\ell(n)^{-1}n^{-(1-\beta)}\int_{a}^{n\pi} \sigma^{-2\beta'} \,d\sigma
\ll \ell(n)^{-1}n^{-(1-\beta)}a^{-(2\beta'-1)}.\end{aligned}$$ Altogether, we obtain $\ell(n)n^{1-\beta}I \ll n^{-\beta}+a^{-\beta'}+a^{-(2\beta'-1)}\ll
a^{-(2\beta'-1)}$ as required.

\[lem-GLeigen\] Let $\beta\in(0,1)$. Let $n\ge1$, $a\in(0,\epsilon n)$. Then $$\lim_{a\to\infty}\lim_{n\to\infty}\ell(n) 
n^{1-\beta}\int_0^{a/n}(1-\lambda(\theta))^{-1} e^{-in\theta}\,d\theta = 
d_\beta',$$ where $d_\beta'=i{{\textstyle}\int}_0^\infty e^{-i\sigma}\sigma^{-\beta}\,d\sigma/{{\textstyle}\int}_0^\infty e^{i\sigma}\sigma^{-\beta}\,d\sigma$.

This is identical to [@GarsiaLamperti62 Lemma 3.4.1] and we give the proof only for completeness. By Lemma \[lem-AD\], we can write $(1-\lambda(\theta))^{-1}=c_\beta^{-1}\ell(1/\theta)^{-1}\theta^{-\beta}h(\theta)$ where $c_\beta=-i{{\textstyle}\int}_0^\infty e^{i\sigma}\sigma^{-\beta}\,d\sigma$ and $\lim_{\theta\to0} h(\theta)=1$. Hence $$\begin{aligned}
\int_0^{a/n}(1-\lambda(\theta))^{-1} e^{-in\theta}\,d\theta
 & =n^{-1}\int_0^a (1-\lambda(\sigma/n))^{-1}e^{-i\sigma}\,d\sigma \\ &
=c_\beta^{-1}n^{-(1-\beta)}\int_0^a e^{-i\sigma}\sigma^{-\beta}\ell(n/\sigma)^{-1}h(\sigma/n)\,d\sigma, \end{aligned}$$ so that $$\ell(n)n^{1-\beta}\int_0^{a/n}(1-\lambda(\theta))^{-1} e^{-in\theta}\,d\theta 
=
c_\beta^{-1}\int_0^a e^{-i\sigma}\sigma^{-\beta}[\ell(n/\ell(n/\sigma)]h(\sigma/n)\,d\sigma.$$ For fixed $a$, it follows from the dominated convergence theorem that $$\lim_{n\to\infty}
\ell(n)n^{1-\beta}\int_0^{a/n}(1-\lambda(\theta))^{-1} e^{-in\theta}\,d\theta 
=c_\beta^{-1}\int_0^a e^{-i\sigma}\sigma^{-\beta}\,d\sigma,$$ and the result follows.

[Theorem \[thm-conv\], $\beta\in(\frac12,1)$.]{} By Section \[sec-Fourier\], $T_n=\frac{1}{\pi}{\operatorname{Re}}\int_0^\pi T(\theta)e^{-in\theta}\,d\theta$. Let $$\begin{aligned}
& D(a,n) =
\int_0^\pi T(\theta)e^{-in\theta}\,d\theta-
\int_0^{a/n}(1-\lambda(\theta))^{-1}Pe^{-in\theta}\,d\theta \\
& \qquad  = \int_0^{a/n}\bigl\{T(\theta)-(1-\lambda(\theta))^{-1}P\bigr\}
e^{-in\theta}\,d\theta+
\int_{a/n}^\pi T(\theta)Pe^{-in\theta}\,d\theta,\end{aligned}$$ so $D(a,n)\ll a/n+\ell(n)^{-1}n^{-(1-\beta)}a^{-(2\beta'-1)}$ by Lemma \[lem-AD\](b) and Lemma \[lem-GL\]. Hence $\lim_{a\to\infty}\lim_{n\to\infty}\ell(n)n^{1-\beta}D(a,n)=0$. By Lemma \[lem-GLeigen\], $\lim_{a\to\infty}\lim_{n\to\infty}\ell(n)n^{1-\beta}T_n=
\frac{1}{\pi}{\operatorname{Re}}d_\beta'=d_\beta$. The result follows since $T_n$ is independent of $a$.

Convergence for $\beta=1$ {#sec-one}
=========================

In this section, we prove Theorem \[thm-conv\] in the case $\beta=1$. There are several differences from the case $\beta\in(\frac12,1)$. First, $T(e^{i\theta})\not\in L^1$; instead it is shown below that ${\operatorname{Re}}T(e^{i\theta})$ is integrable.

Estimating ${\operatorname{Re}}\{(1-\lambda(z))^{-1}\}$ is slightly easier than in Section \[sec-Fourier\] but estimating ${\operatorname{Re}}T(z)$ is harder since ${\operatorname{Re}}\{(1-\lambda(z))^{-1}(P(z)-P)\}$ is 