# Function of bounded variation

I'm not exactly sure how an integral might be useful here. Somehow this question I will be asking is supposed to be related to bounded linear functionals but I'm still not seeing how.
Let a function $f$ belong to $C[a,b]$. We want to show that there's a function $g$ of bounded variation on $[a,b]$ for which $\displaystyle \int_a^b$$f(x)dg(x)$ = $\|f\|_{max}$ and $TV(g)$ = $1$, where $TV$ is the total variation of $g$. Could we begin by the fact that $f$ is continuous on $[a,b]$ and assume we have a function of bounded variation $g$ on $[a,b]$, and hence, $g$ is absolutely continuous on $[a,b]$? I would really appreciate some help here, since I have been struggling with this for quite a while.

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Absolute continuity implies bounded variation, not the other way around. – Jesse Madnick Apr 24 '11 at 19:42

 Would we be considering $g$ as a step function with two values? – Libertron Apr 24 '11 at 22:18 @Jonas I deleted my previous comments because they were misguided and distracted from this fine answer. I hope this is OK with you. – Glen Wheeler Apr 26 '11 at 7:05
As an application of the Hahn-Banach theorem (e.g., see here), there is a bounded linear functional $\phi\in C[a,b]^*$ such that $\|\phi\|=1$ and $\phi(f)=\|f\|_\text{max}$. The dual space $C[a,b]^*$ is isometrically isomorphically identified as the space of left-continuous functions of bounded variation vanishing at $a$ with total variation norm, $\|g\|=\mathrm{TV}(g)$, where the functional corresponding to $g$ is the Riemann-Stieltjes integral $f\mapsto \int_a^bfdg$ (e.g., see pages 12-19 of Douglas's Banach algebra techniques in operator theory). The result follows.