This is a standard consequence of perturbation theory for analytic families of operators [@Kato].

We have defined the bounded linear operators $T_n,R_n:\mathcal{B}\to\mathcal{B}$ given by $$T_nv=1_YL^n(1_Yv),\enspace n\ge0, \qquad
R_nv=1_YL^n(1_{\{\varphi=n\}}v)=R(1_{\{\varphi=n\}}v),\enspace n\ge1.$$ The power series $$T(z)=\sum_{n=0}^\infty T_n z^n,\enspace z\in{{\mathbb D}},\qquad
R(z)=\sum_{n=1}^\infty R_n z^n,\enspace z\in\bar{{\mathbb D}},$$ are analytic on the open unit disk ${{\mathbb D}}$, and $R(z)$ is continuous on $\bar{{\mathbb D}}$ by (H1). We have the usual relation $T_n=\sum_{j=1}^n T_{n-j}R_j$ for $n\ge1$, and it follows that $T(z)=I+T(z)R(z)$ on ${{\mathbb D}}$. Hence $T(z)=(I-R(z))^{-1}$ on ${{\mathbb D}}$. By (H2)(ii), $T(z)$ extends continuously to $\bar{{\mathbb D}}-\{1\}$ via the formula $T(z)=(I-R(z))^{-1}$.

\[prop-PQ\] There exists $\epsilon,C>0$ such that $\|T(z)-(1-\lambda(z))^{-1}P(z)\|\le C$ for $z\in \bar{{\mathbb D}}\cap B_\epsilon(1)$, $z\neq1$, and $\|T(z)\|\le C$ for $z\in\bar{{\mathbb D}}\setminus B_\epsilon(1)$.

Choose $\epsilon>0$ so that the family of simple eigenvalues $\lambda(z)$ is defined on $\bar{{\mathbb D}}\cap B_\epsilon(1)$. For $z\in \bar{{\mathbb D}}\cap B_\epsilon(1)$, we can write $R(z)=\lambda(z)P(z)+R(z)Q(z)$. If in addition $z\neq1$, then we have (in an obvious notation) $$T(z)=(1-\lambda(z))^{-1}P(z)+(I-R(z))^{-1}Q(z).$$ By (H2), the second term is uniformly bounded in the operator norm, and $T(z)$ is uniformly bounded for $z\in\bar{{\mathbb D}}\setminus B_\epsilon(1)$.

Strategy of the proof of Theorem \[thm-conv\] {#sec-strategy}
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Our aim is to compute the operators $T_n$ defined above. Most of the analysis is carried out on 