Theorem 18.41.9313505FAE88E

Proof. For fixed $z\ne 0$, put $\rho =|z|$ and $T=m\rho ^{2r}$. Then $L_ s(q_ m)/T\to \lambda /(r\rho ^{2r})$, and this limit differs from one when $\rho \neq R_\lambda $. Chebyshev's inequality and (18.41.7) therefore give

The selected density is at most the full density, that is, the density with $q=0$. When $T\geq 1$, the same comparison with $q=0$ bounds it by $C_ r m\rho ^{2r-2}$; when $T\leq 1$, (18.41.3) with $q=0$ bounds it by $C_ r m^{1/r}\leq C_ r m$. Thus $m^{-1}B_{m,r}^{(q_ m)}$ is uniformly bounded on each compact set. Since the circle $\rho =R_\lambda $ and the origin have zero area, dominated convergence and $\mathrm{dd}^{\mathrm{c}}\phi =(r^2/\pi )|z|^{2r-2}\mathrm{d}A$ prove (18.41.10).

To compute the limit of the potentials inside the circle, fix $0<\rho <R_\lambda $ and write

The estimate of Gamma ratios from Stirling's formula gives

for all sufficiently large $m$. The right side tends to $(r\rho ^{2r}/\lambda )^{1/r}<1$. Hence the sum of these terms lies between $d_{m,q_ m}$ and a fixed multiple of that term, and Stirling's formula yields

For $\rho >R_\lambda $, the limit of the densities in (18.41.12) is positive. Hence $m^{-1}\log B_{m,r}^{(q_ m)}(z)\to 0$, and therefore $m^{-1}\log K_{m,r}^{(q_ m)}(z,z)$ tends to $\phi (z)$.

At $\rho =R_\lambda $, put $k_ m=\max \{ q_ m,\lfloor rm\rho ^{2r}\rfloor \} $. Then $k_ m/m\to r\rho ^{2r}=\lambda $. The single monomial of degree $k_ m$ and Stirling's formula give $\liminf m^{-1}\log K_{m,r}^{(q_ m)}(z,z)\geq \phi (z)$, and the bound for the full kernel already proved gives the reverse inequality for the upper limit. At zero the selected kernel vanishes for all sufficiently large $m$. This proves pointwise convergence everywhere.

For local $L^1$ convergence, the bounds for the full kernel give a uniform upper bound for $m^{-1}\log K_{m,r}^{(q_ m)}(z,z)$ on each compact set. The single selected monomial of degree $q_ m$ gives the lower bound

where $c_ m$ is bounded as $m\to \infty $, again by Stirling's formula. Since $0\leq q_ m/m\leq \lambda $, the negative part on a fixed compact set is bounded by a constant plus $2\lambda |\log |z||$ near zero, which is locally integrable. Dominated convergence proves the $L^1_{\mathrm{loc}}$ assertion.

We turn to the envelope. The two radial expressions for $V_\lambda $ and their first radial derivatives agree at $R_\lambda $, because $rR_\lambda ^{2r}=\lambda $. Away from zero and that circle, direct differentiation proves (18.41.11). Since the first derivatives match, the circle carries no measure, and $\mathrm{dd}^{\mathrm{c}}\log |z|^2=\delta _0$ gives the atom at zero. The right side is positive, so $V_\lambda $ is subharmonic. For $t=|z|^{2r}/(\lambda /r)<1$, the inequality $\log t+1\leq t$ proves $V_\lambda \leq \phi $ inside the circle; outside it there is equality.

Finally, let $v$ be subharmonic with $v\leq \phi $ and $v\leq \lambda \log |z|^2+C$ near zero. On the punctured disk $0<|z|<R_\lambda $, the function $v-\lambda \log |z|^2$ is subharmonic and bounded above near zero, so it extends subharmonically across zero. On the boundary circle it is at most $\lambda /r-\lambda \log R_\lambda ^2$, since $v\leq \phi =\lambda /r$ there. The maximum principle gives $v\leq V_\lambda $ inside the disk, and outside the disk this follows from $v\leq \phi $. Since $V_\lambda $ itself satisfies the conditions, it is the largest such function. Removing the atom at the pole gives the last assertion. $\square$