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Let X be a Banach space. If $D \subset X^*$ is (weak*ly or strongly?) dense, then does $f(x_n) \to f(x)$ $\forall f \in D$ imply that $x_n \to x$ weakly?

My thoughts: If $g_m \to g$ in the dual, then we can write:

$|g(x) - g(x_n)| \leq |g(x) - g_m(x)| + |g_m(x) - g_m(x_n)| + |g_m(x_n) - g(x_n)|$, where the middle term always vanishes for a fixed $m$ as $n \to \infty$ by assumption, though this is a moving target. The first term will vanish as $m \to \infty$ in either the weak* topology or the strong topology on the dual, and the latter term vanishes if the dual topology was the strong topology and the sequence $x_n$ was bounded in norm.

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and the latter term vanishes if the dual topology was the strong topology and the sequence $x_n$ was bounded in norm.

That is the crucial thing. The sequence needs to be bounded to deduce weak convergence from pointwise convergence on a norm-dense subset.

If $(x_n)$ is a bounded sequence in $X$, it is an equicontinuous sequence as a sequence of functions $X^\ast \to \mathbb{K}$ (via the canonical embedding $X\hookrightarrow X^{\ast\ast}$), and for equicontinuous sequences, pointwise convergence on a dense subset implies pointwise convergence everywhere, that is here weak convergence.

If the sequence $(x_n)$ is not bounded, there still can be dense subsets, even linear subspaces, $D\subset X^\ast$ such that $f(x_n) \to f(x)$ for all $f\in D$, but of course an unbounded sequence cannot be weakly convergent.

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