Proof of Theorem 1.2

Assume first that $\int_{\mathbb{D}} F^{M+1} (z)\, d\nu(z) \lt \infty$ or, equivalently, that $\sum_{n,k}\mu_{n,k}^{M+1} \lt +\infty$. Define the events

\begin{equation*} A_{n,k}=\{X_{n,k} \gt M\}=\{X_{n,k}\ge M+1\}. \end{equation*}

Then,

\begin{equation*} \mathbb{P}(A_{n,k})=1-\sum_{j=0}^M \mathbb{P}(X_{n,k}=j)=1-\mathrm{e}^{-\mu_{n,k}} \bigl(\sum_{j=0}^M \frac{\mu_{n,k}^j}{j!}\bigr). \end{equation*}

By hypothesis $\lim\limits_n(\sup_k\mu_{n,k})= 0$, so we can use Taylor’s formula

(3)\begin{equation} 1-\mathrm{e}^{-x}(\sum_{j=0}^M\frac{x^j}{j!})=\frac{x^{M+1}}{(M+1)!}+o(x^{M+1})\qquad x\to 0 \end{equation}

to deduce that

\begin{equation*} \sum_{n,k} \mathbb{P}(A_{n,k})\lesssim \sum_{n,k}\frac{\mu_{n,k}^{M+1}}{(M+1)!} \lt +\infty. \end{equation*}

By the Borel–Cantelli lemma almost surely $X_{n,k}\leq M$ for all but at most a finite number of $T_{n,k}$.

In principle, this does not imply that $\Lambda_\mu$ can be split into M separated sequences, because it might happen that points of two neighbouring $T_{n,k}$ come arbitrarily close. This possibility is excluded by repeating the above arguments to a new dyadic partition, made of shifted boxes $\tilde{T}_{n,k}$ having the ‘lower vertices’ (those closer to $\mathbb{T}$) at the centre of the $T_{n,k}$’s (see Figure 2); let

\begin{equation*} \tilde T_{n,k}=\Bigl\{z=re^{it}: \frac 32 2^{-(n+2)} \lt 1-r\leq \frac 32 2^{-(n+1)}\, ;\ \frac{k+1/4}{2^n}\le \frac t{2\pi} \lt \frac{k+3/4}{2^n}\Bigr\}. \end{equation*}

Figure 2.

Dyadic partitions: $\{T_{n,k}\}_{n,k}$ in blue, $\{\tilde T_{n,k}\}_{n,k}$ in red.

Since each $\tilde T_{n,k}$ is included in the union of at most four $T_{m,j}$, we still have $\sum_{n,k} \tilde\mu_{n,k}^{M+1} \lt \infty$, and therefore, as before, $\tilde X_{n,k}=N_{\tilde T_{n,k}}$ is at most M, except for maybe a finite number of indices (n, k). This prevents that two adjacent $T_{n,k}$ have more than M points getting arbitrarily close. In conclusion, for all but a finite number of indices $X_{n,k}\leq M$, hence the part of $\Lambda_\mu$ in these boxes can be split into M separated sequences. Adding the remaining finite number of points to any of these sequences may change the separation constant but not the fact that they are separated.

Assume now that $\sum_{n,k}\mu_{n,k}^{M+1}=+\infty$. We shall prove that for every $\delta_{l_0}=2^{-l_0}$, $l_0\in\mathbb{N}$,

\begin{equation*} \mathbb{P}\bigl({{\Lambda}}\,\, {\mathrm{union\,\, of\,\, {\mathit{M}}\,\, \delta_{l_0}-separated\,\, sequences}}\bigr)=0. \end{equation*}

Split each side of $T_{n,k}$ into $2^{l_0}$ segments of the same length. This defines a partition of $T_{n,k}$ in $2^{2l_0}$ small boxes of side length $2^{-n} 2^{-l_0}$, which we denote by

\begin{equation*} T_{n,k}^{l_0,j}\qquad j=1,\dots, 2^{2l_0}. \end{equation*}

Let $X_{n,k}^{l_0,j}=N_{T_{n,k}^{l_0,j}}$ denote the corresponding counting variable, which follows a Poisson law of parameter $\mu_{n,k,l_0,j}=\mu(T_{n,k}^{l_0,j})$.

It is enough to show that for any l ₀,

\begin{equation*} \mathbb{P}(X_{n,k}^{l_0,j} \gt M \text{ for infinitely many }n,k,j)=1. \end{equation*}

By the second part of the Borel–Cantelli lemma, since the $X_{n,k}^{l_0,j}$ are independent, we shall be done as soon as we see that

(4)\begin{equation} \sum_{n,k}\sum_{j=1}^{2^{2 l_0}} \mathbb{P}\bigl(X_{n,k}^{l_0,j}\ge M+1\bigr)=+\infty. \end{equation}

For any Poisson variable X of parameter λ, the probability

\begin{equation*} \mathbb{P}(X\ge M+1)=\mathrm{e}^{-\lambda}\bigl(\sum_{m=M+1}^\infty \frac{\lambda^m}{m!}\bigr)=1-\mathrm{e}^{-\lambda}\bigl(\sum_{m=0}^M \frac{\lambda^m}{m!}\bigr) \end{equation*}

increases in λ. Hence, there is no restriction in assuming that $0\leq\mu_{n,k,l_0,j}\leq \mu_{n,k}\leq 1/2$ for all $n,k,j$. Then, we can use Taylor’s formula (3) to deduce that

\begin{equation*} \mathbb{P}\bigl(X_{n,k}^{l_0,j}\ge M+1\bigr)\simeq \frac{\mu_{n,k,l_0,j}^{M+1}}{(M+1)!}, \end{equation*}

and therefore, (4) is equivalent to

\begin{equation*} \sum_{n,k}\sum_{j=1}^{2^{2l_0}} \mu_{n,k,l_0,j}^{M+1}=+\infty. \end{equation*}

The fact that this sum is infinite is just a consequence of the hypothesis and Jensen’s inequality

\begin{align*} \mu_{n,k}^{M+1}=\Bigl(\sum_{j=1}^{2^{2l_0}} \mu_{n,k,l_0,j}\Bigr)^{M+1} \leq 2^{2l_0 (M+1)} \sum_{j=1}^{2^{2l_0}} \frac{\mu_{n,k,l_0,j}^{M+1}}{2^{2l_0}} \leq 2^{2l_0 M} \sum_{j=1}^{2^{2l_0}} \mu_{n,k,l_0,j}^{M+1}. \end{align*}

Proof of Theorem 1.5

(a) Assume first that $\sum_n \mu_{n,k}^{\gamma} \lt +\infty$ for some γ > 1. Observe that it is enough to check the Carleson condition

\begin{equation*} \sum_{\lambda\in Q(I)} (1-|\lambda|)\leq C |I| \end{equation*}

on the dyadic intervals $I_{n,k}$ given in (2). Let $Q_{n,k}=Q(I_{n,k})$. Decomposing the sum on the different layers A_m, it is enough to show that almost surely there exists C > 0 such that for all $n\geq 0$, $k=0,\dots, 2^{n-1}$

\begin{equation*} \sum_{\lambda\in Q_{n,k}} (1-|\lambda|)\simeq \sum_{m\geq n} \sum_{j: T_{m,j}\subset Q_{n,k}} 2^{-m} X_{m,j}\leq C 2^{-n} . \end{equation*}

