# Is the total variation norm of a measure equal to its norm as a bounded functional?

Let $X$ be a topological space and let $\mu$ be a Borel, regular measure, with finite total variation $\| \mu \| _{TV}$. One may view $\mu$ as a bounded linear functional on the Banach space $C_b (X)$ of the bounded continuous functions on $X$; as such, it has a norm given by $\| \mu \| = \sup _{\| f \| = 1} \left| \int _X f \ \mathrm d \mu \right|$.

Are the two norms of $\mu$ equal?

(In particular, an affirmative answer would clarify why the total variation norm has this slightly unintuitive definition.)

• Is $\mu$ here a signed or complex measure? It seems to be trivial if $\mu$ is a positive measure. – Nate Eldredge Jan 24 '18 at 21:22
• @NateEldredge: Indeed, it is a complex measure. – Alex M. Jan 24 '18 at 21:37

It's easy to see that $||\mu||\le||\mu||_{TV}$. If $X$ is a locally compact Hausdorff space then the Riesz Representation Theorem says that $$||\mu||_{TV}=\sup_{f\in C_0(X),||f||=1 }|\int f\,d\mu|\le||\mu||.$$
I don't have a counterexample, but I tend to doubt that $||\mu||_{TV}\le||\mu||$ holds for a general topological space; you need some hypothesis on $X$ to ensure that there are "enough" continuous functions...
• On the other hand, the assumption that $\mu$ is regular excludes the usual counterexamples. – Nate Eldredge Jan 24 '18 at 21:14