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?
Best Answer
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 $q<p.$