This is equivalent to

(5)\begin{equation} \sup_{n,k}\ 2^n\sum_{m\geq n} \sum_{j: T_{m,j}\subset Q_{n,k}} 2^{-m} X_{m,j} \lt \infty. \end{equation}

Denote

\begin{equation*} X_{n,m,k}=N_{Q_{n,k}\cap A_m}=\#(\Lambda\cap Q_{n,k}\cap A_m)=\sum_{j: T_{m,j}\subset Q_{n,k}} X_{m,j}, \end{equation*}

which is a Poisson variable of parameter

\begin{equation*} \mu_{n,m,k}=\mu(Q_{n,k}\cap A_m)=\sum_{j: T_{m,j}\subset Q_{n,k}} \mu_{m,j}. \end{equation*}

Set

\begin{equation*} Y_{n,k}=2^n\sum_{m\ge n}2^{-m}X_{n,k,m}=\sum_{m\ge n}2^{n-m}X_{n,k,m}, \end{equation*}

so that (5) becomes $\sup_{n,k} Y_{n,k} \lt +\infty$.

Let A > 0 be a big constant to be fixed later on. Again by the Borel–Cantelli Lemma, it is enough to show that

(6)\begin{equation} \sum_{n,k} \mathbb{P}\bigl(Y_{n,k} \gt A\bigr) \lt +\infty, \end{equation}

since then $Y_{n,k}\leq A$ for all but maybe a finite number of $n,k$; in particular, $\sup_{n,k} Y_{n,k} \lt \infty$.

The first step of the following reasoning is an adaptation to the Poisson process of the proof given in [Reference Chalmoukis, Hartmann, Kellay and Wick9, Theorem 1.1] and which allowed to improve the result on Carleson sequences for the probabilistic model with fixed radii and random arguments. However, while in the original proof, the Carleson boxes $Q_{n,k}$ are decomposed into layers $Q_{n,k}\cap A_m$ ($m\ge n$), in this new situation (as well as for (b)), Carleson boxes are decomposed into top-halves $T_{m,j}\subset Q_{n,k}$, which requires more delicate arguments to reach the convergence needed in the Borel–Cantelli lemma.

Recall that the probability generating function of a Poisson variable X of parameter λ is $\mathbb{E}(s^X)=\mathrm{e}^{\lambda(s-1)}$. By the independence of the different $X_{n,k,m}$, $m\ge n$,

\begin{equation*} \mathbb{E}(s^{Y_{n,k}})=\prod_{m\ge n}\mathbb{E}((s^{2^{n-m}})^{X_{n,m,k}}) =\prod_{m\ge n}\mathrm{e}^{\mu_{n,m,k}(s^{2^{n-m}}-1)}. \end{equation*}

Thus for any s > 1, by Markov’s inequality

\begin{equation*} \mathbb{P}(Y_{n,k} \gt A)=\mathbb{P}(s^{Y_{n,k}} \gt s^A) \le \frac{1}{s^A}\mathbb{E}(s^{Y_{n,k}})=\frac{1}{s^A}\prod_{m\ge n}\mathrm{e}^{\mu_{n,m,k}(s^{2^{n-m}}-1)}. \end{equation*}

Using the estimate $x(a^{1/x}-1)\le a$, for $a,x \gt 1$, with a = s and $x=2^{m-n}$,

\begin{align*} \log \mathbb{P}(Y_{n,k} \gt A)&\le -A\,\log s+\sum_{m\ge n}(s^{2^{n-m}}-1)\, \mu_{n,m,k}\\ &\le -A\log s+\sum_{m\ge n} s 2^{n-m} \, \mu_{n,m,k}\\ &=-A\,\log s+s\sum_{m\ge n} 2^{n-m} \sum_{j:T_{m,j}\subset Q_{n,k}}\mu_{m,j}. \end{align*}

We want to optimize this estimate for s > 1. Set

\begin{equation*} B_{n,k}=\sum_{m\ge n} 2^{-(m-n)} \sum_{j:T_{m,j}\subset Q_{n,k}}\mu_{m,j} \end{equation*}

and define

\begin{equation*} \phi (s)=-A\,\log s+s B_{n,k}. \end{equation*}

Let us observe first that the $B_{n,k}$ are uniformly bounded (they actually tend to 0). Indeed, let β denote the conjugate exponent of γ ($\frac 1{\gamma}+\frac 1{\beta}=1$). Since, for $m\geq n$, there are $2^{m-n}$ boxes $T_{m,j}$ in $Q_{n,k}$, by Hölder’s inequality on the sum in the index j we deduce that

\begin{align*} B_{n,k}&\le \sum_{m\ge n}2^{-(m-n)}\Bigl(\sum_{j:T_{m,j}\subset Q_{n,k}} \mu_{m,j}^{\gamma} \Bigr)^{1/\gamma}2^{(m-n)/\beta}\\ &=\sum_{m\ge n}2^{-(m-n)/\gamma}\Bigl(\sum_{j:T_{m,j}\subset Q_{n,k}} \mu_{m,j}^{\gamma} \Bigr)^{1/\gamma} \lt +\infty. \end{align*}

Taking A big enough we see that the minimum of ϕ is attained at $s_0=A/B_{n,k} \gt 1$. Hence,

\begin{equation*} \log \mathbb{P}(Y_{n,k} \gt A) \le \phi(s_0)=-A\,\log\frac{A}{B_{n,k}}+A. \end{equation*}

Therefore,

\begin{equation*} \mathbb{P}\bigl(Y_{n,k} \gt A\bigr)\le \left(\frac{B_{n,k}}{A}\right)^A\mathrm{e}^A, \end{equation*}

and

\begin{equation*} \sum_{n,k} \mathbb{P}(Y_{n,k} \gt A) \le \left(\frac{e}{A}\right)^A\sum_{n,k}B_{n,k}^A. \end{equation*}

The estimate on $B_{n,k}$ obtained previously is not enough to prove that this last sum converges. In order to obtain a better estimate take p > 1, to be chosen later on, its conjugate exponent q (i.e. $\frac 1p+\frac 1q=1$), and apply Hölder’s inequality in the following way:

\begin{align*} B_{n,k}&=\sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-(m-n)}\mu_{m,j} =2^n \sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac mp}2^{-\frac mq}\mu_{m,j}\\ &\le 2^n\Bigl(\sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\beta}p}\Bigr)^{1/\beta} \times\Bigl(\sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\gamma}q}\mu^{\gamma}_{m,j}\Bigr)^{1/\gamma}. \end{align*}

