I am trying to understand this proposition about the dual of $C([a,b])$. I would like some help with the following:
(1) What does the integral with respect to a function of bounded variation mean?
(2) According to Riesz Representation Theorem, the dual of $C([a,b])$ is the space of regular Borel measures (Radon measures) on $[a,b]$. So what is the relation between these measures and the bounded variation functions?
It would be nice to get some explanation in terms of distributions.
Best Answer
In general, the total variation of a measure (which appears in the identification of $C^*$, endowed with the operator norm, with the space of Radon measures, endowed with the total variation), defined for $\mu\in C_0(\Omega)^*$ as $$|\mu|(\Omega) := \sup\left\{\sum_{i=0}^\infty |\mu(X_i)|:\bigcup_{i=0}^\infty X_i = \Omega\right\},$$ is not the same as the total variation of a function (which appears in the definition of the normed space $BV$), defined for $f\in BV(\Omega)$ as $$ TV(f) := \sup \left\{\int_\Omega f (-\mathrm{div} \varphi) \,dx: \varphi\in(C_0^\infty(\Omega))^n,\, \sup_{x\in\Omega}|\varphi(x)|\leq 1\right\}.$$
You could say, however, that if a function has bounded variation, its distributional gradient exists as a Radon measure. The total variation (in the sense of the seminorm on $BV$) of the function is then the same as the total variation (in the sense of measures) of its distributional gradient.
However, in the one-dimensional case, the Riesz representation theorem actually does yield a function of bounded variation (this is in fact Riesz' original statement of 1909). In this case, the integration with respect to a function (of bounded variation) is in the sense of a Riemann-Stieltjes integral.