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I'm stuck on the following exercise

If $f: [a,b]\to \mathbb R$ is of bounded variation, $t(x) = Var(f;[a,x])$ for ($a\le x \le b$), then $t$ is $\lambda$-a.e. differentiable and $t' = |f'|$ almost everywhere. (here $\lambda$ denotes Lebesgue measure).

$t$ is monotonically increasing, so is differentiable almost everywhere. Also, we have for $y<x$:

$$ \begin{align} \frac{t(x)-t(y)}{x-y} &= \frac{Var(f,[y,x])}{x-y} \\ &\ge \frac{|f(x)-f(y)|}{x-y} \\ \end{align} $$

Letting $y\to x$ we obtain - at each point, where both limits exist - that $t'(x) \ge |f'(x)|$. I don't know how to prove the other inequality, though.

There is also a hint in the exercise that one should use the following result:

If $f_n: [a,b]\to \mathbb R$ is a sequence of montonically increasing functions such that $F = \sum_n f_n$ converges, then for almost all $x\in [a,b]$ we have $$F'(x) = \sum_n f_n'(x)$$

But I really don't see how this could be used here...

I have also thought about writing $f(x) = f^+(x) - f^-(x)$ for monotonically increasing $f^+$ and $f^-$. And then $t(x) = f^+(x) + f^-(x)$, so I would need to prove $$(f^+)'(x) + (f^-)'(x) = |(f^+)'(x) - (f^-)'(x)|$$ for almost all $x$. This is the same as saying that for almost all $x$ we either have $(f^+)'(x) = 0$ or $(f^-)'(x) = 0$. This seems like an Ansatz, which might lead to something, but I can't push it to its conclusion.

Any help would be greatly appreciated. Thanks! =)

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up vote 0 down vote accepted

We prove this step by step:

  1. Consider a "bump function" that bridges $f$ to $(-\infty,+\infty)$ such that $f$ is defined everywhere in $\mathbb{R}$, $f(x)\rightarrow 0$ as $x\rightarrow -\infty$. This bump function should not influence the conclusion except we notice that the total variation may be changed by a constant from the original total variation.

  2. We have the modified $f\in NBV$. So there corresponds a unique Borel measure $\mu$ such that $f(x)=\mu(-\infty,x)$. Moreover for this $\mu$ we have $$T_{f}(x)=|\mu|(-\infty,x)$$

  3. By Lesbegue decomposition theorem we have $$|\mu|=\mu_{a}+\mu_{b},\mu_{a}\le \lambda, \mu_{b}\perp \lambda$$ (therefore $|\mu|'=\mu_{a}'$ almost everywhere)

  4. By Lesbegue decomposition theorem we have $$\mu_{a}=\int_{E}hd\lambda$$for some unique $h\in L^{1}(\lambda)$. Now we verify that $h=|f'|$ almost everywhere.

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You are mistaken, I'm afraid. Consider $f(x) = \chi_{[c,b]}(x)$ for $a<c<b$. Then $f' = 0$ a.e. – Sam Jan 11 '13 at 18:17
I do not know what you are talking about. Your function is perfectly fine; it is absolutely continuous. I suggest you prove every statement step by step. – Bombyx mori Jan 11 '13 at 19:17
Ok, step by step then: My function $f$ is differentiable almost everywhere, and $f' = 0\in L^1[a,b]$. $f$ has total variation $=1$, but $g(x) = \int_{-\infty}^x |f'(x)|\, dx = 0$... – Sam Jan 11 '13 at 20:41
I see. I might have skipped a constant. – Bombyx mori Jan 12 '13 at 5:07
Ok, so basically your idea consists in writing $df = f' \, d\lambda +d\mu_{\mathrm{sing}}$ with $d\lambda \perp d\mu_{\mathrm{sing}}$. Then $dt = |f'| \, d\lambda + d|\mu_{\mathrm{sing}}|$ implies that $$t' = \frac{dt}{d\lambda} = \frac{|f'| \, d\lambda}{d\lambda} + \frac{d|\mu_{\mathrm{sing}}|}{d\lambda} = |f'|\quad \text{ a.e.}$$ I'm not sure, whether I already had knowledge of these facts at the time I asked the question, but that's alright. Thanks for your answer! – Sam Jan 12 '13 at 13:04

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