Lemma 5.30.12319B6057130

Proof. The maximum of a plurisubharmonic function over concentric balls is a convex function of the logarithm of the radius. This follows by restricting to each complex line through the center and comparing, on an annulus, with the harmonic function that is affine in $\log |\zeta |$. Fix a radius $r>A\delta $ for which all the balls of radius $r$ about points of the given compact set lie in the larger subset. Comparing secant slopes of this convex function between $\log \delta $, $\log (A\delta )$ and $\log r$ gives

which proves the first assertion.

To compare the maximum with the spherical mean, we use the Poisson kernel in real dimension $2n$. For a point at distance at most $\delta /2$ from the center, the normalized Poisson kernel of the ball of radius $\delta $ is bounded below by a positive dimensional constant $c_ n$. Since $u$ is bounded above by $S_\delta u$ and by its harmonic extension from the sphere of radius $\delta $, we obtain

These inequalities hold for bounded plurisubharmonic functions by local smoothing followed by a decreasing limit. By (5.30.2), $S_\delta u-A_\delta u$ therefore tends uniformly to zero.

For $0<t\leq 1$, the spherical mean is increasing in the radius, and

By logarithmic convexity, the first term is at most $B\log (1/t)/\log (r/(t\delta ))$, and by the preceding estimate the second is at most $C_ nB/\log (r/(t\delta ))$. Write the radial convolution as an average of the spherical means $A_{t\delta }u$ with respect to the probability density $\widehat\rho (t)=|\mathbb S^{2n-1}|t^{2n-1}\rho (t)$ on $[0,1]$, where $\rho (t)$ denotes the value of $\rho $ at radius $t$. Since $\int _0^1(1+\log (1/t))\widehat\rho (t)\, \mathrm{d}t<\infty $, integration gives

Combining the two comparisons proves the second assertion. $\square$