1. ## Uniform boundedness Theorem

Space c_0

let y= (eta_j), eta_j element of C (complex numbers) be such that

sum psi_i eta_j converges for every x=(psi_j0 element of c_0, where

c_0 contained in l^{\infty} is the subspace of all complex sequences converging to zero.

Show that sum |eta_j| < \infty. Use the Uniform Boundedness theorem.

2. Let $\displaystyle a=(a_i)$ be the sequence with the given properties, and let $\displaystyle x=(x_i)$ be a general member of $\displaystyle c_0$.

Consider the sequence of linear functionals $\displaystyle T_n$ on $\displaystyle c_0$ given by $\displaystyle T_nx=\sum_{i=1}^n a_ix_i$.

Using the fact that $\displaystyle ||T_n||=\sup_{||x||=1}|T_nx|$, the norm of $\displaystyle T_n$ is easily shown to be $\displaystyle ||T_n||=\sum_{i=1}^n|a_i|$.

The supremum is attained at $\displaystyle x\in c_0$ where $\displaystyle |x_i|=1$ with $\displaystyle a_ix_i=|a_i|$ for $\displaystyle 1\leq i\leq n$, and where $\displaystyle x_i=0$ for $\displaystyle i>n$. Note that $\displaystyle x\in c_0$ and $\displaystyle ||x||=\sup |x_i|=1$.

Now $\displaystyle \sup_n |T_nx|<\infty$ for all $\displaystyle x\in c_0$, since by hypothesis the series $\displaystyle \sum_{i=1}^\infty a_ix_i$ converges for all such $\displaystyle x$.

By the uniform boundedness principle, $\displaystyle \sup_n||T_n||<\infty$. Therefore $\displaystyle \sum_{i=1}^\infty|a_i|<\infty$, which means that $\displaystyle a\in l^1$.