Let $X$ be a Banach space and $D\subset X^{*}$ be a $\sigma(X^{*},X)$ -dense subset and $(x_{n})_{n}\subset X$ be a bounded sequence such that ${\displaystyle \lim_{n\to \infty}}\left\langle x_{n},x^{*}\right\rangle $ exists for all $x^{*}\in D$. Does $(x_{n})_{n}$ is $\sigma(X,X^{*})$ - convergent?
Topological question
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general-topology
functional-analysis
weak-convergence
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1Is $\sigma(X^*,X)$ a standard notation? I have not seen it... – 2017-01-04
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1@Ian Yes, [it's a standard notation](https://en.wikipedia.org/wiki/Topology_of_uniform_convergence#G-topologies_on_the_continuous_dual_induced_by_X). – 2017-01-04
1 Answers
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It does not follow that $(x_n)$ is $\sigma(X,X^{\ast})$-convergent.
It would follow if $D$ were assumed norm-dense or $\sigma(X^{\ast}, X^{\ast\ast})$-dense in $X^{\ast}$ (these conditions are almost equivalent; of course there are - in the infinite-dimensional case - $\sigma(X^{\ast}, X^{\ast\ast})$-dense subsets that aren't norm-dense, but the existence of $\lim \langle x_n, x^{\ast}\rangle$ follows for $x^{\ast} \in \operatorname{span} D$, and a linear subspace is norm-dense if and only if it is $\sigma(X^{\ast}, X^{\ast\ast})$-dense). Thus the $\sigma(X,X^{\ast})$-convergence follows for reflexive $X$.
For a counterexample, let $X = \ell^1(\mathbb{N})$, and - identifying $\ell^1(\mathbb{N})^{\ast}$ with $\ell^{\infty}(\mathbb{N})$ as usual - $D = c_{00}$ the space of sequences with only finitely many nonzero terms. Then with $x_n = e_n$, we have $\langle x_n, x^{\ast}\rangle \to 0$ for every $x^{\ast} \in D$, but $(e_n)$ is not $\sigma(\ell^1(\mathbb{N}), \ell^{\infty}(\mathbb{N}))$-convergent.
$\ell^1(\mathbb{N})$ has the Schur property, every weakly convergent sequence is norm-convergent, thus it's well-suited for counterexamples concerning weak convergence.