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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.

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1 Answer

up vote 6 down vote accepted

Here's an overkill argument: According to what is usually called the Orlicz-Pettis theorem, a series in a normed space is weakly subseries convergent (if and) only if it is subseries convergent in the norm. Your series clearly isn't norm-convergent.

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.

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Can you define briefly the term "weakly subseries convergence"? –  TCL Jan 7 '11 at 14:01
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
Yes,I know the Orlicz-Pettis theorem and these definitions,but I want to find the direct proof.Your discussion is end with <ε,that shows it is convergent,but I want to a example which is not convergent. –  Strongart Jan 8 '11 at 10:55
@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
I am sorry that my English is poor.Now I know you want to remark that the subseries is convergent in the l^∞.Thank you. –  Strongart Jan 9 '11 at 10:55
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