Theorem 8.3.296D01E7C4F0F

Proof. For $m=1$, the distributional subharmonicity and the lower-test condition say that the function is a viscosity solution of the Laplace equation. The elementary equivalence with harmonicity and bounded harmonic Liouville give the conclusion. For $m=n$, admissibility is distributional plurisubharmonicity. The continuous representative is plurisubharmonic, and bounded plurisubharmonic Liouville applies. For a fixed $m\geq 2$, we therefore argue by induction on $n\geq m$.

The reduction in dimension is particularly simple for this cone. For a Hermitian $(n-1)$-by-$(n-1)$ matrix $A$ and $m<n$,

Consequently $A\oplus 0$ belongs to $C_{n,m}$, respectively its interior, exactly when $A$ belongs to $C_{n-1,m}$, respectively its interior. For the closed cones, the nonnegative inequalities characterize the closure: the expansion of $S_ j(\lambda +\delta \mathbf1)$ has nonnegative coefficients and a strictly positive $\delta ^ j$ term. If an admissible function is independent of $z_ n$, convolving it and using the block identity shows that the function in the remaining variables is admissible. Extending its lower tests independently of $z_ n$ shows that maximality also passes to dimension $n-1$.

Suppose $v$ is nonconstant. Normalize $\inf v=0$, $\sup v=1$, and let $L$ be its Lipschitz constant. Choose a smooth radial nonnegative probability kernel $\eta $, supported in the closed unit ball and positive in its interior. Put

Both $v$ and $v^2$ are subharmonic: for the second assertion use $\Delta (v_\epsilon ^2)=2v_\epsilon \Delta v_\epsilon +2|\nabla v_\epsilon |^2\geq 0$ and pass to the uniform limit. Their radial averages increase with $r$, by monotonicity of spherical means. For either $f=v$ or $f=v^2$,

Thus the limiting average is independent of its center. The submean inequality makes it at least $f(x)$ for every $x$, and the upper bound is $\sup f=1$. Hence

First suppose there are a fixed $\rho >0$, numbers $\epsilon _ k\downarrow 0$, $r_ k\to \infty $, points $x_ k$, and unit complex directions $\xi _ k$, such that

Translate and rotate to make $x_ k$ the origin and $\xi _ k$ the last complex direction. Compactness of uniformly bounded Lipschitz functions and the preceding lemma give a locally uniform limit $a$ that is admissible and maximal. The transformed mollifications have the same limit. Their last derivatives tend to zero in $L^2$ on every fixed ball, so $\partial _{z_ n}a=0$ in distributions. Since $a$ is real and continuous, it is independent of $z_ n$. The induction hypothesis gives $a=c\in [0,1]$. In the displayed gap inequality the first average is at most one, while the fixed-radius average and the value at the origin tend to $c$. This would give $4/3\leq 1-c\leq 1$, a contradiction.

If this case does not occur, then for every fixed $\rho >0$ there is $C_\rho \geq 1$ such that

for every unit complex direction $\xi $, whenever $0<\epsilon <C_\rho ^{-1}$, $r>C_\rho $ and $[v^2]_ r(x)+[v]_\rho (x)-2v(x)\geq 4/3$. Indeed, failure for every integer choice $C_\rho =k$ produces the sequences excluded above. The dependence on the fixed radius $\rho $ is retained. Translate so that $v(0)<1/12$. Choose $\rho $ with $[v]_\rho (0)>3/4$, then $r>C_\rho $ with $[v^2]_ r(0)>3/4$. The open set

contains the origin.

On this set, $[v^2]_ r-c|z|^2$ is smooth and admissible for some fixed $c>0$. To prove this, decompose

The matrix $A_\epsilon $ belongs to $C_{n,m}$ by convexity and $v_\epsilon \geq 0$. For every complex direction the unweighted energy on $B(x,r)$ is at least $C_\rho ^{-1}$. Since $|\nabla v_\epsilon |\leq L$, a sufficiently thin boundary annulus carries at most half that amount, uniformly in $x,\epsilon ,\xi $. On the remaining smaller closed ball $\eta _ r$ has a positive minimum. Therefore $B_\epsilon \geq c_0I$ on $D$, with $c_0>0$ independent of small $\epsilon $. Choose $0<c\leq c_0$. Then $A_\epsilon +B_\epsilon -cI\in C_{n,m}$. Uniform convergence of $v_\epsilon ^2$ and convolution at the fixed radius give convergence of every derivative to those of $[v^2]_ r$, proving the assertion.

The smooth function

is admissible on $D$. The set $D'=\{ v<h\} $ contains the origin, is bounded, and lies in $D$. Continuity gives $v=h$ on its boundary. The comparison lemma gives $h\leq v$ on $D'$, a contradiction. This completes the induction. $\square$