Theorem 6.7.6 (Jensen–Lelong)BD2098F182CC

Proof. Step 1: smooth functions. Suppose first that $u$ is smooth. By proposition 3.1.5 in $\mathbb R^{2n}$ and the substitution $r=\mathrm e^{t/2}$, for which $\mathrm dr/\mathrm dt=r/2$,

and (6.7.7) follows by integration, since $\mu =\Delta u\, \mathrm dV$ has a continuous density, so that the right side is continuous in $t$. To see that $D_ u(a,\cdot )$ is nondecreasing, we show that $t\mapsto M_ u(a,\mathrm e^{t/2})$ is a convex function on $(-\infty ,\log R^2)$; the derivative of a differentiable convex function is nondecreasing. For a unit vector $\omega \in S^{2n-1}$, the function $v_\omega (\zeta )=u(a+\zeta \omega )$ is smooth on a neighborhood of the closed disc $|\zeta |\leq R$ in $\mathbb C$, and by (5.2.9) its Laplacian is

because the complex Hessian of the smooth plurisubharmonic function $u$ is positive semidefinite (proposition 5.9.2(1)). Let $m_\omega (r)=(2\pi )^{-1}\int _0^{2\pi }v_\omega (r\mathrm e^{\mathrm i\theta })\, \mathrm d\theta $ be the circle means of $v_\omega $ about $0$; they are the spherical means of $v_\omega $ in $\mathbb R^2$, because surface measure on the unit circle is arc length (subsection 2.9.20). By proposition 3.1.5 in $\mathbb R^2$, where $s_1=2\pi $, and the same substitution,

which is nondecreasing in $t$ because $\Delta v_\omega \geq 0$. By the mean value theorem, a differentiable function with nondecreasing derivative is convex; so $t\mapsto m_\omega (\mathrm e^{t/2})$ is convex. Surface measure on $S^{2n-1}$ is invariant under the orthogonal transformations $\omega \mapsto \mathrm e^{\mathrm i\theta }\omega $ of $\mathbb R^{2n}$ (subsection 3.1.1), so $\int _{S^{2n-1}}u(a+r\omega )\, \mathrm dS(\omega )=\int _{S^{2n-1}}u(a+r\mathrm e^{\mathrm i\theta }\omega )\, \mathrm dS(\omega )$ for every $\theta $. Averaging over $\theta \in [0,2\pi ]$ and using Fubini's theorem for the continuous integrand, we obtain

An average of convex functions is convex, so $t\mapsto M_ u(a,\mathrm e^{t/2})$ is convex, and $D_ u(a,\cdot )$ is nondecreasing on $(0,R)$.

Step 2: approximation. For general $u$, let $u_\varepsilon =u*\rho _\varepsilon $ be the regularizations of (3.4.13). For small $\varepsilon $ they are defined on a neighborhood of the compact set $\overline{B(a,R)}$, and by proposition 5.9.2(2) they are smooth and plurisubharmonic there and decrease to $u$ as $\varepsilon \downarrow 0$. Subtracting a constant, which changes neither side of (6.7.7), nor $D_ u$, nor the Lelong number, we may assume $u_\varepsilon \leq 0$ on $\overline{B(a,R)}$ for all small $\varepsilon $. For $0\leq \eta <1$ put $\kappa _\eta =(1-\eta )/(1+\eta )^{2n-1}$. If $0<r<R$ and $|x-a|\leq \eta r$, the explicit formula (3.1.17) shows that the Poisson kernel $P$ of $B(a,r)$ satisfies $P(x,\zeta )\geq \kappa _\eta P(a,\zeta )=\kappa _\eta /(s_{2n-1}r^{2n-1})$ for $\zeta \in \partial B(a,r)$.

Fix $0<r<R$ and choose $b\in B(a,r)$ with $u(b)>-\infty $, which is possible since the pole set of $u$ has measure zero; let $\eta =|b-a|/r$. By the maximum principle (proposition 3.4.7), applied to $u_\varepsilon $ minus the Poisson integral of its boundary values (theorem 3.1.18), we have $u_\varepsilon (x)\leq \int _{\partial B(a,r)}P(x,\zeta )u_\varepsilon (\zeta )\, \mathrm dS(\zeta )$ for $x\in B(a,r)$. At $x=b$, since $u_\varepsilon \leq 0$, this gives

