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I'm solving Conway's Functional Analysis. (Weakly Compact Operator) The problem is

Show that every bounded sequence in $c_0$ has a weakly Cauchy subsequence, but not every weakly Cauchy seqence in $c_0$ converges.

I solved second proposition via setting $x_n = (1,1,1, \cdots, 1, 0,0,\cdots)$ (for $i$-th component, $x_{ni} =1$ and then all of them is zero) but I cannot solve first proposition.

I showed that $\langle x^*, x_n \rangle$ has convergence subsequence for each $x^*$, but I need to find "weakly" cauchy subsequence. Can you help me?

probafds123
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2 Answers2

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If $(x^{(n)}_i)_i$ is a bounded sequence in $c_0$ then there exists $M <\infty$ such that $|x^{(n)}_i| \leq M$ for all $i$ and $n$. By Cantor's diagonal procedure there exist a subsequence $(n_k)$ such that $\lim_{k \to \infty} x^{(n_k)}_i=x_i$ exists for each $i$. It remains to show that $\sum a_i x^{(n_k)}_i \to \sum _i a_ix_i$ for all $(a_i) \in \ell^{1}$. For this note that there exist $N$ such that $\sum_{i>N} |a_i x^{(n_k)}_i - a_ix_i|<\epsilon$ for all $k$. Can you finish?

Kavi Rama Murthy
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Another approach for the first part is to note that any ball in $c_0$ embeds in the corresponding ball in $\ell^\infty$, which is both compact and metrizable in the weak topology. Moreover a sequence in the former is Cauchy iff it is Cauchy in the latter.

Ruy
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