Lemma 14.21.6.96D40C4808C2 Let $a=(a^{ij})$ be a measurable symmetric matrix field on a connected domain $\Omega \subset \mathbb R^ m$. For each compact $K\Subset \Omega $, suppose there are constants $0<\lambda _ K\leq \Lambda _ K<\infty $ such that
\[ \lambda _ K I\leq a(x)\leq \Lambda _ KI \]
almost everywhere on $K$, where $I$ is the identity matrix. Repeated indices $i,j=1,\ldots ,m$ are summed, and $z_{ij}=\partial _ i\partial _ jz$. If $z\in W^{2,\infty }_{\rm loc}(\Omega )$, $a^{ij}z_{ij}\geq 0$ almost everywhere, and $z$ attains its supremum at a point of $\Omega $, then $z$ is constant.
Proof.
We first prove strict weak comparison on a relatively compact domain: if $b\in C^{1,1}$, $Lb:=a^{ij}b_{ij}\geq \kappa >0$ almost everywhere, and $b\leq M$ on the boundary, then $b\leq M$ inside. Otherwise an interior maximum point lies in a closed ball on which $a\leq \Lambda I$. Apply 14.21.5 to the set of full measure where both the differential inequality and the Taylor expansion hold. Choosing $2\delta m\Lambda <\kappa $ gives the contradiction $\kappa \leq Lb\leq 2\delta \operatorname {tr}a\leq 2\delta m\Lambda $.
Take the $C^{1,1}$ representative of $z$, write $M=\sup _\Omega z$, and suppose that $K=\{ z=M\} $ is a nonempty proper subset of $\Omega $. Near a relative boundary point of $K$, choose $y\notin K$ so that the ball of radius $r=\operatorname {dist}(y,K)>0$ is compactly contained in $\Omega $ and touches $K$ at $x_0$. On the closed annulus $r/2\leq |x-y|\leq r$, use fixed bounds $\lambda I\leq a\leq \Lambda I$ and set
\[ b(x)=\mathrm{e}^{-\alpha |x-y|^2}-\mathrm{e}^{-\alpha r^2}. \]
For sufficiently large $\alpha $,
\[ Lb=\mathrm{e}^{-\alpha |x-y|^2} \bigl (4\alpha ^2 a^{ij}(x_ i-y_ i)(x_ j-y_ j) -2\alpha \operatorname {tr}a\bigr )\geq \kappa >0 \]
almost everywhere on the annulus. On its inner sphere $z<M$ uniformly; on its outer sphere $b=0$. Choose $\eta >0$ small enough that $z+\eta b\leq M$ on both spheres. Strict weak comparison, applied to $z+\eta b$, gives this inequality throughout the annulus. At $x_0$, however, $z=M$, $b=0$, $\nabla z=0$, and $b$ increases to first order in the inward radial direction, which contradicts that inequality. Thus $K$ has no relative boundary point in $\Omega $, and connectedness proves the assertion. The argument uses no derivatives of the measurable coefficients.
$\square$
Pome
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