Choose now p so that $1 \lt p \lt \beta$; then,

\begin{align*} \sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\beta}p} &=\sum_{m=n}^{\infty} 2^{-\frac{m\beta}p} \ 2^{m-n} =2^{-n}\sum_{m=n}^{\infty}2^{-m(\frac{\beta}p-1)} \simeq 2^{-n}2^{-n(\frac{\beta}p-1)}=2^{-n\frac{\beta}p}. \end{align*}

Thus, from the above estimate,

\begin{equation*} B_{n,k}\le 2^{\frac nq} \Bigl(\sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\gamma}q}\ \mu^{\gamma}_{m,j} \Bigr)^{1/\gamma}. \end{equation*}

Choosing $A=\gamma$ yields

\begin{equation*} \sum_{n,k}B_{n,k}^\gamma \le \sum_{n,k} 2^{\frac{n\gamma}q} \sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\gamma}q}\ \mu^{\gamma}_{m,j} . \end{equation*}

We now apply Fubini’s theorem to exchange the sums. The important observation here is that each $T_{m,j}$ has only one ancestor at each level $n\le m$ (i.e., one $T_{n,k}$ containing $T_{m,j}$). Hence,

\begin{align*} \sum_{n,k} B_{n,k}^{\gamma}&\le\sum_{m,j} 2^{-\frac{m\gamma}q}\ \mu^{\gamma}_{m,j} \sum_{\substack{n\leq m\\ k : Q_{n,k} \supseteq T_{m,j}}} 2^{\frac{n\gamma}p} =\sum_{m,j} 2^{-\frac{m\gamma}q}\ \mu^{\gamma}_{m,j} \sum_{n\le m} 2^{\frac {n\gamma} q}\\ &\le 2\sum_{m,j} 2^{-\frac{m\gamma}q} \mu^{\gamma}_{m,j}\ 2^{\frac{m\gamma}q} =2\sum_{m,j} \mu^{\gamma}_{m,j}. \end{align*}

This finishes the proof of (6), hence of this part of the theorem.

Let us now assume that $\sum_{n,k}\mu_{n,k}^{\gamma} =+\infty$ for every γ > 1. Suppose $M\ge 1$ is an integer. Since the sum diverges for $\gamma=M+1$, Theorem 1.2 implies that the sequence $\Lambda_{\mu}$ is almost surely not a union of M separated sequences. In particular, there is $\lambda_0\in\Lambda_{\mu}$ such that $D_{\lambda_0}=\{z\in \mathbb{D}:\rho(\lambda_0,z) \lt 1/2\}$ contains at least M + 1 points of $\Lambda_{\mu}$. Then, letting $I_{\lambda_0}$ be the interval centred at $\lambda_0/|\lambda_0|$ with length $1-|\lambda_0|$, we have $\sum_{\lambda\in Q(I_{\lambda_0})}(1-|\lambda|) \gt rsim M |I_{\lambda_0}|$, where the underlying constant does not depend on M or λ ₀. This being true for every integer $M\ge 1$, the sequence cannot be 1-Carleson.

(b) Proceeding as in the first implication of (a) we see that it is enough to prove that almost surely

(7)\begin{equation} \sup_{n,k} Y_{n,k} \lt +\infty\ , \end{equation}

where now

(8)\begin{equation} \quad Y_{n,k}=2^{n\alpha}\sum_{m\geq n} 2^{-m\alpha} \sum_{j: T_{m,j}\subset Q_{n,k}} X_{m,j}. \end{equation}

The same estimates as in (a) based on the probability generating function yield, for s > 1,

\begin{equation*} \log \mathbb{P}\bigl(Y_{n,k}\geq A\bigr)\leq \phi(s)=-A\,\log s+ B_{n,k}, \end{equation*}

where now

\begin{equation*} B_{n,k}=\sum_{m\geq n} 2^{-(m-n)\alpha} \sum_{j: T_{m,j}\subset Q_{n,k}} \mu_{m,j}. \end{equation*}

As in (a), the hypotheses imply that $B_{n,k}$ is uniformly bounded: letting β denote the conjugate exponent to γ ($\frac 1{\gamma}+\frac 1{\beta}=1$) and noticing that $\alpha-1/\beta=1/\gamma-(1-\alpha) \gt 0$,

\begin{align*} B_{n,k}&\leq \sum_{m\geq n} 2^{-(m-n)\alpha} \Bigl(\sum_{j: T_{m,j}\subset Q_{n,k}} \mu_{m,j}^\gamma\Bigr)^{1/\gamma} \ 2^{(m-n)/\beta} \\ &= \sum_{m\geq n} 2^{-(m-n)(\alpha-1/\beta)} \Bigl(\sum_{j: T_{m,j}\subset Q_{n,k}} \mu_{m,j}^\gamma\Bigr)^{1/\gamma}. \end{align*}

Therefore, optimizing the estimate for s > 1 exactly as we did in (a), we obtain $\mathbb{P}\bigl(Y_{n,k}\geq A\bigr)\lesssim B_{n,k}^A$, and we are lead to prove that for some A > 0

(9)\begin{equation} \sum_{n,k} \mathbb{P}\bigl(Y_{n,k}\geq A\bigr)\lesssim \sum_{n,k} B_{n,k}^A \lt \infty. \end{equation}

Again, we introduce an auxiliary weight p – to be determined later – and its conjugate exponent q. Split $2^{-m\alpha}=2^{-\frac{m\alpha}p}2^{-\frac{m\alpha}q}$ and use Hölder’s inequality to obtain

\begin{align*} B_{n,k}& \le 2^{n\alpha}\Bigl(\sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\alpha\beta}p}\Bigr)^{1/\beta} \times\Bigl(\sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\alpha\gamma}q}\mu^{\gamma}_{m,j}\Bigr)^{1/\gamma}. \end{align*}

The first sum is finite: since by hypothesis $\alpha\beta=\frac{\alpha \gamma}{\gamma-1} \gt 1$, there exists $1 \lt p \lt \frac{\alpha \gamma}{\gamma-1}$ and

\begin{align*} \sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\alpha\beta}p}&=\sum_{m\geq n} 2^{-\frac{m\alpha\beta}p} 2^{m-n} =2^{-n} \sum_{m\geq n} 2^{-m(\frac{\alpha\beta}p-1)}\simeq 2^{-n\frac{\alpha\beta}p}. \end{align*}

This implies that

\begin{align*} B_{n,k}^\gamma\lesssim 2^{n\alpha\frac{\gamma}{q}} \sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\alpha\gamma}{q}}\mu^{\gamma}_{m,j} \end{align*}

and we can conlude the proof of (9) as before:

\begin{align*} \sum_{n,k} B_{n,k}^\gamma &\lesssim\sum_{n,k} 2^{n\alpha\frac{\gamma}{q}} \sum_{\substack{m\geq n\\ j:T_{m,j}\subset Q_{n,k}}} 2^{-\frac{m\alpha\gamma}q} \mu^{\gamma}_{m,j} =\sum_{m,j} \mu^{\gamma}_{m,j} 2^{-m\alpha\frac{\gamma}{q}} \sum_{\substack{n\leq m\\ k: Q_{n,k} \supseteq T_{m,j} }} 2^{n\alpha\frac{\gamma}{q}}\\ &=\sum_{m,j} \mu^{\gamma}_{m,j} 2^{-m\alpha\frac{\gamma}{q}} \sum_{n\leq m} 2^{-n\alpha\frac{\gamma}{q}}\simeq \sum_{m,j} \mu^{\gamma}_{m,j} \lt +\infty. \end{align*}

