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Let $K$ be a compact Hausdorff space and $X$ be a Banach space. By the Riesz-Singer representation theorem, we know that there exists a linear isometry from $C(K,X)^*$ onto $rcabv(K,X^*)$, the Banach space of all regular, countably additive, Borel $X^*$-valued measures in $K$ with bounded variation, equipped with the norm $$\|\mu\| = |\mu|(K), \forall \mu \in rcabv(K,X^*).$$ (Here, $|\mu|$ denotes the variation of $\mu$)

Suppose $(\mu_n)_{n \in \mathbb{N}}$ is a weak$^*$-null sequence in $rcabv(K,X^*)$, that is, for each $f \in C(K,X)$, $$\lim_{n \to \infty} \int_K f d\mu_n =0.$$

Let $U$ be an open set in $K$ and $x \in X$. Is it true that $$\lim_{n \to \infty} \mu_n(U)(x) = 0?$$

Vinícius Morelli
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1 Answers1

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If I missed nothing, this is not even true in much simpler situations:

Let $K = (0,1)$ and $X = \mathbb{R}$. Then, the sequence of Dirac differences $$\delta_0 - \delta_{1/n}$$ converges weak-* to $0$, but $$(\delta_0 - \delta_{1/n})(\{0\}) = 1.$$

gerw
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  • That is correct; however, ${0}$ is not open. – Vinícius Morelli Mar 04 '16 at 21:53
  • I apologize, your example does settle the question. It suffices to notice that $$(\delta_0 - \delta_{\frac{1}{n}}) (]0,1[) = -1, \forall n \geq 2.$$

    Thank you very much for the example. Again, I apologize for not noticing and accepting your answer earlier.

     – Vinícius Morelli Mar 05 '16 at 18:22
  • I missed the assumption of an open set. Luckily, you fixed it :) – gerw Mar 05 '16 at 18:33