zero for $y\in(0,1/\delta]$.

Also, by Potter’s bounds for any $\beta'\in(\beta,1)$, $$\begin{aligned}
A_{(ii)} & 
 \le 2|g_z|_\infty \int_0^{\pi}e^{-\sigma y}
\sigma^{-\beta}h(\sigma/\theta)\ell(1/\theta)^{-1}\,d\sigma 
 \ll |g_z|_\infty \int_0^{\pi}\sigma^{-\beta'}\,d\sigma\ll |g_z|_\infty=o(1).\end{aligned}$$ Hence $A_{(ii)}$ is negligible relative to $A_{(i)}$ and we obtain $1-\lambda(z)\gg 
\theta\ell(1/\theta)\theta^{\beta-1}=\ell(1/\theta)\theta^\beta$ completing the proof.

It is convenient in the next result (and crucial in Section \[sec-one\]) to discuss the real part of an operator. We recall that the operators $T_n$ are defined on the real Banach space $\mathcal{B}$. Passing to the complexification, there is a natural conjugation $u+iv\mapsto u-iv$ on $\mathcal{B}$. Given an operator $A:\mathcal{B}\to\mathcal{B}$, define the conjugate $\bar A:\mathcal{B}\to\mathcal{B}$ by setting $\bar A v = \overline{A\bar v}$, and the real part ${\operatorname{Re}}A=\frac12(A+\bar A)$. In the case of the operator $T(z)=\sum_{n=0}^\infty T_nz^n$, this coincides with the definitions $\overline{T(z)}=T(\bar z)$ and ${\operatorname{Re}}T(z)=\sum_{n=0}^\infty T_n{\operatorname{Re}}(z^n)$.

\[cor-Fourier\] $T_n=\widehat T_n=
\frac{1}{\pi}{\operatorname{Re}}\int_0^\pi T(e^{i\theta})e^{-in\theta}\,d\theta$ for all $n\ge0$.

Since $T(z)=\sum_{j=0}^\infty T_jz^j$ is analytic on the open unit disk ${{\mathbb D}}$, $T_n =\frac{1}{2\pi}\rho^{-n}\int_0^{2\pi}T(\rho e^{i\theta}) 
e^{-in\theta}\,d\theta$, for all $\rho\in(0,1)$.

By Proposition \[prop-PQ\], $T(z)=O(1)$ on $\bar{{\mathbb D}}\setminus B_\epsilon(1)$. Further, on $\bar{{\mathbb D}}\cap B_\epsilon(1)$, $T(z)=(1-\lambda(z))^{-1}P(z)+O(1)\ll (1-\lambda(z))^{-1}+O(1)$. By Lemma \[lem-zestimate\], $\|T(\rho e^{i\theta})\|\ll 
\ell(1/\theta)^{-1}\theta^{-\beta}$ for $z=\rho e^{i\theta}\in
\bar{{\mathbb D}}$ uniformly in $\rho$. Since $\ell(1/\theta)^{-1}\theta^{-\beta}$ is integrable, it follows from the dominated convergence theorem (as $\rho\to1$) that $T_n= \frac{1}{2\pi}\int_0^{2\pi}T(e^{i\theta}) e^{-in\theta}\,d\theta=
\widehat T_n$. 