# A formal series in the Banach space $c_0$

Let $(e_n)$ be the standard unit vector basis for $c_0$, let $x_1=e_1$, $x_n=e_n-e_{n-1}$ when $n\gt1$.

Prove that the formal series $\sum x_n$ is not weakly subseries convergent.

-
Please don't formulate your requests in the imperative mode. Somewhat unrelated, your profile "no university ask me to do some research or even help their fool students." is arrogant, off-putting, and ill-advised. –  Alex B. Jan 6 '11 at 13:04

More seriously: Note that if there is a weak limit of a sequence in $c_{0}$, it also is its weak$^{\ast}$-limit in $\ell^{\infty} = (c_{0})^{\ast\ast}$ (here $c_{0}$ is viewed as a subspace of $\ell^{\infty}$ via the canonical inclusion). Since the weak$^{\ast}$-topology is Hausdorff, it suffices to exhibit a subseries which is weak$^{\ast}$-convergent to some $s \in \ell^{\infty} \smallsetminus c_{0}$. The subseries $s_{k} = \sum_{n = 1}^{k} x_{2n} = (-1,1,-1,1,\cdots,-1,1,0,0,0,\cdots)$ weak$^{\ast}$-converges to $s = (-1,+1,\cdots)$: For all $(y_{n}) \in \ell^{1}$ and all $\varepsilon > 0$ there is $N$ such that $\sum_{n \geq N} |y_{n}| < \varepsilon$, therefore $|\langle s - s_{k}, (y_{n}) \rangle_{\ell^{\infty}, \ell^{1}}|< \varepsilon \quad \text{for k \geq N}$ by the Hölder inequality.
Consider the series $\sum_{n = 1}^{\infty} x_{n}$, that is to say the sequence $s_{k} = \sum_{n = 1}^{k} x_{n}$. A subseries is of the form $\sum_{j = 1}^{\infty} x_{n_{j}}$ for some subsequence $(x_{n_{j}})$ of $(x_{n})$. A series is "subseries convergent" if every subseries converges. "Weakly subseries convergent" means that the series and each of its subseries converges in the weak topology. Admittedly, the terminology is somewhat unfortunate. The standard proof of Orlicz-Pettis uses that a sequence is weakly convergent in $\ell^{1}$ only if it's norm-convergent, see e.g. Diestel-Uhl. –  t.b. Jan 7 '11 at 14:16
@Strongart: I don't know exactly what you mean. The subseries $(s_{k})$ only converges in the (much) larger space $\ell^{\infty}$ with the weak$^{\ast}$-topology. In the space $c_{0}$ with the weak topology it diverges. –  t.b. Jan 8 '11 at 11:08