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.



Weak basis of normed linear space
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 $|||\cdot|||$.


