PomeLemma 5.30.1.2319B6057130 For each fixed $A>1$,
\[ S_{A\delta }u-S_\delta u\longrightarrow 0, \qquad S_\delta u-u_\delta \longrightarrow 0 \]
uniformly on compact subsets as $\delta \downarrow 0$.
5.30 Comparing local regularizationsDB8E63271D73
DB8E63271D73
To glue local convolutions, we need them to agree on overlaps up to an error tending uniformly to zero. A bounded plurisubharmonic function need not be continuous, so this comparison cannot come from uniform convergence to the original function. Instead, the logarithmic convexity of its suprema over balls controls both the change of scale and the change of coordinates.
Let $U\subset \mathbb C^ n$ be a domain and $u$ a bounded plurisubharmonic function on $U$. For $z$ in a fixed relatively compact subset of $U$ and small $\delta >0$, put
\[ S_\delta u(z)=\sup _{|\zeta -z|\leq \delta }u(\zeta ),\qquad A_\delta u(z)=\fint _{|\zeta -z|=\delta }u(\zeta )\, \mathrm{d}\sigma (\zeta ), \qquad u_\delta =u*\rho _\delta , \]
where $\mathrm{d}\sigma $ is the Euclidean surface measure and $\rho $ is a smooth nonnegative radial mollifier of integral one, supported in the unit ball, with $\rho _\delta (\zeta )=\delta ^{-2n}\rho (\zeta /\delta )$. All balls below lie in a fixed larger relatively compact subset of $U$, and $B$ denotes a bound for the oscillation of $u$ on that subset.
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
4BCD629A10F1
5.30.2
\begin{equation} 0\leq S_{A\delta }u-S_\delta u \leq \frac{B\log A}{\log (r/\delta )}, \end{equation}
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
\[ S_\delta u-S_{\delta /2}u \geq c_ n(S_\delta u-A_\delta u). \]
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
\[ 0\leq A_\delta u-A_{t\delta }u \leq (S_\delta u-S_{t\delta }u) +(S_{t\delta }u-A_{t\delta }u). \]
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
\[ 0\leq A_\delta u-u_\delta \leq \frac{C_{n,\rho }B}{\log (r/\delta )}. \]
Combining the two comparisons proves the second assertion. $\square$
Corollary 5.30.3.ACA94F989A1E Let $U,V\subset \mathbb C^ n$ be domains, let $F:U\to V$ be biholomorphic, and let $u$ be bounded and plurisubharmonic on $U$. Then
\[ u_\delta -\bigl ((u\circ F^{-1})*\rho _\delta \bigr )\circ F \]
tends locally uniformly to zero as $\delta \downarrow 0$.
Proof. On a compact subset, $F$ and its inverse have uniformly bounded first derivatives. Thus, for a fixed $A>1$, the inverse image under $F$ of the ball of radius $\delta $ about $F(z)$ lies between the balls about $z$ of radii $\delta /A$ and $A\delta $. Consequently, $S_\delta (u\circ F^{-1})(F(z))$ and $S_\delta u(z)$ both lie between $S_{\delta /A}u(z)$ and $S_{A\delta }u(z)$, and the first assertion of lemma 5.30.1 shows that their difference tends to zero uniformly. The second assertion of the lemma, applied to $u$ and to $u\circ F^{-1}$, compares each of the two convolutions with the corresponding supremum regularization. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.