Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
Absolute continuity implies bounded variation, not the other way around. –  Jesse Madnick Apr 24 '11 at 19:42

2 Answers 2

up vote 5 down vote accepted

Hint: try a step function with two values.

share|improve this answer
    
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

Robert Israel has already indicated a much better answer than that which I am writing. Here is an overkill answer, just for fun.

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.