Theorem 14.15.2AC61FE141521

Proof. Let $v$ be a bounded globally Lipschitz maximal function on $\mathbb C^ n$. We argue by induction on $n$. For $n=2$, $P_2(H)\geq 0$ is equivalent to $H\geq 0$, so $v$ is plurisubharmonic, and lemma 5.24.5 shows that it is constant.

We first explain the reduction in dimension for $n\geq 3$. If a function is independent of $z_ n$, its complex Hessian has the block form $H\oplus 0$. The first block of $P_ n(H\oplus 0)$ is

and the last entry is $\operatorname {tr}H/(n-1)$. Nonnegativity of $P_{n-1}(H)$ implies $\operatorname {tr}H\geq 0$, and positive definiteness implies $\operatorname {tr}H>0$. Thus $P_ n(H\oplus 0)$ is nonnegative, respectively positive definite, exactly when $P_{n-1}(H)$ is. The distributional version follows from the same linear identities. A test function in $n-1$ variables extends to $\mathbb C^ n$ independently of $z_ n$, so maximality also passes to $n-1$ variables. Consequently the induction hypothesis applies to bounded Lipschitz maximal functions that are independent of one complex variable.

Suppose now that $v$ is nonconstant. Adding a constant and multiplying by a positive constant, we may assume that $\inf v=0$ and $\sup v=1$. Let $L$ be the Lipschitz constant of $v$. Choose a smooth radial nonnegative probability kernel $\eta $, supported in the closed unit ball and strictly positive in its interior. For a bounded measurable function $f$ and $r>0$, write

Both $v$ and $v^2$ are subharmonic. For $v^2$, use $v_\epsilon \geq 0$ and $\Delta (v_\epsilon ^2)=2v_\epsilon \Delta v_\epsilon +2|\nabla v_\epsilon |^2\geq 0$, and pass to the uniform limit as $\epsilon \to 0$.

For $f=v$ and for $f=v^2$, the averages $[f]_ r(x)$ increase with $r$, by the monotonicity of spherical means of subharmonic functions. Their limit as $r\to \infty $ is independent of $x$, since

By the submean inequality this limit is at least $f(x)$ for every $x$, while it is at most $\sup f=1$. Therefore

We distinguish two cases. First suppose that there exist $\rho >0$, sequences $\epsilon _ k\downarrow 0$ and $r_ k\to \infty $, points $x_ k$ and unit complex directions $\xi _ k$ such that

Translate $x_ k$ to the origin and make $\xi _ k$ the last complex direction by a unitary change of coordinates. The resulting functions have the same bounds and Lipschitz constant, so by the Arzelà–Ascoli theorem and lemma 14.15.1 a subsequence converges locally uniformly to a maximal function $a$. The corresponding mollifications at scale $\epsilon _ k$ have the same local uniform limit, and by the integral condition their derivatives in the last complex direction tend to zero in $L^2$ on every fixed ball. Hence $\partial a/\partial z_ n=0$ in the sense of distributions. Since $a$ is real and continuous, it is independent of $z_ n$.

The dimension reduction and the induction hypothesis show that $a$ is a constant $c\in [0,1]$. In the first inequality of (14.15.4), the first term is at most one, while the average at the fixed radius $\rho $ and the value at the origin both tend to $c$. It would follow that $4/3\leq 1-c\leq 1$, a contradiction.

In the remaining case, for every fixed $\rho >0$ there is $C_\rho \geq 1$ such that, whenever $0<\epsilon <C_\rho ^{-1}$, $r>C_\rho $, and

one has

Indeed, if this failed for $C_\rho =k$ for every integer $k$, a diagonal choice would give sequences as in (14.15.4).

After a translation we may assume that $v(0)<1/12$. By (14.15.3), choose $\rho $ with $[v]_\rho (0)>3/4$, and then $r>C_\rho $ with $[v^2]_ r(0)>3/4$. The open set

contains the origin. We claim that, for some $c>0$,

is smooth and admissible on $D$.

To prove the claim, differentiate the square of the smooth mollification:

Denote by $A_\epsilon $ and $B_\epsilon $ the convolutions at the fixed radius $r$ of these two terms. The first satisfies $P_ n(A_\epsilon )\geq 0$, because $v_\epsilon \geq 0$, $v_\epsilon $ is admissible, and $P_ n$ is linear.

The second satisfies $B_\epsilon \geq c_0I$ on $D$ for small $\epsilon $, with a constant $c_0>0$. Here the passage from the unweighted integral in (14.15.5) to the convolution needs care. The bound $|\nabla v_\epsilon |\leq L$ makes the energy of a sufficiently thin boundary annulus of $B(x,r)$ at most $C_\rho ^{-1}/2$, uniformly in $x,\epsilon ,\xi $. On the remaining smaller closed ball, the fixed kernel $\eta _ r$ has a positive minimum. For points $x\in D$, (14.15.5) then gives a uniform positive lower bound for the weighted integral in every direction, which is the asserted matrix inequality for $B_\epsilon $.

Choose $0<c\leq c_0$. Since $P_ n$ preserves nonnegative matrices,

As $\epsilon \downarrow 0$, $v_\epsilon ^2\to v^2$ uniformly, and convolution at the fixed positive radius $r$ gives convergence of all derivatives. This proves the claim.

The smooth function

is therefore admissible on $D$. The set $D'=\{ v<h\} $ contains the origin, it is bounded because $v$ and its averages are bounded, and it lies in $D$. Moreover $v=h$ on $\partial D'$. Lemma 14.15.1, applied on $D'$, gives $h\leq v$ there, which contradicts the definition of $D'$. This excludes the second case and proves the theorem. $\square$