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Let $f$ be of bounded variation on $[a,b]$, and define $v(x) = TV(f_{[a,x]})$ for all $x \in [a,b]$.

  • show that $|f'| \leq v'$ a.e. on $[a,b]$, and infer from this that $$\int_a^b |f'|\leq TV(f).$$

Proof:

Since $f$ is of bounded variation, then it is differentiable a.e. and $f'$ is integrable. Since $v$ is increasing, then it is differentiable a.e. as well. Now, since both $f$ and $g$ are differentible almost everywhere, there exist a subset $E$ of $[a,b]$ with $m(E) = 0$ such that $f'$ and $v'$ exists for $x \in [a,b]\sim E$. Thus, for $x \in [a,b] \sim E$, we have

$$f'(x) = \lim_{h \to 0+} \frac{f(x+h) - f(x)}{h},$$ and $$v'(x) = \lim_{h \to 0+} \frac{TV(f_{[a,x+h]}) - TV(f_{[a,x]})}{h} = \lim_{h \to 0+} \frac{TV(f_{[x,x+h]})}{h},$$ thus, $$|f'(x)| = \lim_{h \to 0+}\frac{1}{h} |f(x+h) - f(x)| \leq \lim_{h \to 0+}\frac{1}{h}TV(f_{[x,x+h]})) = v'(x),$$ here we used the fact that $$TV(f_{[x,x+h]})) = \sup\{V(f,P): P \text{ is a partition of } [x,x+h]\},$$ where $$V(f,p) = \sum_{i=1}^n |f(x_i) - f(x_{i-1)}|.$$ Now, $v$ is increasing, therefore, $$\forall x \in [a,b] : \int_a^x v' \leq v(x) - v(a) = v(x).$$ In particular, we have $$\int_a^b |f'| \leq \int_a^b v' \leq v(b) = TV(f_{[a,b]}),$$ as required.

Is this proof correct?

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1 Answer 1

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There may be a problem with the line

$$|f'(x)| = \lim_{h \to 0}\frac{1}{h} |f(x+h) - f(x)| \leq \lim_{h \to 0}\frac{1}{h}TV(f_{[a,x]})) = v'(x)$$

$h$ can approach zero from the left or from the right, yet you seem to assume that $x+h$ is contained in the interval $[a,x].$

Worse, the expression $\frac{1}{h}TV(f_{[a,x]}))$ does not converge.

I do not think that the proof is beyond repair but I would have to think carefully. It may be sufficient to replace $\frac{1}{h}TV(f_{[a,x]}))$ with $\frac{1}{h}TV(f_{[x,x+h]}))$ using the convention that the interval $[p,q]$ really denotes the interval $[q,p]$ if $q<p.$

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  • $\begingroup$ I thnik it was a typo, i corrected what i wanted to say. It is correct right now? $\endgroup$
    – okie
    Commented Dec 8, 2015 at 21:17
  • $\begingroup$ The new expression still diverges as $h$ goes to 0. $\endgroup$ Commented Dec 8, 2015 at 21:25
  • $\begingroup$ still had another error, I fixed again what i wanted. $\endgroup$
    – okie
    Commented Dec 8, 2015 at 21:51
  • $\begingroup$ Have you thought about the line starting with $v'(x)$, and how it behaves when $h$ is negative? $\endgroup$ Commented Dec 8, 2015 at 22:02
  • $\begingroup$ It is not enough to let $h \to 0+$ instead of $h \to 0$? $\endgroup$
    – okie
    Commented Dec 8, 2015 at 22:17

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