(c) Here, we give a measure µ for which $\sum_{n,k}\mu_{n,k}^{1/(1-\alpha)} \lt +\infty$ but $\mathbb{P}(\Lambda_\mu\ \textrm{is } \alpha\textrm{-Carleson})=0$. Let

\begin{equation*} \mathrm{d}\mu(z)= \frac{\mathrm{d}m(z)}{(1-|z|^2)^{\alpha+1}\,\log\bigl(\frac {\mathrm{e}}{1-|z|^2}\bigr)}= \frac{\mathrm{d}\nu(z)}{(1-|z|^2)^{\alpha-1}\,\log\bigl(\frac {\mathrm{e}}{1-|z|^2}\bigr)}, \end{equation*}

which is the measure $\mu=\mu(\alpha+1,1)$ given in the family of examples of § 5. By a simple computation (see (11)),

\begin{equation*} \mu_{n,k}\simeq \frac{2^{-n(1-\alpha)}}n\qquad n\geq 1,\ k=0, \dots, 2^n-1, \end{equation*}

and therefore, since k ranges over 2ⁿ terms,

\begin{align*} \sum_{n,k}\mu_{n,k}^{1/(1-\alpha)}\simeq \sum_{n\geq 1} \frac{1}{n^{1/(1-\alpha)}} \lt +\infty. \end{align*}

On the other hand, letting $Y_{n,k}$ be as in the proof of part (b) (see (8)), we get

\begin{eqnarray*} \mathbb{E}(Y_{n,k})&=&2^{n\alpha} \sum_{m\ge n}2^{-m\alpha}\sum_{j:T_{m,j}\subset Q_{n,k}}\mu_{m,j} \simeq 2^{n\alpha} \sum_{m\ge n}2^{-m\alpha} 2^{m-n}\frac{2^{-(1-\alpha)m}}{m}\\ &=&2^{-(1-\alpha)n}\sum_{m\ge n}\frac{1}{m}=+\infty \end{eqnarray*}

Thus, the expected weight of any single Carleson window $Q_{n,k}$ is infinite and $\Lambda_{\mu}$ cannot be α-Carleson.

(d) One could think of considering a divergent series $\sum_{n,k}\mu_{n,k}^{\gamma}=+\infty$ such that $\sum_{n,k}\mu_{n,k}^{\gamma'} \lt +\infty$ for every $\gamma' \gt \gamma$, and then apply (b), showing that $\Lambda_{\mu}$ is α-Carleson when $\gamma' \lt \frac{1}{1-\alpha}$, i.e. when $\alpha \gt 1-\frac{1}{\gamma'}=\frac{\gamma'-1}{\gamma'}$. However, this does not yield the whole range $\alpha\in (0,1)$ for a fixed measure, as required by the statement.

In order to construct an example working for all $\alpha\in (0,1)$, we pick a measure µ supported in a Stolz angle of vertex 1, i.e. let, for $n\geq 1$,

\begin{equation*} \mu_{n,k}=\left\{\begin{array}{ll} \dfrac{1}{n^{1/\gamma}}&\textrm{if }k=0\\ \quad 0 &\textrm{if } k \gt 1. \end{array}\right. \end{equation*}

(We could equivalently take the measure $\tau(2,1/\gamma)$ given in § 5, Example 3). Then,

(10)\begin{equation} \sum_{n,k}\mu_{n,k}^{\gamma}=\sum_n \frac{1}{n}=\infty \end{equation}

but for every $\gamma' \gt \gamma$,

\begin{equation*} \sum_{n,k}\mu_{n,k}^{\gamma'}=\sum_n \frac{1}{n^{\gamma'/\gamma}} \lt +\infty. \end{equation*}

To prove that $\Lambda_\mu$ is almost surely α-Carleson we will argue as before. Set $Y_{n,k}$ as in the proof of (b) (see (8)) and follow the same steps to prove that

\begin{equation*} \mathbb{P}(Y_{n,k}\ge A)\lesssim B_{n,k}^A, \end{equation*}

where

\begin{equation*} B_{n,k}=\sum_{m\ge n} \sum_{j:T_{m,j}\subset Q_{n,k}} \mu_{m,j}2^{-(m-n)\alpha}. \end{equation*}

By construction, $B_{n,k}=0$ for all k > 0. On the other hand,

\begin{equation*} B_{n,0}=2^{n\alpha}\sum_{m\ge n}2^{-m\alpha}\mu_{m,0} =2^{n\alpha}\sum_{m\ge n}\frac{2^{-m\alpha}}{m^{1/\gamma}} \le \frac{1}{n^{1/\gamma}}. \end{equation*}

(Observe that this last expression is independent of α.) Hence,

\begin{equation*} \sum_{n,k}B_{n,k}^{\gamma'}=\sum_nB_{n,0}^{\gamma'} \le \sum_n \frac{1}{n^{\gamma'/\gamma}} \lt +\infty, \end{equation*}

and as in the proof of (b) the Borel–Cantelli lemma allows to conclude that Λ is almost surely α-Carleson.

4.1. Hardy (and Bergman) spaces

In this section, we characterize the measures µ for which the associated Poisson process $\Lambda_\mu$ is almost surely an interpolating sequence for the Hardy spaces.

Recall that a sequence $\Lambda=\{\lambda_n\}_{n\in\mathbb{N}}\subset\mathbb{D}$ is interpolating for

\begin{equation*} H^\infty=\bigl\{f\in H(\mathbb{D}) : \|f\|_\infty=\sup_{z\in D} |f(z)| \lt \infty\bigr\} \end{equation*}

whenever for every sequence of bounded values $\{w_n\}_{n\in\mathbb{N}}\subset\mathbb{C}$ there exists $f\in H^\infty$ such that $f(\lambda_n)=w_n$, $n\in\mathbb{N}$. According to a famous theorem by Carleson, Λ is $H^\infty$-interpolating if and only if it is separated and 1-Carleson [Reference Carleson8]. This characterization extends to all Hardy spaces

\begin{equation*} H^p=\Bigl\{f\in H(\mathbb{D}) : \|f\|_p=\sup_{r \lt 1}\Bigl(\int_0^{2\pi} |f(r\mathrm{e}^{\mathrm{i}t})|^p\, \frac{dt}{2\pi}\Bigr)^{1/p} \lt +\infty\Bigr\}\qquad 0 \lt p \lt \infty, \end{equation*}

for which the interpolation problem is defined in a similar manner (the data w_n to be interpolated should satisfy $\sum_n(1-|\lambda_n|^2)|w_n|^p \lt +\infty$, see e.g. [Reference Duren14, Chapter 9]).

