18.41 Degenerate curvature and growing vanishing orders99460DD0DBB7

Proof. The monomials are orthogonal by rotational invariance. For every integer $k\geq 0$, polar integration with radial variable $\rho $ and the substitution $t=m\rho ^{2r}$ give

For an entire function $f(z)=\sum _{k\geq 0}a_ kz^ k$, Parseval's identity on circles and Tonelli's theorem yield $\| f\| _{m,r}^2=\sum _{k\geq 0}|a_ k|^2\| z^ k\| _{m,r}^2$. Thus the normalized monomials form an orthonormal basis, and those with $k\geq q$ form an orthonormal basis of the selected subspace. This proves (18.41.2). By Stirling's formula, the reciprocal Gamma coefficients decay faster than every fixed geometric progression, even after multiplication by any fixed power of $k$; this gives uniform convergence on compact sets together with all derivatives. Substituting $z=w=m^{-1/(2r)}\zeta $ gives (18.41.3). On a disk $|\zeta |\leq R$ of fixed radius, the series in (18.41.3) is bounded by the tail, from the index $q$ on, of the convergent series with $|\zeta |$ replaced by $R$. This tail tends to zero as $q\to \infty $, proving the last assertion. $\square$

Proof. Recall that $\mathbb E N_ T=T$ and $\operatorname {Var}(N_ T)=T$. Stirling's formula, applied to the ratio of Gamma functions, gives

The second bound follows from the first after the constant is increased to account for finitely many small integers. Since $0\leq 1-\alpha <1$, it implies

For $T\geq 2$, consider the event $E_ T=\{ |N_ T-T|\leq T/2\} $. On this event, the expansion of the Gamma ratio and the bounded derivative of $s\mapsto s^{1-\alpha }$ on $[1/2,3/2]$ give

By the Cauchy–Schwarz inequality, its expectation is at most $C_\alpha T^{-1/2}$. On the complement, Chebyshev's inequality gives $\mathbb P(E_ T^ c)\leq 4/T$, and the preceding second-moment bound together with the Cauchy–Schwarz inequality gives the same estimate for the contribution of $E_ T^ c$. Enlarging the constant handles $1\leq T\leq 2$. $\square$

Proof. Let $\rho =|z|>0$ and put $\alpha _ s=(s+1)/r$ for integers $0\leq s<r$. Grouping the positive series (18.41.2) by the residues $s$ and using $T=m\rho ^{2r}$, we obtain the exact identity

Divide by $mr^2\rho ^{2r-2}/\pi =r^2T/(\pi \rho ^2)$ and apply lemma 18.41.5 to each of the finitely many residues. Since the indicators are bounded by one independently of their thresholds, the resulting error is uniform in $q$. $\square$

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. $\square$

Proof. We first recall the central limit theorem for Poisson variables in the form used here. For a fixed real $M>0$, Stirling's formula for $\ell !$ gives, uniformly over integers $\ell $ with $|\ell -T|\leq M\sqrt T$,

Indeed, insert the expansion of the factorial in $\mathrm{e}^{-T}T^\ell /\ell !$ and expand the logarithm in powers of $(\ell -T)/T$; the cubic remainder is $O_ M(T^{-1/2})$. Riemann sums over bounded intervals therefore converge to Gaussian integrals, and Chebyshev's bound $\mathbb P(|N_ T-T|>M\sqrt T)\leq M^{-2}$ then allows $M\to \infty $. Consequently, if real thresholds $a_ T$ satisfy $(a_ T-T)/\sqrt T\to a\in \mathbb R$, then $\mathbb P(N_ T\geq a_ T)\to 1-\Phi (a)$; rounding changes the normalized threshold by at most $T^{-1/2}$, and $\Phi $ is continuous.

Set $T_ m=m|z_ m|^{2r}$. Then $T_ m\to \infty $ and, for every integer $0\leq s<r$,

Apply the preceding Poisson limit in (18.41.7). The uniform error there tends to zero, and each of the finitely many probabilities tends to $1-\Phi (-\tau )=\Phi (\tau )$. $\square$