Let $X$ be a normed space. If $(x_n)$ is a Cauchy sequence with $x_n \overset{w}{\rightarrow} 0$, then $x_n \to 0$

Question

The following is the problem 3.35 of the book Banach Space Theory from Fabian, Habala, et al.

Let $X$ be a normed space. If $(x_n)$ is a Cauchy sequence with $x_n \overset{w}{\rightarrow} 0$, then $x_n \to 0$.

$x_n \overset{w}{\rightarrow} 0$ means that $(x_n)$ is a weakly null sequence.

There is the following hint: $x_n \in x_m + \epsilon B_X$ and $x_m + \epsilon B_X$ is weakly closed.

My points:

1. Since $(x_n)$ is Cauchy, for a given $\epsilon$, there exists $n_0$ such that $x_n \in x_m + \epsilon B_X$ for all $n \geq n_0$.

2. The set $x_m + \epsilon B_X$ is weakly closed by Mazur theorem, since $B_X$ is norm closed and convex.

However, I cannot see how this joint with the fact that $(x_n)$ is weakly null imply that $x_n \to 0$.

Answer 1

By embedding $X$ into its completion, without loss of generality, we may assume that $X$ is a Banach space. Since $(x_{n})$ is Cauchy, there exists $x\in X$ such that $x_{n}\rightarrow x$. Suppose the contrary that $x\neq0$. By Hahn-Banach theorem, there exists $f\in X^{\ast}$ such that $f(x)\neq0$. Since $x_{n}\stackrel{w}{\rightarrow}0$, we have $f(x_{n})\rightarrow f(0)=0$. On the other hand, $x_{n}\rightarrow x$ implies that $f(x_{n})\rightarrow f(x)$, so $f(x)=0$, which is a contradiction.