The separation condition given in Theorem 1.2 implies immediately that $\Lambda_\mu$ is 1-Carleson, by Theorem 1.5; hence, the following result follows.

Theorem 4.1. Let $\Lambda_\mu$ be the Poisson process associated with a positive, σ-finite, locally finite measure µ. Then, for any $0 \lt p\leq \infty$,

\begin{equation*} \mathbb{P}\bigl(\Lambda_\mu\ {is\,\, {H^{{p}}}-interpolating}\bigr) = \left\{\begin{array}{ll} 1\quad &\textrm{if $\ \int_{\mathbb{D}} F_\mu^2(z)\, d\nu(z) \lt \infty$} \\ 0\quad &\textrm{if $\ \int_{\mathbb{D}} F_\mu^2(z)\, d\nu(z)=\infty$}. \end{array}\right. \end{equation*}

To complete the picture, we discuss zero sequences Λ for H^p, $0 \lt p\leq \infty$. These are deterministically characterized by the Blaschke condition $\sum_{\lambda\in\Lambda}(1-|\lambda|) \lt \infty$. Noticing that $\{\sum_{\lambda\in\Lambda_\mu}(1-|\lambda|) \lt \infty\}$ is a tail event and using Kolmogorov’s 0-1 law we get:

Proposition 4.2. Let $\Lambda_\mu$ be the Poisson process associated with a positive, σ-finite, locally finite measure µ. Then, for any $0 \lt p\leq \infty$,

\begin{equation*} \mathbb{P}\bigl(\Lambda_\mu {\,is\, a\, zero\, set\, for\, {H^{{p}}}}\bigr) = \left\{\begin{array}{ll} 1\quad &\textrm{if $\ \int_{\mathbb{D}} (1-|z|)\, d\mu(z) \lt \infty$} \\ 0\quad &\textrm{if $\ \int_{\mathbb{D}} (1-|z|)\, d\mu(z)=\infty$}. \end{array}\right. \end{equation*}

Observe that

\begin{equation*} \int_{\mathbb{D}} (1-|z|)\,\mathrm{d}\mu(z)=\sum\limits_{n,k} \int_{T_{n,k}} (1-|z|)\,\mathrm{d}\mu(z)\simeq\sum\limits_{n,k}2^{-n}\mu_{n,k} \end{equation*}

and that the condition is just

\begin{equation*} \mathbb{E}\bigl[\sum_{\lambda\in\Lambda_\mu}(1-|\lambda|)\bigr]=\mathbb{E}\bigl[\sum_{n,k}\sum_{\lambda\in T_{n,k}}(1-|\lambda|)\bigr] \simeq \sum\limits_{n,k}2^{-n}\mathbb{E}\bigl[X_{n,k}\bigr] =\sum\limits_{n,k}2^{-n}\mu_{n,k} \lt \infty. \end{equation*}

Observe also that $\sum_{k=0}^{2^n-1}\mu_{n,k}=\mu(A_n)$ for all $n\in\mathbb{N}$; hence,

\begin{equation*} \sum\limits_{n,k}2^{-n}\mu_{n,k}=\sum_n 2^{-n}\mu(A_n). \end{equation*}

Proof of Proposition 4.2

Denote $X_n=N_{A_n}=\sum_{k=0}^{2^n-1} X_{n,k}$ and denote $\mu_n=\mathbb{E}[X_n]=\mu(A_n)$.

Assume first that $\sum_{n} 2^{-n} \mu_n \lt +\infty$. Set $Y=\sum_n 2^{-n} X_n$ and observe that, by the independence of the different X_n,

\begin{align*} \mathbb{E}[Y]&=\sum_n 2^{-n} \mu_n \lt +\infty\qquad && {\rm Var}(Y)=\sum_n 2^{-2n}\mu_n \lt +\infty. \end{align*}

Then, by Markov’s inequality,

\begin{equation*} \mathbb{P}(Y\ge 2\mathbb{E}(Y))\le \frac{1}{2}. \end{equation*}

Since $\{Y=\infty\}$ is a tail event, Kolmogorov’s 0-1 law implies that $\mathbb{P}(Y=+\infty)=0$, and in particular, the Blaschke sum is finite almost surely.

Assume now that $\sum_{n} 2^{-n} \mu_n=+\infty$. Split the sum in two parts:

\begin{equation*} \sum_n 2^{-n}\mu_n=\sum_{n:\mu_n\le 2^n/n^2}2^{-n}\mu_n+ \sum_{n:\mu_n \gt 2^n/n^2}2^{-n}\mu_n. \end{equation*}

It is enough to consider the second sum, since the first one obviously converges. Since ${\rm Var}[X_n]=\mu_n$, Chebyshev’s inequality yields,

\begin{equation*} \mathbb{P}(X_n\le \frac{1}{2}\mu_n) =\mathbb{P}(X_n\le \mu_n-\frac{\mu_n}{2})\leq \mathbb{P}(|X_n- \mu_n|\geq \frac{\mu_n}{2}) \le \frac{4}{\mu_n}. \end{equation*}

Hence,

\begin{equation*} \sum_{n:\mu_n \gt 2^n/n^2} \mathbb{P}(X_n\le \frac{1}{2}\mu_n) \le \sum_{n:\mu_n \gt 2^n/n^2}\frac{4}{\mu_n}\le \sum_{n:\mu_n \gt 2^n/n^2}\frac{4n^2}{2^n} \lt +\infty. \end{equation*}

Now, by the Borel–Cantelli lemma, $X_n \gt \frac{1}{2}\mu_n$ for all but maybe a finite number of the n with $\mu_n \gt 2^n/n^2$; hence,

\begin{equation*} \sum_{n:\mu_n \gt 2^n/n^2} 2^{-n}X_n \succsim\frac{1}{2}\sum_{n:\mu_n \gt 2^n/n^2}2^{-n}\mu_n, \end{equation*}

which diverges, by hypothesis.

4.1.1. Remark. Interpolation in Bergman spaces

Interpolating sequences Λ for the (weighted) Bergman spaces

\begin{equation*} B_\alpha^p=\Bigl\{f\in H(\mathbb{D}) : \|f\|_{\alpha,p}^p=\int_{\mathbb{D}} |f(z)|^p (1-|z|^2)^{\alpha p-1}\,\mathrm{d}m(z) \lt \infty\Bigr\}, \end{equation*}

with $0 \lt \alpha$, $0 \lt p\leq \infty$ are characterized by the separation together with the upper density condition

\begin{equation*} D_+(\Lambda):=\limsup_{r\to 1^-} \sup_{z\in\mathbb{D}} \frac{\sum\limits_{1/2 \lt \rho(z,\lambda)\leq r}\,\log\frac 1{\rho(z,\lambda)}}{\,\log(\frac 1{1-r})} \lt \alpha \end{equation*}

(see [Reference Seip23] and [Reference Hedenmalm, Korenblum and Zhu17, Chapter 5] for both the definitions and the results).

