On weakly unconditional convergence

Question

Suppose that $X$ is a Banach space and $T\colon X\to X$ is a bounded linear operator. If the series $\sum\limits_{n=1}^{\infty} x_n$ ($x_n\in X$) is weakly unconditional convergence then is it true that the series $\sum\limits_{n=1}^{\infty} T(x_n)$ is also weakly unconditional convergence? If so how to prove?

Answer 1

The answer is yes.

Assume that $w-\sum_{n=1}^\infty x_n$ exists. If $f \in X'$ then $f \circ T \in X'$ as well so

$$f\left(\sum_{n=1}^NTx_n\right) = (f\circ T)\left(\sum_{n=1}^Nx_n\right)\xrightarrow{N\to\infty}(f\circ T)\left(w-\sum_{n=1}^\infty x_n\right)$$

so $w-\sum_{n=1}^\infty Tx_n$ exists.

To obtain the result for weak unconditional convergence, apply the proof above to $w-\sum_{n=1}^\infty x_{\sigma(n)}$ for every permutation $\sigma : \mathbb{N} \to \mathbb{N}$.