Mazur's Weak Basis Theorem

Question

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.

Answer 1

This answer is not correct, see the comment by Nate Eldregde. I do not see how to repair it.

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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 $f(x)+\epsilon<0$.

Answer 2

The authors suggest in the exercise that we try to imitate the proof of Theorem 1.1.3. Let us follow this kind suggestion. Let $(e_k)_{k=1}^{\infty }$ be a weak basis. For $n \in \mathbb{N}$, define the linear maps $e_n^{\#}\colon X\to \mathbb{K}$ by $e_n^{\#}(x)=a_n$ whenever $x=\sum_{k=1}^{\infty}a_k e_k$ in the weak topology. Note that $e_n^{\#}(e_m)=\delta _{nm}$, where $\delta _{nm}$ denotes the usual Kronecker delta, so if we show that:

1. $e_n^{\#}$ are bounded
2. $x=\sum_{n=1}^\infty e_n^\#(x) e_n $ in the usual norm.

Then, it will follow that $(e_k)_{k=1}^{\infty }$ is a Schauder basis.

Let us prove 1:

For this purpose, it clearly suffices to show that the natural projections $$ S_n\bigg( \sum_{k=1}^{\infty}a_ke_k \bigg) =\sum_{k=1}^{n}a_ke_k $$ are bounded for all $n\ge 1$. By hypothesis $S_nx\xrightarrow{w}x$ for all $x \in X$, so the sequence $(S_nx)_{n=1}^{\infty }$ is norm bounded, and it makes sense to define $|||x|||=\sup_{n\ge 1} \|S_nx\|$. It is direct to check that $|||{\cdot }|||$ is a norm. Also, since $x^{*}(S_nx)\to x^{*}(x)$ for all $x^{*} \in X^{*}$, we have $$ |x^{*}(x)|\leq \sup_{n\ge 1}|x^{*}(S_nx)| $$ which implies $$ \|x\|=\sup_{\|x^{*}\|\le 1}|x^{*}(x)|\le \sup_{\|x^{*}\|\le 1}\sup_{n\ge 1} |x^{*}(S_nx)|=|||x|||. $$ We have thus shown that $\|{\cdot }\|\le |||{\cdot }|||$.

Our next goal is to prove that the space $(X,|||{\cdot }|||)$ is complete. To do this, let $(x_n)_{n=1}^{\infty }$ be a Cauchy sequence in $|||{\cdot }|||$. By the inequality shown above, $(x_n)$ is Cauchy in $\|{\cdot }\|$, and so it converges to some $x \in X$. We will show that $(x_n)$ also converges to $x$ in $|||{\cdot }|||$. Fixed any $k \in \mathbb{N}$, $\|S_kx_n-S_kx_m\|\le |||x_n-x_m|||$, so the sequence $(S_kx_n)$ converges to a certain $y_k \in X$ in the original norm $\|{\cdot}\|$. Also, by its very definition, $S_kx_n \in \text{span}\{e_1,\ldots ,e_k\}$, and since finite-dimensional subspaces are closed, $y_k \in \text{span}\{e_1,\ldots ,e_k\}$. Another important property of finite-dimensional spaces is that any linear operator is continuous, so we must have $$e_j^{\#}(x_n)=e_j^{\#}(S_kx_n)\to e_j^{\#}(y_k)\quad\text{for }1\le j\le k.$$

Now we claim that $y_k\xrightarrow{w} x$. To show this, fix $x^{*}\in X^{*}$ and $\varepsilon >0$. Choose $n \in \mathbb{N}$ such that $|||x_n-x_m|||<\varepsilon$ for every $m\ge n$ (this is possible because $(x_n)$ is Cauchy in $|||{\cdot}|||$), and choose $k_0\in \mathbb{N}$ such that $|x^{*}(x_n)-x^{*}(S_kx_n)|<\varepsilon$ for all $k\ge k_0$ (this is possible because $S_kx_n\xrightarrow{w}x_n$). Then for $k\ge k_0$: \begin{align} |x^{*}(y_k)-x^{*}(x)|&\le \lim_{m\to \infty }|x^{*}(S_kx_m)-x^{*}(S_kx_n)|\\&+|x^{*}(S_kx_n)-x^{*}(x_n)|+\lim_{m\to \infty } |x^{*}(x_n)-x^{*}(x_m)|\\ &\le \|x^{*}\|\lim_{m\to \infty } |||x_m-x_n|||+ \varepsilon +\|x^{*}\|\lim_{m\to \infty }|||x_m-x_n|||\\ &\le (2\|x^{*}\|+1)\varepsilon . \end{align} So $x^{*}(y_k)\to x^{*}(x)$ for all $x^{*} \in X^{*}$. In other words, $$ y_k=\sum_{j=1}^{k} e_j^{\#}(y_k)e_j\xrightarrow{w} x, $$ and since the expansion of $x$ in the weak basis $(e_j)$ is unique, it follows that $y_k=S_kx$. With this we can finally show that \begin{align} |||x_n-x|||&=\sup_{k\ge 1} \|S_kx_n-S_kx\|\\ &=\sup_{k\ge 1} \|S_k x_n-y_k\|\\ &=\sup_{k\ge 1}\lim_{m\to \infty } \|S_kx_n-S_kx_m\|\\ &\le \limsup_{m\to \infty } |||x_n-x_m|||\xrightarrow[n\to \infty ]{}0, \end{align} and therefore $(X,|||{\cdot }|||)$ is complete.

We are almost done. We have already shown that the identity map $id_X\colon (X,\|{\cdot}\|)\to (X,|||{\cdot }|||)$ is continuous (this is precisely the inequality we proved at the beginning), and so it is an isomorphism by the Bounded Inverse Theorem. Hence there exists a constant $C>0$ such that $|||x|||\le C\|x\|$, that is, such that $\|S_k\|\le C$ for all $k\in\mathbb{N}$. And this finishes the proof.

Let us prove 2:

We first notice $X=\overline{span\{e_n\:|\: n \in \mathbb{N}\}}$. Indeed, if $x\in X\setminus \overline{span\{e_n\:|\: n \in \mathbb{N}\}} $, by geometric Hahn Banach, there would exist $f$ that separates them strictly. Therefore we arrive at the following contradiction:

$$0\not=f(x)=f(\sum_{k=1}^\infty a_k e_k)=\sum_{k=1}^\infty a_k f(e_k)=0$$

Because it is dense, we may find a sequence $x_n\rightarrow x$ such that $x_n \in span\{e_1,...,e_n\}$ (going to a subsequence if necessary and repeating terms if necessary). Clearly:

$$\lVert S_n(x)-x\rVert = \lVert S_n(x)-S_n(x_n)+x_n-x\rVert \leq (\sup_n \lVert S_n \rVert+1)\lVert x-x_n \rVert\rightarrow 0$$

Indeed, $S_n(x)$ converges weakly to $x$, so it is norm bounded and by Banach Steinhaus, $\sup_n \lVert S_n\rVert<\infty$. Thus, $x=\sum_{k=1}^{\infty}a_k e_k$ in the usual norm.