Since every 1-Carleson sequence has density $D_+(\Lambda)=0$, the same conditions of Theorem 4.1 also characterize a.s. Bergman interpolating sequences, regardless of the indices α and p. Again, because of the big fluctuations of the Poisson process, the conditions required to have separation a.s. are so strong that they can only produce sequences of zero upper density.

Another indication of the big fluctuations of the Poisson process is the following. For the invariant measure $\mathrm{d}\nu(z)=\frac{\mathrm{d}m(z)}{(1-|z|^2)^2}$, which obviously satisfies $\nu_{n,k}\simeq 1$ for all n, k, it is not difficult to see that almost surely,

\begin{equation*} D_+(\Lambda_{\nu})=+\infty\qquad\textrm{and}\quad D_-(\Lambda_{\nu}):=\liminf_{r\to 1^-} \inf_{z\in\mathbb{D}} \frac{\sum\limits_{1/2 \lt \rho(z,\lambda)\leq r}\,\log\frac 1{\rho(z,\lambda)}}{\log(\frac 1{1-r})}=0. \end{equation*}

Therefore, there are way too many points for $\Lambda_\nu$ to be interpolating for any $B_\alpha^p$, but there are too few for it to be sampling, since these sets must have strictly positive lower density $D_-(\Lambda)$ (see [Reference Hedenmalm, Korenblum and Zhu17, Chapter 5]).

4.2. Interpolation in the Bloch space

We consider now interpolation in the Bloch space $\mathcal B$, consisting of functions f holomorphic in $\mathbb{D}$ such that

\begin{equation*} \|f\|_{\mathcal B}:=|f(0)|+\sup_{z\in\mathbb{D}}|f'(z)|(1-|z|^2) \lt +\infty. \end{equation*}

Since Bloch functions satisfy the Lipschitz condition $|f(z)-f(w)|\leq \|f\|_{\mathcal B}\, \delta(z,w)$, where $\delta(z,w)=\frac{1}{2}\,\log\frac{1+\rho(z,w)}{1-\rho(z,w)}$ denotes the hyperbolic distance, Nicolau and Bøe defined interpolating sequences for $\mathcal B$ as those $\Lambda=\{\lambda_n\}_{n\in\mathbb{N}}$ such that for every sequence of values $\{v_n\}_{n\in\mathbb{N}}$ with $\sup\limits_{n\neq m}\frac{|v_n-v_m|}{\delta(\lambda_n,\lambda_m)} \lt \infty$ there exists $f\in\mathcal B$ with $f(\lambda_n)=v_n$, $n\in\mathbb{N}$ [Reference Bøe and Nicolau5].

As explained in [Reference Bøe and Nicolau5], condition (b) can be replaced by:

• (b)’ for some $0 \lt \gamma \lt 1$ and C > 0, and for all Carleson windows Q(I),

  \begin{equation*} \#\bigl\{\lambda\in Q(I) : 2^{-(l+1)}|I| \lt 1-|\lambda| \lt 2^{-l}|I|\bigr\}\leq C 2^{\gamma l}\ , \qquad l\geq 0. \end{equation*}

In [Reference Seip24, Corollary 2], it is mentioned that it can also be replaced by:

• (b)” , there exist $0 \lt \gamma \lt 1$ and such that Λ is γ-Carleson.

In view of conditions (a) and (b)”, the following characterization of Poisson processes which are a.s. Bloch interpolating sequences follows from Theorems 1.2 and  1.5(b) (with $\gamma\in (2/3,1)$).

Theorem 4.3. Let $\Lambda_\mu$ be the Poisson process associated with a positive, σ-finite, locally finite measure µ. Then,

\begin{equation*} \mathbb{P}\bigl(\Lambda_\mu\,\, {\mathrm{is\,\, \mathcal B-interpolating}}\bigr) = \left\{\begin{array}{ll} 1\quad &\textrm{if $\ \int_{\mathbb{D}} F_\mu^3(z)\, d\nu(z) \lt \infty$} \\ 0\quad &\textrm{if $\ \int_{\mathbb{D}} F_\mu^3(z)\, d\nu(z)=\infty$}. \end{array}\right. \end{equation*}

Note. In case $\int_{\mathbb{D}} F_\mu^3(z)\,\mathrm{d}\nu(z)$, it is also possible to prove (b)’ directly, with the same methods employed in the proof of Theorem 1.5. It is enough to prove the estimate for dyadic arcs $I_{n,k}$, and for those

\begin{equation*} \#\bigl\{\lambda\in Q(I_{n,k}) : 2^{-(l+1)}|I_{n,k}| \lt 1-|\lambda| \lt 2^{-l}|I_{n,k}|\bigr\}\simeq \sum_{j : T_{n+l,j}\subset Q_{n,k}} X_{n+l, j}. \end{equation*}

In the above, the left-hand side corresponds essentially to the number of points in the layer $Q(I_{n,k})\cap A_{n+l}$. Thus, with $m=n+l$, (b)’ is equivalent to

\begin{equation*} \sup_{n,k}\sup_{m\geq n} 2^{-\gamma(m-n)} \sum_{j : T_{m,j}\subset Q_{n,k}} X_{m,j} \lt +\infty. \end{equation*}

Letting

\begin{equation*} Y_{n,k,m}=2^{-\gamma(m-n)} \sum_{j : T_{m,j}\subset Q_{n,k}} X_{m,j}\ ,\qquad \mathbb{E}[Y_{n,k,m}]=2^{-\gamma(m-n)} \sum_{j : T_{m,j}\subset Q_{n,k}} \mu_{m,j} \end{equation*}

and proceeding as in the first part of the proof of Theorem 1.5(a) we get (taking A = 3):

\begin{align*} \sum_{n,k}\sum_{m\geq n} \mathbb{P}\bigl(Y_{n,k,m}\geq 3\bigr)&\lesssim\sum_{n,k}\sum_{m\geq n} \Bigl[2^{-\gamma(m-n)} \sum_{j : T_{m,j}\subset Q_{n,k}} \mu_{m,j}\Bigr]^3\\ &\leq \sum_{n,k}\sum_{m\geq n} 2^{-3\gamma(m-n)} \sum_{j : T_{m,j}\subset Q_{n,k}} \mu_{m,j}^3\ 2^{2(m-n)}\\ &=\sum_{m,j} \mu_{m,j}^3 \sum_{n\leq m} \sum_{k : Q_{n,k} \supseteq T_{m,j}} 2^{-(3\gamma-2)(m-n)}. \end{align*}

For any $\gamma \gt 2/3$, this sum is bounded by $\sum_{m,j} \mu_{m,j}^3$, so we can conclude with the Borel–Cantelli lemma.

4.3. Interpolation in Dirichlet spaces

Our last set of results concerns interpolation in the Dirichlet spaces,

\begin{equation*} \mathcal D_\alpha=\bigl\{f\in H(\mathbb{D}) : \|f\|_{\mathcal D_\alpha}^2=|f(0)|^2+\int_{\mathbb D} |f'(z)|^2 (1-|z|^2)^{\alpha}\,\mathrm{d}m(z) \lt \infty\bigr\}, \end{equation*}

with $\alpha\in(0,1)$. The limiting case α = 1 can be identified with the Hardy space H ².

