Riesz-Markov-Kakutani Theorem: Total variation Norm and weak*

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?

• See Kavi's answer (I was about to say the same thing when I saw it had already been said). Aug 8, 2019 at 14:08
• 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! Aug 8, 2019 at 17:28

The dual of $$C(X)$$ is $$M(X)$$ with the total variation norm. You are quoting RMK theorem wrongly.
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.
• 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*... Aug 8, 2019 at 11:35
• 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. Aug 8, 2019 at 12:54
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 $$\frac{1}{n}\sum_{j=0}^{(n-1)/2} [f(\frac{2j}{n})-f(\frac{2j+1}{n})] \to 0$$.