 Solution

For every finite scalar sequence $a=(a_n)$ with $|a_n|\le 1$, define a bounded linear functional on the continuous dual space $X^{\ast }$ byFor each fixed $f\in X^{\ast }$, the hypothesis giveswhere the supremum is over all finite choices and the phases $a_n$ may be chosen to align the complex numbers $f(x_n)$.

The dual $X^{\ast }$ is a Banach space even when $X$ is incomplete. The Uniform boundedness principle therefore gives a constant $C$ such thatFor each $N$, choose $a_n$ for $n\le N$ so that $a_{n}f(x_n)=|f(x_n)|$. ThenLetting $N\to \infty$ provesThis is the uniform bound from pointwise absolute summability of dual evaluations.