Is $\sup_{\| f \| \leq 1}{\left| \int f d\mu \right|} = \sup_{\| f \| \leq 1}{\{ |\mu(f)| \}}?$

Question

The answer given by t.b. mentioned the following

One of the most convenient way of writing a total variation norm is

$$\| \mu \| = \sup_{\| f \| \leq 1}{\left| \int f d\mu \right|}$$

In this paper, page $123$, the author mentioned that element of the dual space of real-valued lipschitz functions $Lip_0(X)^*$ can be identify as a measure $\mu$, with the above norm.

By definition of a dual space $Lip_0(X)^*$, every element is an evaluation function $\mu : Lip_0(X) \rightarrow \mathbb{F}$. Then its operator norm is $$\| \mu \| = \sup_{\| f \| \leq 1}{\{ |\mu(f)| \}}$$ where $f \in Lip_0(x)$.

Question: Are the two norms the same? In other words, is $$\sup_{\| f \| \leq 1}{\left| \int f d\mu \right|} = \sup_{\| f \| \leq 1}{\{ |\mu(f)| \}}?$$

Answer 1

This is not an answer to your question, but merely an attempt to encourage you to formulate the question in a way that will encourage more attention and answers.

You write:

By definition of a dual space $Lip_0(X)^*$, every element is an evaluation function $\mu : Lip_0(X) \rightarrow \mathbb{F}$.

This is not true. The norm-closure in $Lip_0(X)^{*}$ of the evaluations gives the predual, which is denoted $\mathcal{F}(X)$ in the paper you mentioned. The predual space is in general not equal to the dual space. (That's why it has a different name). In fact, a functional $\phi\in Lip_0(X)^{*}$ belongs to the predual of $Lip_0(X)$ if and only if it is continuous in the $w^*$ topology. (See "Lipschitz Algebras" by N. Weaver, p. 36). Also note that on page $123$ the authors do not claim that

All elements of the dual space $Lip_0(X)^{*}$ can be identified with measures

as readers may understand from your first statement. What is claimed by the authors on page $123$ of the paper you mentioned is the following:

It is sometimes convenient to think of $\mathcal{F}(X)$ as the completion of the set of Borel measures $\mu$ on $X$ with finite support under the norm $$||\mu||_{\mathcal{F}}=\sup_{||f||_{Lip}\leq 1}\int f\,d\mu$$

Note the words finite support. A measure has finite support if an only if it is a linear combination of a finite number of measures concentrated at single points. Not all measures belong to the predual, but only those that are obtained as norm-limits of Cauchy sequences of measures of finite support. The authors then add the following comment on page $123$:

It now follows that any finite Borel measure supported on a compact subset $K$ of $X$ can be identified with a member of $\mathcal{F}(X)$.

Not every Borel measure, but only finite measures with compact support.

So now, you may want to re-formulate your question. What is it exactly that you are asking? I can think of two possible questions:

1. Are you asking whether all Borel measures that belong to the predual $\mathcal{F}(X)$ have a norm given by the supremum over integrals?

2. Are you asking whether that norm, being a supremum taken over Lipschitz functions, coincides with their total variation?