Wiener algebra and generating function

Question

Let $W^+$ be the algebra of function $f:\overline{\mathbb{D}}\mapsto \mathbb{C}$ such that $f(z)=\sum _{n\geq 0}c_nz^n, \forall |z|\leq 1,$ equipped with the norm $\|f\|_{W^+}=\sum _{n\geq 0}|c_n|$. Let $0\leq p_n \leq 1$ and $P(z)=\sum _{n\geq 0}p_nz^n$ and assume that there is a function $R\in W^+$ such that $R(z)\neq 0$ for all $z$ and $R(1)=m$. If $(1-z)P(z)R(z)=1$ for $|z|<1$ how can we deduce that $p_n \to 1/m$ ? Since $R(z)\neq 0$ and the spectrum of $W^+$ can be identified with $\overline{\mathbb{D}}$, $R$ is invertible. So we can extend $(1-z)P(z)$ to a function in $W^+$. Then $\lim _{z\to 1} (1-z)P(z)=1/m$. But I don't see how we can deduce informations about $(p_n)$. In order to have information about the coefficient of a given generating function we can derive it, but in this case it doesn't work.

Answer 1

Write $\frac{1}{R(z)}=\sum b_nz^n$. The equation $\frac{1}{R(z)}=(1-z)P(z)$ shows that $b_n=p_n-p_{n-1}$ for $n\geq 1$. Since by hypothesis $p_0=1=b_0$ we have $p_n=\sum _{k=0}^nb_k$ so $\lim p_n=\sum _{k=0}^\infty b_k=1/m$.

This result is related to Erdos-Feller-Pollard's theorem.