Corollary 5.12.22C1A82803889

Proof. It suffices to prove the assertion on every ball $B(z_0,r)\Subset D$: since $D$ is connected, lemma 5.9.4 then gives it on $D$. We construct a function $\psi $ as in theorem 5.12.1 for the domain $B(z_0,r)\times (V+\mathrm{i}\mathbb R^ k)$, which is invariant under imaginary translations and has real fibers $V$.

We first need a continuous convex function $b\geq 0$ on $V$ with compact sublevel sets. If $V=\mathbb R^ k$, take $b(t)=|t|^2$. Otherwise let $\delta (t)=\operatorname {dist}(t,\mathbb {R}^ k\setminus V)$, which is continuous and positive on $V$. It is concave on $V$: if the balls $B(t_1,r_1)$ and $B(t_2,r_2)$ lie in $V$ and $0\leq \lambda \leq 1$, then by convexity $V$ contains $\lambda B(t_1,r_1)+(1-\lambda )B(t_2,r_2)=B(\lambda t_1+(1-\lambda )t_2,\lambda r_1+(1-\lambda )r_2)$, so $\delta (\lambda t_1+(1-\lambda )t_2)\geq \lambda r_1+(1-\lambda )r_2$, and letting $r_ i\uparrow \delta (t_ i)$ gives concavity. Consequently $1/\delta $ is convex: since $x\mapsto 1/x$ is decreasing and convex on $(0,\infty )$,

Put $b(t)=|t|^2+1/\delta (t)$. For $c>0$ the set $\{ b\leq c\} $ is closed in $V$ and contained in the compact set $\{ t:|t|^2\leq c,\ \delta (t)\geq 1/c\} \subset V$, hence compact.

A continuous convex function $f$ on $V$ defines a plurisubharmonic function $f(\operatorname{Re}w)$ on every domain of the form $D'\times (V+\mathrm{i}\mathbb R^ k)$. Indeed, $f(\operatorname{Re}w)$ is continuous, and its restriction to an affine complex line is $f\circ A$ for a real affine map $A$ from an open subset of $\mathbb {C}$ to $V$. If the closed disc of radius $r$ about $\zeta _0$ lies in the domain of $A$, then, since $A$ is affine, $A(\zeta _0)$ is the midpoint of $A(\zeta _0+r\mathrm{e}^{\mathrm{i}\theta })$ and $A(\zeta _0-r\mathrm{e}^{\mathrm{i}\theta })$ for every $\theta $, and the convexity of $f$ gives

Averaging over $\theta \in [0,2\pi ]$, we obtain the submean inequality for the continuous function $f\circ A$ on this circle.

Now put $\psi (z,w)=(r^2-|z-z_0|^2)^{-1}+b(\operatorname{Re}w)$. This function is continuous, positive and invariant under imaginary translations. It is plurisubharmonic as a sum of two plurisubharmonic functions (proposition 5.9.5). The second term is plurisubharmonic by the preceding paragraph. For the first, $x\mapsto (r^2-x)^{-1}$ is convex and increasing on $(-\infty ,r^2)$ and $|z-z_0|^2$ has the complex Hessian $I$, so by the chain rule the complex Hessian of $(r^2-|z-z_0|^2)^{-1}$ is positive semidefinite, and proposition 5.9.2(1) applies. For $c>0$, the set $\{ (z,t)\in B(z_0,r)\times V:\psi (z,t)\leq c\} $ is closed in $B(z_0,r)\times V$ and, since both terms of $\psi $ are nonnegative, contained in the compact set $\{ |z-z_0|^2\leq r^2-1/c\} \times \{ b\leq c\} $; it is therefore compact, and for $c\leq 0$ it is empty. Hence theorem 5.12.1 applies on $B(z_0,r)\times (V+\mathrm{i}\mathbb R^ k)$, and shows that $\inf _{t\in V}u(z,t)$ is plurisubharmonic or identically $-\infty $ on $B(z_0,r)$.

Finally, if $\ell $ is real affine, then $\ell (\operatorname{Re}w)$ is the real part of a complex affine function of $w$. Hence $-\ell (\operatorname{Re}w)$ is smooth with vanishing complex Hessian, so it is plurisubharmonic by proposition 5.9.2(1), and $u-\ell (\operatorname{Re}w)$ is plurisubharmonic by proposition 5.9.5. It is invariant under imaginary translations, and the first part applies to it. $\square$