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Let $X$ be a compact Hausdorff space, the Riesz-Markov-Kakutani theorem states that the topological dual of $C(X)$ is the space $M(X)$ of regular countably additive complex Borel measure on $X$ equipped with the weak* topology. It also states that given $\mu \in M(X)$ as linear functional its norm is equal to the total-variation norm of $\mu$ as complex measure.

Does this mean that the topology induced by the total-variation norm coincides with the weak* topology?

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  • $\begingroup$ See Kavi's answer (I was about to say the same thing when I saw it had already been said). $\endgroup$ Aug 8, 2019 at 14:08
  • $\begingroup$ As it seems i was reading some lecture notes full of typos and got confused with a few statements and identifications... Thank you all for your help. And yes @mathworker21, indeed it was very helpfull! $\endgroup$ Aug 8, 2019 at 17:28

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The dual of $C(X)$ is $M(X)$ with the total variation norm. You are quoting RMK theorem wrongly.

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  • $\begingroup$ Yes, indeed! As I commented above, I was reading some wrong notes and got confused over some identifications. $\endgroup$ Aug 8, 2019 at 17:30
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No, it does not. The norm topology on the dual space $M(X)$ is not the same as the weak-star topology. For example, according to the Banach-Alouglu theorem, the unit ball of the dual of any Banach space is compact in the weak-star topology, but unless the space is finite dimensional, a Banach space equipped with its norm topology never has a compact unit ball.

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  • $\begingroup$ I am clearly missing something here (any help is appreciated)... Isn't the topology on the dual of $C(X)$ dictated by the norm of linear functionals (that coincide with the total variation norm)? On the other hand the RMK Thm says that such topology is the weak*... $\endgroup$ Aug 8, 2019 at 11:35
  • $\begingroup$ Yes, you are missing the following point: The topological dual of a Banach space $X$ is the space of continuous linear functionals on $X$ equipped with the norm topology. Your addition of "weak-star topology" to the theorem is wrong. $\endgroup$ Aug 8, 2019 at 12:54
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Take $X = [0,1]$ and $\mu_n = \sum_{j=0}^{n-1} (-1)^j\delta_{\frac{j}{n}}$. I claim $\mu_n \to 0$ weak* but $||\mu-\mu_n||_{TV} \not \to 0$. The latter is obvious, since $||\mu_n||_{TV} = 1$ for each $n$ whereas $||\mu||_{TV} = 0$. To see that $\mu_n \to 0$ weak*, take any $f \in C(X)$ and use uniform continuity to see that $\frac{1}{n}\sum_{j=0}^{(n-1)/2} [f(\frac{2j}{n})-f(\frac{2j+1}{n})] \to 0$.

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