Weakly convergent series

Question

Consider sequence $\{x_n\}_{n = 0}^\infty $ in complex Banach space X. Suppose that for all $z \in \mathbb{C}$ such that $|z| < 1$ the series $\sum_{n = 0}^\infty z^nx_n$ is weakly convergent. How to show that the series $\sum_{n = 0}^\infty z^nx_n$ is norm convergent for all $z \in \mathbb{C}$ such that $|z| < 1$?

Answer 1

Let $|z| < 1$ and choose $|z| < |w| < 1$. By assumption, the series $\sum_n w^n x_n$ is weakly convergent. In particular, $w^n x_n \to 0$ weakly, so that $(w^n x_n)_n$ is a bounded sequence (essentially, this is the uniform boundedness principle).

But then $$ \|z^n x_n\| = \left|\frac{z}{w}\right|^n\|w^n x_n\| \leq C \cdot \left|\frac{z}{w}\right|^n, $$ with $\sum_n |z/w|^n < \infty$ (geometric series). Hence, the series $\sum_n z^n x_n$ is (absolutely) convergent and thus convergent, since $X$ is complete.

EDIT: Note that the sequence $(x_n)_n$ itself is not necessarily bounded; consider e.g. $X = \Bbb{R}$ and $x_n = n$.