# weakly cauchy sequence is norm compact

A sequence $(x_n)$ is weakly cauchy if for every $x^*\in X^*$, $(x^*(x_n))$ converges. Let $c$ denote the space of convergent functions.

Theorem: A weakly cauchy sequence is norm-bounded.

Proof: Let $(x_n)$ be a weakly cauchy sequence in $X$. Define $T:X^*\to c$ by $T(x^*)=(x^*(x_n))$. By the closed graph theorem, $T$ is continuous, so $||T||=\text{sup}_n||x_n||<\infty$.

Question 1: Why is the graph of $T$ is closed in $X^*\times c$?

Question 2: Why is $||T||=\text{sup}_n||x_n||$? I have $||T||=\text{sup}_{||x^*||=1}||Tx^*||_{\infty}=\text{sup}_{||x^*||=1}||(x^*(x_n))||_{\infty}\leq\text{sup}_{||x^*||=1}||x^*|||(x_n)||_{\infty}=||(x_n)||_{\infty}$

but I don't see the other inequality.

• Why not just prove it using the Uniform boundedness principle? Much easier. – Aweygan Sep 12 '17 at 1:34

If $x \in X$ then $$||x||=\sup\{|x^*(x)|:||x^*|| \leq 1\}$$
Note that $$||T||=\sup\{||T(x^*)||:||x^*||=1\}=\sup\{||T(x^*)||:||x^*||\leq1\}$$