Weak basis of normed linear space Let X be an infinite dimensional normed linear space. A sequence $(e_n)$ in $X$ is called a basis if for every $x \in X$ there is a unique sequence of scalars $(a_n)$ such that $x=\sum_n a_n e_n $ (equality in norm).
We say that $(e_n)$ is a Weak basis if for every $x \in X$ there is a unique sequence of scalars $(a_n)$ such that $\sum_{k=1}^n a_k e_k $ converges weakly to x as $n \to \infty$.
I am looking for a reference to the following theorem, which i have seen in several places, yet was not able to find a proof, nor to come up with one myself: if $(e_n)$ is a weak basis, then it is a basis.
 A: We will show that if $F$ is a Banach space, and $\{f_n\}_{n\in\mathbb{N}}$ is a weak basis, it is a basis. For $n\in\mathbb{N}$ define $P_{n}$ on $F$ by $P_{n}f=\sum\limits_{k=1}^{n}a_{k}e_{k}$, where $f=\sum\limits_{n\in\mathbb{N}}a_{n}e_{n}$ (weak limit).
Define a norm $|||\cdot|||$ on $F$ by $|||f|||=\sup\limits_{n\in\mathbb{N}} \|P_{n}f\|$. Since the norm is weakly lower semi-continuous, it follows that $|||\cdot|||\ge\|\cdot\|$. It is enough to show that $|||\cdot|||$ is equivalent to $\|\cdot\|$. From the Closed Graph theorem, this amounts to showing that $|||\cdot|||$ is a complete norm.
To that end, let $\left\{f_{n}\right\}_{n\in\mathbb{N}}\subset F$ be such that $|||f_{n}|||\le\frac{1}{2^{n}}$, for every $n\in\mathbb{N}$. We need to show that $\sum\limits_{n\in\mathbb{N}}f_{n}$ converges to $f$ in $|||\cdot|||$, where $f=\sum\limits_{n\in\mathbb{N}}f_{n}$ in $\|\cdot\|$. For every $m,n\in\mathbb{N}$ we have $\|P_{m}f_{n}\|\le|||f_{n}||| \le\frac{1}{2^{n}}$, and so $\sum\limits_{n\in\mathbb{N}}P_{m}f_{n}$ converges to $g_{m}$. Note that $P_{m}f_{n}\in span\left\{e_{1},...,e_{m}\right\}$, for every $m,n\in\mathbb{N}$, and so $g_{m}\in span\left\{e_{1},...,e_{m}\right\}$. Since $P_{k}$ is continuous on $span\left\{e_{1},...,e_{m}\right\}$, for $m,k\in \mathbb{N}$, it follows that $P_{k}g_{m}=\sum\limits_{n\in\mathbb{N}}P_{k}P_{m}f_{n}=\sum\limits_{n\in\mathbb{N}}P_{k\wedge m}f_{n}=g_{k\wedge m}$. Hence, there are $\left\{a_{n}\right\}_{n\in\mathbb{N}}\subset \mathbb{R}$ such that $g_{m}=\sum\limits_{k=1}^{m}a_{k}e_{k}$.
If we show that $g_{m}\xrightarrow[m\to\infty]{w}f$, then due to uniqueness of decomposition, $g_{m}=P_{m}f$, for every $m\in\mathbb{N}$. Recall that $P_{m}f_{n}\xrightarrow[m\to\infty]{w}f_{n}$, for every $n\in\mathbb{N}$. For $\nu\in S_{E^{*}}$ and $l\in\mathbb{N}$ let $k\in\mathbb{N}$ be such that $\left|\nu\left(f_{n}-P_{m}f_{n}\right)\right|<\frac{1}{l2^{l}}$, for every $n\in\overline{1,l}$ and $m\ge k$. Then,
we have that $$\left|\nu\left(f-g_{m}\right)\right|= \left|\nu\left(\sum\limits_{n\in\mathbb{N}}f_{n}-\sum\limits_{n\in\mathbb{N}}P_{m}f_{n}\right)\right|\le \left\|\sum\limits_{n>l}f_{n}\right\|+ \sum\limits_{n=1}^{l}\left|\nu\left(f_{n}-P_{m}f_{n}\right)\right|+ \left\|\sum\limits_{n>l}P_{m}f_{n}\right\|\le \frac{3}{2^{l}}.$$
Since $\nu$ and $l$ were chosen arbitrarily, it follows that $g_{m}\xrightarrow[m\to\infty]{w}f$. Finally, for every $m,l\in\mathbb{N}$ we have that $$\left\|P_{m}\left(f-\sum\limits_{n=1}^{l}f_{n}\right)\right\|=\left\|g_{m}-\sum\limits_{n=1}^{l}P_{m}f_{n}\right\|= \left\|\sum\limits_{n\in\mathbb{N}}P_{m}f_{n}-\sum\limits_{n=1}^{l}P_{m}f_{n}\right\|\le \sum\limits_{n>l}\left\|P_{m}f_{n}\right\|\le\frac{1}{2^{l}},$$
and so $|||f-\sum\limits_{n=1}^{l}f_{n}|||=\sup\limits_{m\in\mathbb{N}}\left\|P_{m}\left(f-\sum\limits_{n=1}^{l}f_{n}\right)\right\|\le\frac{1}{2^{l}}$. Thus, $\sum\limits_{n\in\mathbb{N}}f_{n}$ converges to $f$ in $|||\cdot|||$.
