Proposition 6.7.943A1E607F822

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 $\nu (\log |z-a|^2,a)=1$. $\square$