Thus the integrals of the nonnegative functions $-u_\varepsilon $ over the sphere are bounded independently of $\varepsilon $. As $-u_\varepsilon $ increases to $-u$, the monotone convergence theorem shows that $M_ u(a,r)$ is finite and $M_{u_\varepsilon }(a,r)\to M_ u(a,r)$. For fixed $x\in B(a,r)$ the kernel $P(x,\cdot )$ is positive and bounded on $\partial B(a,r)$, so the same argument, applied to the functions $-P(x,\cdot )u_\varepsilon $, lets us pass to the limit in the Poisson inequality, and we obtain $u(x)\leq \int _{\partial B(a,r)}P(x,\zeta )u(\zeta )\, \mathrm dS(\zeta )$ for $x\in B(a,r)$.

Since the functions $u_\varepsilon $ decrease to the locally integrable function $u$, the monotone convergence theorem, applied to $u_{\varepsilon _0}-u_\varepsilon $ for a fixed $\varepsilon _0$, shows that $u_\varepsilon \to u$ in $L^1_{\mathrm{loc}}$; hence the positive measures $\Delta u_\varepsilon $ converge to $\mu $ in the sense of distributions, hence vaguely (proposition 2.6.1). By proposition 2.6.2, $\Delta u_\varepsilon (B(a,r))\to \mu (B(a,r))$ for every radius $r$ with $\mu (\partial B(a,r))=0$; since the spheres $\partial B(a,r)$, $0<r<R$, are disjoint and $\mu $ is finite on $B(a,R)$, only countably many radii are excluded (lemma 2.2.5). Moreover, if $r_2<R$ and $\chi \in C^\infty _ c$ is a cutoff function with values in $[0,1]$, equal to one on $B(a,r_2)$, with support in a neighborhood of $\overline{B(a,R)}$ on which all $u_\varepsilon $ with small $\varepsilon $ are defined, then $\Delta u_\varepsilon (B(a,r_2))\leq \int \chi \, \Delta u_\varepsilon \to \int \chi \, \mathrm d\mu $, so these masses are bounded uniformly in $\varepsilon $.

By Step 1 each $D_{u_\varepsilon }(a,\cdot )$ is nondecreasing; letting $\varepsilon \downarrow 0$, we see that $D_ u(a,\cdot )$ is nondecreasing on the dense set of radii with $\mu (\partial B(a,r))=0$, and hence on $(0,R)$, because $r\mapsto \mu (B(a,r))$ is left continuous. Finally, (6.7.7) holds for each $u_\varepsilon $ by Step 1. As $\varepsilon \downarrow 0$, its left side converges to $M_ u(a,r_2)-M_ u(a,r_1)$, while the integrands on the right converge for all but countably many $t$ and are bounded uniformly in $\varepsilon $; dominated convergence gives (6.7.7) for $u$.

Step 3: the limits. The nondecreasing nonnegative function $D_ u(a,\cdot )$ has a limit $\ell =\lim _{r\downarrow 0}D_ u(a,r)\in [0,\infty )$. Put $g(t)=M_ u(a,\mathrm e^{t/2})$ and fix $t_0<\log R^2$. Given $\delta >0$, choose $s_\delta <\min \{ t_0,0\} $ with $D_ u(a,\mathrm e^{s/2})\leq \ell +\delta $ for $s\leq s_\delta $. For $t<s_\delta $, (6.7.7) and the monotonicity of $D_ u(a,\cdot )$ give

Dividing by $-t>0$ and letting $t\to -\infty $, we see that all limit points of $g(t)/t$ lie in $[\ell ,\ell +\delta ]$. Hence $M_ u(a,r)/\log r^2\to \ell $ as $r\downarrow 0$. It remains to show that $\ell =\nu (u,a)$. Let $0<\eta <1$. For $x\in B(a,\eta r)$, the Poisson inequality of Step 2 and the kernel bound give, since $u\leq 0$,

Together with $M_ u(a,r)\leq \sup _{\partial B(a,r)}u=N(\log r^2)$, where $N$ is as in proposition 6.7.1, we obtain

We divide by $\log r^2<0$ and let $r\downarrow 0$. Since $\log (\eta ^2r^2)/\log r^2\to 1$, this gives $\nu (u,a)\geq \kappa _\eta \ell $ and $\ell \geq \nu (u,a)$. Letting $\eta \downarrow 0$, so that $\kappa _\eta \to 1$, we conclude that $\ell =\nu (u,a)$, which proves (6.7.8). $\square$