# Functions of bounded variation as the dual of $C([a,b])$

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.

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Sorry for my english. (1) The integral respect to a bounded variation function is in the sense of Riemann-Stieltjes. – user82002 Jun 11 '13 at 20:46

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.

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Thank you for the answer. – Cantor Dec 1 '12 at 11:43
could I kindly ask you to write the statements in your answer in mathematical form because then it would be more clear what you mean by variation of a measure and of a function. – Cantor Dec 1 '12 at 12:13
@Cantor: I have added the definition. – Christian Clason Dec 1 '12 at 12:24
I think I don't really understand the definition of $TV(f)$. $Df$ is the distributional derivative of $f$. Is there a general definition for absolute value of a distribution and the integral of it? How do I make sense of $\int_\Omega|Df|$? The next part (with the $\sup$) looks like the linear operator norm of $Df$ as a functional from $C_c^\infty(\Omega)$ into $\mathbb{R}$ (or $\mathbb{C}$). But then, what is the norm on $C_c^\infty(\Omega)$? Is it even a normed vector space, or just a topological vector space? Perhaps the definition has nothing to do with operator norm and stands on its own? – Cantor Dec 1 '12 at 13:03
The second part is the definition of the norm of the distributional derivative (I have deleted the first part, since it's not really necessary here). It is not the operator norm on $C^\infty$, since you take the supremum of all those $\varphi$ with $C$-norm equal to one (as opposed to $C^\infty$-norm). I've made that explicit now. – Christian Clason Dec 1 '12 at 14:35