A Banach space has an absolutely convergent basis $\iff$ it is topologically isomorphic to $\ell^1$

Question

$X$: Banach space. Then,

$X$ has an absolutely convergent basis $\{x_n\}$ $\iff$ $X$ is tolopogically isomorphic to $\ell^1$

This is Exercise 4.14 in A Basis Theory Primer, Christopher Heil. I am just reading this book and not a homework.

Proof:

$(\implies)$ Let $\{x_n\}$ be an absolutely convergent basis for $X$. Let $T\colon \ell^1\to X$ be defined by $$y\mapsto Ty:=\sum_{n=1}^\infty\frac{y_n}{\|x_n\|_X}x_n,$$ for $y=(y_n)\in\ell^1$. We show that $T$ is linear continuous and $\mathrm{range}(T)=X$. Then, as a consequence of the Open Mapping Theorem the assertion will follow.

(linearity) Take $y^{(1)},y^{(2)}\in\ell^1$ arbitrarily. Let $P_Ny:=(y_1,\dotsc,y_n,0,0,\dotsc)$. Then, we have $\alpha_N:=T(P_Ny^{(1)}+P_Ny^{(2)})=T(P_Ny^{(1)})+T(P_Ny^{(2)})$, but it is easy to check both $T(y^{(1)}+y^{(2)})$ and $T(y^{(1)})+T(y^{(2)})$ are the limit of $\alpha_N$, and thus, checking the scaler multiplication similarly, the linearity follows.

(continuity) We have $\|Ty\|_X\le \sum_{n=1}^\infty |y_n|\cdot1=\|y_n\|_{\ell^1}$.

(onto) Let $x\in X$. Since $\{x_n\}$ is an absolutely convergent basis, we have the representation $x=\sum_{n}a_n(x)x_n$ with a coefficient functional $a_n$ such that $\sum_{n}|a_n(x)|\|x_n\|<\infty$. Letting $y_n:=a_n(x)\|x_n\|$ we have $y\in \ell^1$ and $Ty=x$.

($\impliedby$) With the standard basis $\{\delta_n\}$ where only the $n$-th component of $\delta_n=(0,\dotsc,0,1,0,\dotsc)$ is non-zero, clearly $y=\sum_{n}\delta_ny_n\in\ell^1$ is absolutely convergent, i.e., $\{\delta_n\}$ is an absolutely convergent basis.

Let $\mathcal{T}\colon\ell^1\to X$ be an homeomorphism, and let $x_n:=\mathcal{T}y_n$. Then, $\{x_n\}$ is a basis (Lemma 4.18, Heil's book).

Since $\mathcal{T}^{-1}x\in\ell^1$, because $\delta_n$ is a basis there exist a unique scalars such that $\mathcal{T}^{-1}x=\sum_n c_n \delta_n\in \ell^1$. Since $\delta_n$ is an absolutely convergent basis with $\|\delta_n\|=1$ we have $\sum_n|c_n|<\infty$. From the continuity of $\mathcal{T}$ we have $$ x=\sum_n{c_n}\mathcal{T}\delta_n=\sum_n{c_n}x_n. $$ But $$\sum_n|{c_n}|\|\mathcal{T}\delta_n\|\le \sup_{y\in\ell^1\, \|y\|=1}\|\mathcal{T}y\|\sum_n|{c_n}|<\infty,$$ and thus indeed $\{x_n\}$ is an absolutely convergent basis.

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I think the proof is alright, but what I don't really like about it is that I used Lemma 4.18 (p.140), which appears after Exercise 4.14 (p.139) in the book. I don't think "I wonder if I could show this just by using knowledge prior to Exercise 4.14" is a good way of asking a question, as the book isn't available to everybody even though it is what I wish to ask. So, I wonder if I could show this using, results from Section 2, say Heil, Contents (pdf) where results from elementary functional analysis are introduced.

Answer 1

You might be satisfied by the following alternative proof of Lemma 4.18, which uses only elementary results from functional analysis.

Let $T : X \to Y$ be a topological isomorphism. Suppose $\{x_n\}_{n = 1}^{\infty}$ is a Schrauder basis for $X$. Given $x \in X$, it holds that $\sum_{n = 1}^{\infty} \alpha_n x_n$ for some $\{\alpha_n\}_{n = 1}^{\infty} \subset \mathbb{C}$. Since for all $y \in Y$, there exists an $x \in X$ such that $y = Tx$, it follows that $$ y = T(\sum_{n = 1}^{\infty} \alpha_n x_n) = \sum_{n = 1}^{\infty} \alpha_n Tx_n, $$ which shows that $\{Tx_n\}_{n = 1}^{\infty}$ is a Schrauder basis for $Y$.