Proposition 6.7.1C79BAEA70FC1

Proof. We may assume $a=0$. Let $R_*\in (0,\infty ]$ be the supremum of the radii $r$ for which $\overline{B(0,r)}$ lies in the domain of $u$; since closed balls are compact, $\overline{B(0,r)}$ lies in the domain exactly when $0<r<R_*$. For such $r$ put $S(r)=\sup _{|z|=r}u(z)$. This number is less than $+\infty $ because $u$ is upper semicontinuous and hence bounded above on the compact sphere. It is also greater than $-\infty $. Otherwise, for every unit vector $e$, the function $\zeta \mapsto u(\zeta e)$, which is subharmonic or identically $-\infty $ near the closed disc $|\zeta |\leq r$, would be $-\infty $ on the circle $|\zeta |=r$ and hence, by upper semicontinuity, have upper limits $-\infty $ there; the maximum principle (proposition 3.4.7), applied to $\zeta \mapsto u(\zeta e)+k$ for every $k>0$, would give $u=-\infty $ on the disc. Thus $u$ would be identically $-\infty $ on $B(0,r)$, which is impossible because $u$ is locally integrable (theorem 5.9.1).

Let $0<r_1<r_2<r_3<R_*$, and let $h(\zeta )=A+B\log |\zeta |^2$ be the harmonic function on $\mathbb C\setminus \{ 0\} $ equal to $S(r_1)$ on $|\zeta |=r_1$ and to $S(r_3)$ on $|\zeta |=r_3$. For every unit vector $e$, the function $\zeta \mapsto u(\zeta e)-h(\zeta )$ is subharmonic or identically $-\infty $ near the closed annulus $r_1\leq |\zeta |\leq r_3$, and by upper semicontinuity its upper limits at the two boundary circles are at most $0$; by the maximum principle (proposition 3.4.7) it is at most $0$ on the annulus. Taking the supremum over $e$ on the circle $|\zeta |=r_2$ gives $S(r_2)\leq h(r_2)$. Since $h(r_2)$ is the value at $\log r_2^2$ of the affine function of $\log r^2$ that equals $S(r_1)$ and $S(r_3)$ at $\log r_1^2$ and $\log r_3^2$, this says that $S$ is a convex function of $\log r^2$ on $(0,R_*)$.

Applying the maximum principle in the same way to $\zeta \mapsto u(\zeta e)-S(r)$ on the disc $|\zeta |\leq r$, we obtain $u(\zeta e)\leq S(r)$ for $|\zeta |\leq r$; hence $S$ is nondecreasing and $u(0)\leq S(r)$. A finite convex function on an open interval is continuous, since on a compact subinterval its difference quotients are bounded between two fixed difference quotients taken outside the subinterval. Hence $\sup _{B(0,r)}u=\sup _{\rho <r}S(\rho )=S(r)$, so that $N(t)=S(\mathrm e^{t/2})$ for $t<\log R_*^2$; this proves the assertions about $N$.

Fix $r_0$ as in the statement and put $t_0=\log r_0^2$. Since $N$ is convex and nondecreasing on $(-\infty ,t_0]$, the difference quotients $(N(t_0)-N(t))/(t_0-t)$ are nonnegative and decrease as $t\downarrow -\infty $; let $c\in [0,\infty )$ be their limit. Writing $N(t)=N(t_0)-(t_0-t)\cdot \frac{N(t_0)-N(t)}{t_0-t}$ and dividing by $t$, we obtain $N(t)/t\to c$ as $t\to -\infty $. In particular $c$ does not depend on $t_0$, so for every fixed $s\leq t_0$ the quotients $(N(s)-N(s'))/(s-s')$ also tend to $c$ as $s'\to -\infty $. If $s'<s<t\leq t_0$, convexity gives $(N(t)-N(s))/(t-s)\geq (N(s)-N(s'))/(s-s')$; letting $s'\to -\infty $, we see that every difference quotient of $N$ on $(-\infty ,t_0]$ is at least $c$. Hence $N(t)-ct$ is nondecreasing, and $N(t)\leq c(t-t_0)+N(t_0)$ for $t<t_0$. Since $u(z)\leq S(|z|)=N(\log |z|^2)$, this is (6.7.3).

Since $u(0)\leq S(r)$ for all $r$, we have $N(t)\geq u(0)$, hence $N(t)/t\leq u(0)/t$ for $t<0$, and $c=0$ if $u(0)>-\infty $. Conversely, if $c'\geq 0$ and $u(z)\leq c'\log |z|^2+C$ for $0<|z|<r'$, then $N(t)=S(\mathrm e^{t/2})\leq c't+C$ for all sufficiently negative $t$, and division by $t<0$ gives $c\geq c'$. This proves the characterization of $\nu (u,a)$ as the largest admissible coefficient.

Now let $c\geq 0$, and suppose that $u(z)\leq c\log |z|^2+C$ for all $z\in B(0,r)\setminus A$, where $\overline{B(0,r)}$ lies in the domain of $u$ and $A$ is a Lebesgue null set. Let $0<|z|<r/2$ and $0<\delta <|z|/2$. The function $u$ is subharmonic (theorem 5.9.1), so theorem 3.4.5 gives $u(z)\leq |B(z,\delta )|^{-1}\int _{B(z,\delta )}u\, \mathrm dV$. Every $y\in B(z,\delta )\setminus A$ satisfies $u(y)\leq c\log (|z|+\delta )^2+C$, because $c\geq 0$ and $|y|\leq |z|+\delta $; since $A$ is a null set, we obtain $u(z)\leq c\log (|z|+\delta )^2+C$. Letting $\delta \downarrow 0$, we get $u(z)\leq c\log |z|^2+C$ for $0<|z|<r/2$, and the converse part of the characterization gives $\nu (u,0)\geq c$.

Let $F$ be a biholomorphic map from a neighborhood of $a$ onto a neighborhood of $F(a)$; by proposition 5.9.3, $u\circ F^{-1}$ is plurisubharmonic near $F(a)$. The maps $F$ and $F^{-1}$ are continuously differentiable, hence Lipschitz near $a$ and $F(a)$, so $\log |F(z)-F(a)|^2-\log |z-a|^2$ is bounded near $a$, $z\neq a$. Since adding a bounded function also changes $u$ by a bounded amount, in both cases the characterization just proved shows that the Lelong number is unchanged. $\square$