In these spaces, interpolating sequences are characterized by the separation and a Carleson type condition. This was initially considered by W.S. Cohn, see [Reference Cohn11]; we refer also to the general result [Reference Aleman, Hartz, McCarthy and Richter1]. While separation is a simple condition, that in our random setting is completely characterized by Theorem 1.2, the characterization of Carleson measures in these spaces is much more delicate. This was achieved by Stegenga using the so-called α-capacity [Reference Stegenga25]. In our setting, it is, however, possible to use an easier sufficient one-box condition that can be found in K. Seip’s book, see [Reference Seip24, Theorem 4, p. 38], which we recall here for the reader’s convenience.

The reader should be alerted that in Seip’s book the space $\mathcal D_{\alpha}$ is defined in a slightly different way and that the above statement is adapted to our definition.

For these spaces, Theorems 1.2 and 1.5 lead to less precise conclusions. Indeed, in view of Theorem 1.5(c),(d), we cannot hope for complete characterizations if we do not impose additional conditions on the measure µ .

Clearly, the condition $\int_{\mathbb{D}} F_\mu^2(z)\,\mathrm{d}\nu(z) \lt +\infty$ is also necessary in the case (b) (if the integral diverges, then $\Lambda_{\mu}$ is almost surely not separated).

Proof. (a) If $\sum_{n,k}\mu_{n,k}^2=+\infty$, then $\Lambda_{\mu}$ is almost surely not separated by Theorem 1.2; hence, it is almost surely not interpolating.

If $\sum_{n,k}\mu_{n,k}^2 \lt +\infty$, Theorem 1.2 shows again that the sequence $\Lambda_{\mu}$ is almost surely separated. By Seip’s theorem, it remains to show that $\Lambda_{\mu}$ is almost surely $\alpha'$-Carleson for some $\alpha' \lt \alpha$. Pick $1/2 \lt \alpha' \lt \alpha \lt 1$, so that $1/(1-\alpha') \gt 2$. Choosing $\gamma\in (2,1/(1-\alpha'))$, we get

\begin{equation*} \sum_{n,k}\mu_{n,k}^{\gamma}\lesssim \sum_{n,k}\mu_{n,k}^2 \lt +\infty, \end{equation*}

and by Theorem 1.5(b), we conclude that $\Lambda_{\mu}$ is almost surely $\alpha'$-Carleson.

(b) If $\alpha \lt 1/2$, then $1/(1-\alpha) \lt 2$ and the value γ given by the hypothesis satisfies $1 \lt \gamma \lt 2$. Therefore,

\begin{equation*} \sum_{n,k}\mu_{n,k}^2\lesssim \sum_{n,k}\mu_{n,k}^{\gamma} \lt +\infty, \end{equation*}

which allows to deduce from Theorem 1.2 that $\Lambda_{\mu}$ is almost surely separated.

Since the inequality $\gamma \lt 1/(1-\alpha)$ is strict, we also have $\gamma \lt 1/(1-\alpha')$ for some $\alpha' \lt \alpha$ sufficiently close to α. Again, Theorem 1.5(b) shows that $\Lambda_{\mu}$ is almost surely $\alpha'$-Carleson, and Seip’s theorem implies that $\Lambda_{\mu}$ is almost surely interpolating.

4.4. Additional remarks and comments

The above results show several applications of our Theorems 1.2 and 1.5, but they also give rise to many challenging questions. Is it possible to get a necessary counterpart of Theorem 1.5(b) under reasonable conditions on µ (more general than the class considered in § 5 below)? Is it possible to get precise statements when $\alpha=1/2$? Also, the case of the classical Dirichlet space seems to be largely unexplored for Poisson point processes, while the situation regarding interpolation, separation and zero-sets for the radial probabilistic model is completely known for all $\alpha\in [0,1]$ (see [Reference Bogdan6, Reference Chalmoukis, Hartmann, Kellay and Wick9]).

5.1. Radial measures

Define

\begin{equation*} \mathrm{d}\mu(a,b)(z)= \frac{\mathrm{d}m(z)}{(1-|z|^2)^{a} \,\log^b\bigl(\frac{\mathrm{e}}{1-|z|^2}\bigr)}= \frac{\mathrm{d}\nu(z)}{(1-|z|^2)^{a-2} \,\log^b\bigl(\frac{\mathrm{e}}{1-|z|^2}\bigr)}, \end{equation*}

where either a > 1, $b\in\mathbb{R}$, or a = 1 and $b\leq 1$ (so that $\mu(a,b)(\mathbb{D})=+\infty$).

Observe that

\begin{equation*} \mu(a,b)_{n,k}\simeq \frac{2^{-n(2-a)}}{n^b}\qquad n\geq 1,\ k=0, \dots, 2^n-1, \end{equation*}

and therefore, for γ > 0,

(11)\begin{equation} \sum_{n,k} \mu(a,b)_{n,k}^\gamma\simeq \sum_n 2^n \frac{2^{-n(2-a)\gamma}}{n^{b\gamma}}= \sum_n\frac{2^{-n[(2-a)\gamma-1]}}{n^{b\gamma}}. \end{equation}

Proof. (a) is immediate from Theorem 1.2 and (11) with $\gamma=M+1$ and the usual equivalence of Proposition 2.1.

(b) If $a\geq 2$ the series in (11) diverges for all γ > 1, thus by Theorem 1.5(a) $\Lambda_{a,b}$ is a.s. not 1-Carleson.

On the other hand, if a < 2, there exists γ such that $(2-a)\gamma-1 \gt 0$ (i.e, such that $\gamma \gt \frac 1{2-a})$. For that γ, the series in (11) converges, and we can conclude again by Theorem 1.5(a).

(c) Suppose first that $a \lt 1+\alpha$. As in the previous case, since $2-a \gt 1-\alpha$, there exists $\gamma\in(\frac 1{2-a},\frac 1{1-\alpha})$. For this γ, the series in (11) converges and we can apply Theorem 1.5(b).

If $a \gt 1+\alpha$ and $b\in\mathbb{R}$, then $\Lambda_{\mu(a,b)}$ contains in the mean more points than $\Lambda_{\mu(1+\alpha,1)}$ for which we have shown in Theorem 1.5(c) that it is almost surely not α-Carleson.

It thus remains the case $a=1+\alpha$. Again, when b = 1 – and thus also when b < 1 since then we have more points in the mean – the proof of Theorem 1.5(c) shows that the corresponding sequence is almost surely not α-Carleson.

