6.7 Logarithmic growth and the Jensen formula4939444F8FDD

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. $\square$

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). $\square$

Proof. By (6.7.8), $\nu (w,a)=\lim _{r\downarrow 0}M_ w(a,r)/\log r^2$ for every plurisubharmonic $w$ near $a$, and the spherical means involved are finite and linear in $w$; this proves the formulas for $u+v$ and $bu$. If $w_1\leq w_2+O(1)$ near $a$, then $\nu (w_1,a)\geq \nu (w_2,a)$ by the characterization in proposition 6.7.1; hence $\nu (\max \{ u,v\} ,a)\leq \min \{ \nu (u,a),\nu (v,a)\} $. Conversely, choose $r_0>0$ with $\overline{B(a,r_0)}\subset U$. By (6.7.3), and since the logarithm $\log (|z-a|^2/r_0^2)$ is negative on $B(a,r_0)\setminus \{ a\} $, both $u$ and $v$, and hence $\max \{ u,v\} $, are bounded there by $c\log (|z-a|^2/r_0^2)+C$ with $c=\min \{ \nu (u,a),\nu (v,a)\} $ and a constant $C$. This gives the reverse inequality.

For the family, $u$ is plurisubharmonic by theorem 5.10.2. Choose $r_0$ as above, a common upper bound $C$ for the $u_ i$ on $B(a,r_0)$, and put $c=\inf _ i\nu (u_ i,a)$. By (6.7.3), again using that the logarithm is negative, $u_ i(z)\leq c\log (|z-a|^2/r_0^2)+C$ for $0<|z-a|<r_0$ and all $i$. The supremum of the $u_ i$ satisfies the same bound, and so does its upper semicontinuous regularization $u$ on $B(a,r_0)\setminus \{ a\} $, because the bound is continuous there. Hence $\nu (u,a)\geq c$ by the characterization in proposition 6.7.1. Conversely $u\geq u_ i$ gives $\nu (u,a)\leq \nu (u_ i,a)$ for every $i$.

For (6.7.10), note that $\Delta u_ j\to \Delta u$ in the sense of distributions, because $u_ j\to u$ in $L^1_{\mathrm{loc}}(U)$; hence the positive measures $\Delta u_ j$ converge vaguely to $\Delta u$ (proposition 2.6.1). Fix $0<r<R$ with $\overline{B(a,R)}\subset U$ and $\Delta u(\partial B(a,r))=0$, and let $0<\rho <r$. Then $B(a_ j,\rho )\subset B(a,r)$ for all large $j$. Since $D_{u_ j}(a_ j,\cdot )$ is nondecreasing with limit $\nu (u_ j,a_ j)$ at $0$ (theorem 6.7.6), and $\varlimsup _ j\Delta u_ j(K)\leq \Delta u(K)$ for the compact set $K=\overline{B(a,r)}$ (proposition 2.6.2), we obtain

Let $\rho \uparrow r$; as $\Delta u(\partial B(a,r))=0$, the right side becomes $D_ u(a,r)$. By lemma 2.2.5, applied to the finite measure $\Delta u$ on $B(a,R)$ and the disjoint spheres $\partial B(a,r)$, all but countably many radii $r\in (0,R)$ satisfy $\Delta u(\partial B(a,r))=0$. Letting $r\downarrow 0$ through such radii and using (6.7.8) proves (6.7.10).

Finally, let $k=\operatorname {ord}_ af$, and let $P_ k\neq 0$ be the homogeneous term of degree $k$ in the expansion of $f$ at $a$; the function $\log |f|^2$ is plurisubharmonic near $a$ by proposition 5.10.3. By (5.2.2) there is $C$ with $|f(a+w)-P_ k(w)|\leq C|w|^{k+1}$ for small $w$. Put $c_1=\max _{|\omega |=1}|P_ k(\omega )|$, which is positive because a homogeneous polynomial vanishing on the unit sphere vanishes identically. Since $|P_ k(w)|\leq c_1|w|^ k$, with equality for some $w$ on every sphere $|w|=r$, we obtain $r^ k(c_1-Cr)\leq \sup _{|w|=r}|f(a+w)|\leq r^ k(c_1+Cr)$. By proposition 6.7.1, $N(\log r^2)$ is the supremum of $\log |f|^2$ over the sphere $|z-a|=r$, so $N(\log r^2)=k\log r^2+O(1)$ as $r\downarrow 0$, and $\nu (\log |f|^2,a)=k$. For the last assertion, recall that $\log |z-a|^2$ is plurisubharmonic (proposition 5.10.3). Since $\sup _{B(a,r)}\log |z-a|^2=\log r^2$, we have $N(t)=t$ for this function, and (6.7.2) gives $\nu (\log |z-a|^2,a)=1$. $\square$