Proposition 5.13.112FF7CC0E8132

Proof. Write $M(r)=\sup_{B_n(a,r)}\varphi$. This is finite: upper semicontinuity bounds $\varphi$ above on the closure of each smaller ball, and a plurisubharmonic function not identically $-\infty$ cannot equal $-\infty$ on a ball. It is increasing by inclusion of balls. In dimension zero it is constant.

Assume $n\geq1$ and fix $0<r_1<r_2<r_0$. Let $A\log r+B$ interpolate $M(r_1)$ and $M(r_2)$. On every complex line through $a$, comparison with this harmonic function on the annulus gives \[\varphi(z)\leq A\log|z-a|+B\qquad(r_1\leq|z-a|\leq r_2).\] The boundary bounds also hold on the spheres: if its value at a sphere point exceeded the interior supremum by $\delta>0$, upper semicontinuity would bound nearby values by that value plus $\delta/4$. A sufficiently small centered circle has at least one third of its length on the interior side, so its average would be strictly below the center value, contradicting the submean inequality. For $|z-a|<r_1$, the bound $\varphi(z)\leq M(r_1)$ suffices, because $A\geq0$. Taking the supremum over $|z-a|<r$, where $r_1<r<r_2$, yields \[M(r)\leq \frac{\log r_2-\log r}{\log r_2-\log r_1}M(r_1)+\frac{\log r-\log r_1}{\log r_2-\log r_1}M(r_2).\] This is convexity in $\log r$.