# Mazur's Weak Basis Theorem

It is the Exercise 1.1 in Topics in Banach Space Theory by Albiac and Kalton to prove Mazur's Weak Basis Theorem, which states that every weak basis in a Banach space $X$ is a Schauder basis, where weak basis is defined as follows:

A sequence $(e_n)_n$ in a Banach space $X$ is called a weak basis for $X$ if for every $x\in X$ there is an unique sequence of scalars $(a_n)$ such that $x=\sum_{k=1}^\infty a_ke_k$ in the weak topology.

I tried to show, to no avail, that $\lVert\sum^n a_kx_k\rVert\rightarrow\lVert x\rVert$ so that it would immediately follow that $\sum^n a_ne_n\overset{\lVert\cdot\rVert}{\rightarrow}x$. I'd like suggestions on how to proceed.

Define the closure of the linear hull of the $(x_k)$ in the strong topology by $$C = \operatorname{cl \ span}\{x_k, k\in\mathbb N\}.$$ It is convex and closed. We have to show that $C=X$ under the condition of weak basis. Let $x\in X\setminus C$ be given. Then we can separate $x$ and $C$ by $f\in X^*$: There is $\epsilon>0$ such that $$f(x) + \epsilon \le f(y) \quad\forall y\in C.$$ Since $C$ is a linear space, $f(y)=0$ for all $y\in C$. In particular, $f(x_k)=0$ for all $k$. By assumption, there is a sequence $(a_k)$ such that $\sum_{k=1}^n a_kx_k$ converges weakly to $x$ for $n\to\infty$. This implies $$f(x) = \lim_{n\to\infty} f(\sum_{k=1}^n a_kx_k) = \lim_{n\to\infty} \sum _{k=1}^n a_k f(x_k)=0,$$ which is a contradiction to $f(x)+\epsilon<0$.