Finally, suppose that $a=1+\alpha$ and b > 1. Recall from (8), the notation

\begin{equation*} \quad Y_{n,k}=2^{n\alpha}\sum_{m\geq n} 2^{-m\alpha} \sum_{j: T_{m,j}\subset Q_{n,k}} X_{m,j}. \end{equation*}

In the proof of Theorem 1.5(b), we have shown that

\begin{equation*} P(Y_{n,k}\ge A)\le B_{n,k}^A, \end{equation*}

where

\begin{equation*} B_{n,k}=\sum_{m\geq n} 2^{-(m-n)\alpha} \sum_{j: T_{m,j}\subset Q_{n,k}} \mu_{m,j}. \end{equation*}

From the explicit form of $\mu_{m,j}$, we get

\begin{equation*} B_{n,k}\simeq \sum_{m \ge n}2^{-(m-n)\alpha}\times 2^{m-n}\times \frac{2^{-m(2-a)}}{m^b} =2^{n(\alpha-1)}\sum_{m\ge n}\frac{2^{-m(\alpha+1-a)}}{m^b}, \end{equation*}

which converges exactly when $a \lt 1+\alpha$ or when $a=1+\alpha$ and b > 1, which is the case we are interested in here. In this situation, we get

\begin{equation*} B_{n,k}\simeq \frac{2^{-n(2-a)}}{n^b}. \end{equation*}

Clearly, when $A\ge 1/(2-a)=1/(1-\alpha)$, then $\sum_{n,k}B_{n,k}^A$ converges, and the Borel–Cantelli lemma shows that $Y_{n,k}\ge A$ can happen for an at most finite number of Carleson windows $Q_{n,k}$. Hence, $\Lambda_{\mu(a,b)}$ is a.s. α-Carleson.

5.2. Measures with a singularity on $\mathbb{T}$

Define now

\begin{equation*} \mathrm{d}\sigma(a,b)(z)= \frac{\mathrm{d}m(z)}{|1-z|^{a}\,\log^b\bigl(\frac{\mathrm{e}}{|1-z|}\bigr)}, \end{equation*}

where either a > 2, $b\in\mathbb{R}$, or a = 2 and $b\leq 1$ (so that $\sigma(a,b)(\mathbb{D})=+\infty$). Here,

\begin{align*} \sigma (a,b)_{n,k}=\sigma (a,b)(T_{n,k})\simeq \frac{2^{-2n}}{[(k+1)2^{-n}]^a\,\log^b\bigl(\frac{\mathrm{e}}{(k+1)2^{-n}}\bigr)},\quad n\in\mathbb{N},\ k=0,\ldots, 2^{n-1}. \end{align*}

Hence, for γ > 1,

(12)\begin{equation} \sum_{n,k} \sigma (a,b)_{n,k}^\gamma\simeq \sum_{n} {2^{-n\gamma(2-a)}} \sum_{k=1}^{2^n}\frac 1{\Big(k^a\,\log^b\bigl(\frac {\mathrm{e}}{k2^{-n}}\bigr)\Big)^{\gamma}}. \end{equation}

Let us examine the growth of the sum in k. For that, set

\begin{equation*} S_n(a,b,\gamma)=\sum_{k=1}^{2^n}\frac 1{k^{a\gamma}\,\log^{b\gamma}\bigl(\frac {\mathrm{e}}{k2^{-n}}\bigr)} \simeq \int_1^{2^n}\frac{\mathrm{d}x}{x^{a\gamma}\,\log^{b\gamma}\bigl(\frac {\mathrm{e}}{x2^{-n}}\bigr)}. \end{equation*}

The change of variable $t=\log\bigl(\frac {\mathrm{e}}{x2^{-n}}\bigr)$ leads to

\begin{equation*} S_n(a,b,\gamma)\simeq \int_{\log(2^n\mathrm{e})}^1 \left(\frac{\mathrm{e}^t}{\mathrm{e}2^n}\right)^{a\gamma-1}\frac{-\mathrm{d}t}{t^{b\gamma}} =\frac{2^{-n(a\gamma-1)}}{\mathrm{e}^{a\gamma -1}}\int_{1}^{\log(2^ne)}\mathrm{e}^{t(a\gamma-1)}\frac{\mathrm{d}t}{t^{b\gamma}}. \end{equation*}

Our standing assumption being a > 2 or a = 2 and $b\le 1$, we only need to consider these two cases. In both cases, $\mathrm{e}^{t(a\gamma-1)}/t^{b\gamma}\to +\infty$ as $t\to+\infty$, and the last integral behaves essentially as the value in the upper bound of the integration interval

\begin{equation*} \int_{1}^{\log(2^n\mathrm{e})}\mathrm{e}^{t(a\gamma-1)}\frac{\mathrm{d}t}{t^{b\gamma}} \simeq \frac{2^{n(a\gamma-1)}}{n^{b\gamma}}. \end{equation*}

Hence,

\begin{equation*} S_n(a,b,\gamma)\simeq \frac{1}{n^{b\gamma}}, \end{equation*}

and

(13)\begin{equation} \sum_{n,k} \sigma (a,b)_{n,k}^\gamma\simeq \sum_n 2^{-n\gamma(2-a)}\times \frac{1}{n^{\gamma b}}=\sum_n \frac{2^{-n\gamma(2-a)}}{n^{\gamma b}}. \end{equation}

We are now in a position to prove the following result.

5.3. Measures in a cone

Given a point $\zeta\in\mathbb{T}$, consider the Stolz region

\begin{equation*} \Gamma(\zeta)=\bigl\{z\in \mathbb{D} :\frac{|\zeta-z|}{1-|z|} \lt 2\bigr\}. \end{equation*}

We discuss the previous measures restricted to $\Gamma(\zeta)$. With no loss of generality, we can assume that ζ = 1. Let thus

\begin{equation*} \mathrm{d}\tau(a,b)(z)= \chi_{\Gamma(1)}(z)\,\mathrm{d}\mu(a,b)(z)=\chi_{\Gamma(1)}(z)\frac{\mathrm{d}m(z)}{(1-|z|^2)^{a}\,\log^b\bigl(\frac {\mathrm{e}}{1-|z|^2}\bigr)}, \end{equation*}

where now either a > 2 and $b\in\mathbb{R}$ or a = 2 and $b\leq 1$ (so that $\nu(a,b)(\mathbb{D})=+\infty$). Since, in $\Gamma(1)$, the measures $\mathrm{d}\mu(a,b)$ and $\mathrm{d}\sigma(a,b)$ behave similarly, we could replace $\mathrm{d}\mu(a,b)$ by $\mathrm{d}\sigma(a,b)$ in the definition of $d\tau(a,b)$ and obtain the same results.

Observe that $\nu(a,b)_{n,k}$ is non-zero only for a finite number N of k at each level n, and that for those k

\begin{equation*} \tau(a,b)_{n,k}\simeq \frac{2^{-n(2-a)}}{n^b}\qquad n\geq 1,\ k=0, \dots, N. \end{equation*}

Hence,

(14)\begin{equation} \sum_{n,k} \tau(a,b)_{n,k}^\gamma\simeq \sum_n \frac{2^{-n(2-a)\gamma}}{n^{b\gamma}}, \end{equation}

which is exactly the same estimate as in (13) and thus immediately leads to the same result as Proposition 5.2. This might look surprising since $\sigma(a,b)$ (and a fortiori $\mu(a,b)$) puts infinite mass outside $\Gamma(\zeta)$ (actually outside Stolz angles at ζ with arbitrary opening).