Since $T(\bar z)=\overline{T(z)}$, we obtain the expression $\frac{1}{\pi}{\operatorname{Re}}\int_0^\pi T(e^{i\theta})e^{-in\theta}\,d\theta$.

Convergence for $\beta\in(\frac12,1)$ {#sec-half}
=====================================

In this section, we prove Theorem \[thm-conv\] for $\beta\in(\frac12,1)$.

\[lem-GL\] Let $\beta\in(\frac12,1)$. Let $n\ge1$, $a\in[1,n]$. Then for any $\beta'\in(0,\beta)$, $$\ell(n) n^{1-\beta}
\int_{a/n}^\pi T(\theta) e^{-in\theta}\,d\theta\ll a^{-(2\beta'-1)}.$$ If $\ell$ is asymptotically constant, then the result holds with $\beta'=\beta$.

By Lemma \[lem-AD\](c), we have the estimate $\|T(\theta)\|\ll 
\ell(1/\theta)^{-1}\theta^{-\beta}$. The proof uses this fact together with Proposition \[prop-H1\], and follows Garsia & Lamperti [@GarsiaLamperti62 p. 231]. We give the details partly for completeness and partly because we want to make explicit certain estimates that will be used in Section \[sec-second\].

First, write $$I=\int_{a/n}^\pi T(\theta) e^{-in\theta}\,d\theta
=-\int_{(a+\pi)/n}^{\pi+\pi/n}T(\theta-\pi/n) e^{-in\theta}\,d\theta,$$ so $$2I=\int_{a/n}^\pi T(\theta) e^{-in\theta}\,d\theta
-\int_{(a+\pi)/n}^{\pi+\pi/n}T(\theta-\pi/n) e^{-in\theta}\,d\theta=I_1+I_2+I_3,$$ where $$\begin{aligned}
I_1  & = \int_\pi^{\pi+\pi/n}T(\theta-\pi/n)e^{-in\theta}\,d\theta, \qquad
I_2  = \int_{a/n}^{(a+\pi)/n}T(\theta-\pi/n)e^{-in\theta}\,d\theta, \\
I_3 & =\int_{(a+\pi)/n}^\pi \{ T(\theta)-T(\theta-\pi/n)\}e^{-in\theta}\,d\theta.\end{aligned}$$ Clearly, $I_1\ll 1/n$, while $$\begin{aligned}
I_2 & \ll \int_{a/n}^{(a+\pi)/n}\ell(1/\theta)^{-1}\theta^{-\beta}\,d\theta \ll
\ell(n)^{-1}n^{-(1-\beta)}\int_a^{a+\pi}[\ell(n)/\ell(n/\sigma)]\sigma^{-\beta}
\,d\sigma \\ & 
\ll \ell(n)^{-1}n^{-(1-\beta)}\int_a^{a+\pi}\sigma^{-\beta'} \,d\sigma 
=\ell(n)^{-1}n^{-(1-\beta)}a^{1-\beta'}\{(1+\pi/a)^{1-\beta'}-1\} \\ &
\ll \ell(n)^{-1}n^{-(1-\beta)}a^{-\beta'}.\end{aligned}$$ By the resolvent identity and Proposition \[prop-H1\] (with $m(x)=\ell(x)$), $$\begin{aligned}
I_3 & \ll \int_{(a+\pi)/n}^\pi \|T(\theta)\|\|T(\theta-\pi/n)\|\|R(\theta)-R(\theta-\pi/n)\|\,d\theta \\
& \ll \ell(n/\pi)n^{-\beta}\int_{(a+\pi)/n}^\pi \ell(1/\theta)^{-1}\ell(1/(\theta-\pi/n))^{-1}\theta^{-\beta} (\theta-\pi/n)^{-\beta}\,d\theta \\
& = \ell(n/\pi)n^{-\beta}\int_{a/n}^{\pi-\pi/n} \ell(1/(\theta+\pi/n))^{-1}\ell(1/\theta)^{-1} (\theta+\pi/n)^{-\beta}\theta^{-\beta}\,d\theta.\end{aligned}$$ By Potter’s bounds, $\ell(1/(\theta+\pi/n))^{-1}\ll\ell(1/\theta)^{-1}$ for $n\theta\ge 1$. Hence, $$\begin{aligned}
I_3 & \ll \ell(n)n^{-\beta}\int_{a/n}^\pi \ell(1/\theta)^{-2}\theta^{-2\beta} \,d\theta =
\ell(n)^{-1}n^{-(1-\beta)}\int_{a}^{n\pi} [\ell(n)/\ell(n/\sigma)]^2\sigma^{-2\beta} \,d\sigma
\\
& \ll 