# $T:X\to X$ is absolutely summing if and only if $\sum_{n=1}^\infty \|Tx_n\|<\infty$ whenever $\sum_{n=1}^\infty x_n$ is unconditionally convergent.

The following proposition is extracted from Topics in Banach Space Theory by Albiac and Kalton.

Proposition $8.2.2$: An operator $T:X\to X$ is absolutely summing if and only if $\sum_{n=1}^\infty \|Tx_n\|<\infty$ whenever $\sum_{n=1}^\infty x_n$ is unconditionally convergent.

In the book, the authors do not provide a proof of the above theorem and they say the proof is routine. However, I do not know how to prove it. Any hint is appreciated. If anyone knows of any reference containing a proof of the proposition, may I have it?

Definitions: Let $X$ and $Y$ be Banach spaces. $T$ is said to be absolutely summing if there is a constant $C$ such that for all choices of $(x_k)_{k=1}^n$ in $X,$ $$\sum_{k=1}^n\|Tx_k\|\leq C\sup\bigg\{ \sum_{k=1}^n|x^*(x_k)|:x^*\in X^*, \|x^*\|\leq 1 \bigg\}.$$

We say that $\sum_{n=1}^\infty x_n$ is unconditionally convergent if one of the following holds:

$(1)$ $\sum_{n=1}^\infty x_{\pi(n)}$ converges for every permutation $\pi$ of $\mathbb{N}.$

$(2)$ The series $\sum_{k=1}^\infty x_{n_k}$ converges for every increasing sequence $(n_k)_{k=1}^\infty$

$(3)$ The series $\sum_{n=1}^\infty \varepsilon_n x_n$ converges for every choice of signs $(\varepsilon_n).$

$(4)$ For every $\varepsilon>0,$ there exists $n$ such that if $F$ is a finite subset of $\{n+1,n+2,...\},$ then $$\bigg\| \sum_{j\in F} x_j \bigg\|<\varepsilon.$$

I'm going to assume real scalars for convenience (any complex Banach space is also a real Banach space). We need another equivalent form of unconditional convergence, namely

(5) For every $\varepsilon>0$, there exists $n$ such that $$\sum_{k>n} |x^*(x_k)| < \epsilon$$ for all $x^*\in X^*$ with $\|x^*\|\le 1$.

(Remark: thinking of each $x$ as a function on the unit ball of $X^*$, we can interpret (5) as saying that unconditional convergence is uniform absolute convergence of the series of these functions.)

Let's show that (5) is equivalent to (4). First, (5) implies (4) because for any finite set $F$ as in (4) we have $$\bigg\| \sum_{j\in F} x_j \bigg\| = \sup_{\|x^*\|=1} x^*\bigg( \sum_{j\in F} x_j\bigg) = \sup_{\|x^*\|=1} \sum_{j\in F} x^*(x_j) \le \sup_{\|x^*\|=1} \sum_{k>n} |x^*(x_k)|$$

Conversely, suppose (4) holds. Given $\varepsilon$, apply (4) to $\varepsilon/3$ and get $n$ from it. Given $x^*$ with $\|x^*\|\le 1$ and a number $N>n$, let $$F^+ = \{n < k\le N : x^*(x_k)>0\}, \quad F^- = \{n < k \le N : x^*(x_k)<0\}$$ According to (4), $$\sum_{k\in F^+} |x^*(x_k)| = \sum_{k\in F^+} x^*(x_k) = x^*\left(\sum_{k\in F^+} x_k\right) < \varepsilon/3$$ Similarly, $\sum_{k\in F^-} |x^*(x_k)|<\varepsilon/3$. We thus obtain $$\sum_{k=n+1}^N |x^*(x_k)|<2\varepsilon/3$$ Since $N$ was arbitrary, $$\sum_{k>n} |x^*(x_k)|\le 2\varepsilon/3 < \varepsilon.$$

### Absolutely summing operators

Suppose $T$ is absolutely summable and $\sum x_n$ is unconditionally convergent. Given $\varepsilon>0$, let $n$ be as in property (5) above. Then for every $N>n$ we have $$\sum_{k=n+1}^N \|Tx_k\| \leq C\sup\bigg\{ \sum_{k=n+1}^N |x^*(x_k)|:x^*\in X^*, \|x^*\|\leq 1 \bigg\} < C\varepsilon$$
Since $\varepsilon$ is arbitrarily small, this shows $\sum \|Tx_k\|$ converges.

Conversely, suppose $T$ is not absolutely summing. Then for each $j\in\mathbb{N}$ there is a finite sequence of vectors $(x_k)_{k=1}^n$ such that $$\sum_{k=1}^n\|Tx_k\| > 2^j \sup\bigg\{ \sum_{k=1}^n|x^*(x_k)|:x^*\in X^*, \|x^*\|\leq 1 \bigg\}$$ Multiply all these vectors by a suitable constant so that $\sum_{k=1}^n\|Tx_k\| =1$, hence
$$\sup\bigg\{ \sum_{k=1}^n|x^*(x_k)|:x^*\in X^*, \|x^*\|\leq 1 \bigg\} < 2^{-j}$$ Repeat the above for $j=1, 2, 3, \dots$, getting a finite sequence of vectors each time, and put all these vectors into one series: (vectors from $j=1$), (vectors from $j=2$), (vectors from $j=3$), etc. By construction this series satisfies property (5) and therefore is unconditionally convergent. Also by construction, the series of $\|Tx_j\|$ is divergent, as each $j$-th block contributes $1$ to it.

• Thanks for your answers. May I know whether this kind of proof is considered routine? – Idonknow May 20 '18 at 23:07
• With sufficient experience, it is. – user357151 May 20 '18 at 23:28