\mathbb{R}^N)$. Then, we have the equivalence \[ \Delta_\infty u = 0 \text{ on }\Omega \; \Longleftrightarrow\; \left\{ \begin{array}{l} \text{For all } \Omega'\Subset \Omega \text{ and } A\in \big(\mathcal{A}^{\top,\infty}_{\Omega'} \cup \mathcal{A}^{\bot,\infty}_{\Omega'}\big)(u), \\[3pt] \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}, \end{array} \right. \] where $\mathcal{A}^{\top,\infty}_{\Omega'}(u)$, $\mathcal{A}^{\bot,\infty}_{\Omega'}(u)$ are the sets of affine maps \begin{align*} \mathcal{A}^{\top,\infty}_{\Omega'}(u) = \Big\{&A : \mathbb{R}^n \to \mathbb{R}^N : D^2A \equiv 0 \text{ and there exist } \xi\in \mathbb{R}^N, \text{and }x\in \Omega'(u) \\ & \text{ s. t. $A$ is parallel to the tangent of $\xi|Du|^2$ at }x \Big\}, \end{align*} and \begin{align*} \mathcal{A}^{\bot,\infty}_{\Omega'}(u) = \Big\{&A : \mathbb{R}^n \to \mathbb{R}^N : D^2A \equiv 0 \text{ and there exists }x\in \Omega'(u) \text{such that} \\ & \text{$A$ is normal to $Du$ at $x$ and $A^\top \!Du$ is divergenceless at $x$ } \Big\}. \end{align*} \end{corollary} \begin{proof}[Proof of Theorem \ref{theorem11}] We begin by a general observation about the notion of $\mathcal{D}$-solutions $u:\Omega\subseteq \mathbb{R}^n \to \mathbb{R}^N$ in $C^1(\Omega,\mathbb{R}^N)$ to a homogeneous 2nd order quasilinear system of the form \[ \mathbf{A}(Du): D^2u= 0, \quad \text{on }\Omega, \] when $\mathbf{A}$ is Borel measurable. By definition, every diffuse hessian $\mathcal{D}^2u \in \mathscr{Y}\big(\Omega, \overline{\mathbb{R}}^{Nn^2}_s \big)$ of a candidate solution $u$ is defined a.e.\ on $\Omega$ as a weakly* measurable probability valued map $\Omega \to \smash{{\mathbb{R}}}^{Nn^2}_s\cup\{\infty\}$. Hence, we may modify each $\mathcal{D}^2u$ on a Lebesgue nullset and choose from each equivalence class the representative which is redefined as $\delta_{\{0\}}$ at points where $\mathcal{D}^2u(x)$ does not exist. Moreover, let $u$ be a fix map in $C^1(\Omega,\mathbb{R}^N)$. Since $Du(x)$ exists for all $x\in \Omega$, by perhaps a further re-definition of every $\mathcal{D}^2u$ on a Lebesgue nullset, it follows that $u$ is $\mathcal{D}$-solution to the system if and only if for (any fixed such representative of) any diffuse hessian, we have \[ \mathbf{A}\big(Du(x)\big): \mathbf{X}_x= 0, \quad \text{for all }x\in\Omega \text{ and } \mathbf{X}_x \in \operatorname{supp}_*\big(\mathcal{D}^2u(x)\big). \] (We remind that at points $x\in \Omega$ for which $\mathcal{D}^2u(x) =\delta_{\{\infty\}}$ and hence $\operatorname{supp}_*\big(\mathcal{D}^2u(x)\big)$ $= \emptyset$, the above condition is understood as being trivially satisfied.) We will apply this observation to the two independent systems \[ Du\otimes Du :D^2u = 0,\, |Du|^2\big([Du]^\bot \!\otimes I \big):D^2u = 0 \] comprising the $\infty$-Laplace system. Suppose now that for some $\Omega'\Subset \Omega$ and some affine mapping $A\in \mathcal{A}^{\top,\infty}_{\Omega'}(u)$, we have \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \] Fix any $x\in \Omega$ and any diffuse hessian $\mathcal{D}^2u \in \mathscr{Y}\big(\Omega, \overline{\mathbb{R}}^{Nn^2}_s \big)$ such that $\operatorname{supp}_*\big(\mathcal{D}^2u(x)\big)$ $\neq \emptyset$ and pick any $\mathbf{X}_x \in \operatorname{supp}_*\big(\mathcal{D}^2u(x)\big)$. Fix also $\xi \in \mathbb{R}^N$ and consider the affine map which is defined by \[ A(z):= \xi \otimes \big(\mathbf{X}_x : Du(x)\big) \cdot (z-x),\quad z\in \mathbb{R}^n. \] In index form this means \[ A_\alpha(z)= \xi_\alpha \sum_{\beta=1}^N\sum_{i,j=1}^n\Big((\mathbf{X}_x)_{\beta ji} D_ju_\beta (x)\Big) (z-x)_i,\quad \alpha=1,\dots ,N. \] For $\varepsilon>0$ small, let $\Omega_\varepsilon(x)$ be as in Lemma \ref{lemma1}. Then, $\lambda A \in \mathcal{A}^{\top,\infty}_{\Omega_\varepsilon(x)}(u)$ for any $\lambda >0$. Thus, \[ \|Du\|_{L^\infty(\Omega_\varepsilon(x))} \leq \|Du+\lambda DA\|_{L^\infty(\Omega_\varepsilon(x))} \] and by applying Lemma \ref{lemma1} to $u$ and $A$, we have \begin{align*} 0 &\leq \max_{z\in \overline{\Omega_\varepsilon(x)}} \Big\{ Du(z) : \big(\xi \otimes \mathbf{X}_x :Du(x)\big) \Big\} \\ & = \max_{z\in \overline{\Omega_\varepsilon(x)}} \Big\{ \sum_{\alpha=1}^N\sum_{i=1}^n D_i u_\alpha(z) \xi_\alpha \sum_{\beta=1}^N \sum_{j=1}^n (\mathbf{X}_x)_{\beta ji} D_ju_\beta (x) \Big\} \\ &\leq \max_{z\in \overline{\Omega_\varepsilon(x)}} \Big\{ \sum_{\alpha,\beta=1}^N\sum_{i,j=1}^n \xi_\alpha D_i u_\alpha(z) D_ju_\beta (x) (\mathbf{X}_x)_{\beta ji} \Big\} \\ & \to \sum_{\alpha,\beta=1}^N\sum_{i,j=1}^n \xi_\alpha \Big(D_i u_\alpha(x) D_ju_\beta (x) (\mathbf{X}_x)_{\beta ji}\Big) \end{align*} as $\varepsilon \to 0$, and hence \[ \xi \cdot \big(Du(x) \otimes Du(x) :\mathbf{X}_x\big) \geq 0, \] for any $\xi \in \mathbb{R}^N$. By the arbitrariness of $\xi$ we deduce that $Du(x) \otimes Du(x) :\mathbf{X}_x=0$. As a consequence, $Du \otimes Du :D^2u = 0$ in the $\mathcal{D}$-sense on $\Omega$. Now we argue similarly for the normal component of the system. Suppose that for any $\Omega'\Subset \Omega$ and any $A\in \mathcal{A}^{\bot,\infty}_{\Omega'}(u)$, we have \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}. \] We fix as before $x\in \Omega$ and $\mathbf{X}_x \in \operatorname{supp}_*\big(\mathcal{D}^2u(x)\big)$. If $Du(x)=0$, then the system $|Du|^2[Du]^\bot \Delta u=0$ is trivially satisfied at $x$. If $Du(x)\neq 0$, then we choose any direction normal to $Du(x)$; that is, \[ n_x \in R\big(Du(x)\big)^\bot \subseteq \mathbb{R}^N, \] which means that $n_x^\top Du(x)=0$. We note that if $Du(x) : \mathbb{R}^n \to \mathbb{R}^N$ is surjective, then we can find only the trivial $n_x=0$, but the system $|Du|^2[Du]^\bot \Delta u=0$ is satisfied at $x$ anyhow because $[Du(x)]^\bot =0$. We also fix any matrix $N_x$ in the affine space $\mathscr{L}^{\mathbf{X}_x}(n_x)$. By the definition of $\mathscr{L}^{\mathbf{X}_x}(n_x)$, this means that \[ N_x : Du(x)= -(n_x \otimes I):\mathbf{X}_x. \] We consider the affine map which is defined by \[ A(z):= n_x + N_x (z-x),\quad z\in \mathbb{R}^n. \] We now claim that $\lambda A\in \mathcal{A}^{\bot,\infty}_{\Omega'}(u)$ for any $\lambda \in\mathbb{R}$. Indeed, this is a consequence of our choices and of the following homogeneity property of the space $\mathscr{L}^{\mathbf{X}_x}(a)$: \[ \mathscr{L}^{\mathbf{X}_x}(\lambda a) = \lambda \mathscr{L}^{\mathbf{X}_x}(a), \quad \lambda \in \mathbb{R}. \] Hence, we have \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+\lambda DA\|_{L^\infty(\Omega')}. \] By applying Lemma \ref{lemma1} to $u$ and $A$, we have \[ 0 \leq \max_{z\in \overline{\Omega_\varepsilon(x)}} \big\{ Du(z) : N_x \big\} \to Du(x) : N_x = -(n_x \otimes I):\mathbf{X}_x, \] as $\varepsilon \to 0$. Hence, we have $(n_x \otimes I):\mathbf{X}_x\leq 0$ and by the arbitrariness of the direction $n_x \bot R\big(Du(x)\big)$, we obtain that $(n_x \otimes I):\mathbf{X}_x=0$. Thus, $\big([Du(x)]^\bot \otimes I \big):\mathbf{X}_x =0$ and as a consequence $|Du|^2[Du]^\bot \Delta u=0$ in the $\mathcal{D}$-sense on $\Omega$. Conversely, we fix $\Omega'\Subset \Omega$ and $x\in \Omega'(u)$ and any $A\in \mathcal{A}^{\top,\infty}_{\Omega'}(u)$ corresponding to a diffuse hessian $\mathcal{D}^2u \in \mathscr{Y}\big(\Omega, \overline{\mathbb{R}}^{Nn^2}_s \big)$ and some $\mathbf{X}_x \in \operatorname{supp}_*(\mathcal{D}^2u(x))$ and $\xi \in \mathbb{R}^N$. We take as $h$ to be the function of Lemma \ref{lemma2}. By applying Lemma \ref{lemma2} to this setting, we have \begin{align*} \underline{D}h(0^+) &\geq \max_{y\in {\Omega'(u)}} \big\{ 2 Du(y) : DA \big\} \\ & \geq 2 Du(x) : DA \\ &\geq 2 \sum_{\alpha,\beta=1}^N\sum_{i,j=1}^n D_i u_\alpha(x) \xi_\alpha (\mathbf{X}_x)_{\beta ji} D_ju_\beta (x) \end{align*} and hence \[ \underline{D}h(0^+) \geq 2\xi \cdot \big(Du(x) \otimes Du(x) :\mathbf{X}_x\big) = 0, \] since by assumption $Du \otimes Du :D^2u= 0$ on $\Omega$ in the $\mathcal{D}$-sense. In view of the fact that $h(0)=0$ and $h$ is convex, it follows that \[ h(t) \geq h(0) + \underline{D}h(0^+) t \geq 0, \quad t\geq0, \] and hence \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}, \quad A\in \mathcal{A}^{\top,\infty}_{\Omega'}(u),\; \Omega'\Subset \Omega. \] The case of $A\in \mathcal{A}^{\bot,\infty}_{\Omega'}$ is completely analogous: any such nonconstant $A$ satisfies $A(x) \bot R(Du(x))$ and $DA \in \mathscr{L}^{\mathbf{X}_x}\big(A(x)\big)$ for some $\mathbf{X}_x \in \operatorname{supp}_*(\mathcal{D}^2u(x))$ and some $x\in \Omega'(u)$. By applying Lemma \ref{lemma2} again, we have \[ \underline{D}h(0^+) \geq \max_{y\in {\Omega'(u)}} \big\{ 2 Du(y) : DA \big\} \geq 2 Du(x) : DA . \] If $Du(x)\neq 0$, then by the definition of $\mathscr{L}^{\mathbf{X}_x}\big(A(x)\big)$ we have \begin{align*} \underline{D}h(0^+) &\geq 2 DA : Du(x)\\ &= -2 (n_x \otimes I):\mathbf{X}_x \\ &= - 2 n_x ^\top \Big(\big( [Du(x)]^\bot \otimes I \big) : \mathbf{X}_x \Big) = 0 \end{align*} because by assumption $|Du|^2[Du]^\bot \Delta u= 0$ on $\Omega$ in the $\mathcal{D}$-sense. If $Du(x)=0$, then again $\underline{D}h(0^+) \geq 0$. In either cases, we obtain \[ h(t) \geq h(0) + \underline{D}h(0^+) t \geq 0, \quad t\geq0, \] and hence \[ \|Du\|_{L^\infty(\Omega')} \leq \|Du+DA\|_{L^\infty(\Omega')}, \quad A\in \mathcal{A}^{\bot,\infty}_{\Omega'}(u),\; \Omega'\Subset \Omega. \] The proof is complete. \end{proof} \begin{proof}[Proof of Corollary \ref{corollary12}] If $u\in C^2(\Omega,\mathbb{R}^N)$, then it is an immediate consequence of Lemma \ref{lemma0} that any diffuse hessian of $u$ satisfies \[ \mathcal{D}^2u(x)= \delta_{D^2u(x)},\quad x\in\Omega, \] and by the remarks in the beginning of the proof of Theorem \ref{theorem11}, this happens for all $x\in \Omega$. Hence, the only possible $\mathbf{X}_x$ in the reduced support of $\mathcal{D}^2u(x)$ is $\mathbf{X}_x = D^2u(x)$. For $\mathcal{A}^{\top,\infty}_{\Omega'}$, we have that any possible $A$ satisfies $DA\equiv D\big( \xi |Du|^2)(x)$. For $\mathcal{A}^{\bot,\infty}_{\Omega'}$, we have that any possible $A$ satisfies \[ A(x)^\top Du(x)= 0, \quad DA\in \mathscr{L}^{D^2u(x)}\big(A(x)\big), \] which gives \[ DA : Du(x)= -\big(A(x) \otimes I\big): D^2u(x)= -A(x) \cdot \Delta u(x). \] Thus, \[ \operatorname{div} \big( A^\top Du \big)(x) = DA : Du(x) + A(x) \cdot \Delta u(x)= 0. \] The proof is complete. \end{proof} \subsection*{Acknowledgments} The author has been financially supported by the EPSRC grant EP/N017412/1. The author would like to thank Craig Evans, Robert Jensen, Jan Kristensen and Juan Manfredi for inspiring scientific discussion relevant to the content of this particle, as well as for their encouragement. He is also indebted to the anonymous referee for the careful reading of the manuscript and for preparing their report so swiftly. \begin{thebibliography}{00} \bibitem{AK} H. Abugirda, N. 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