Suppose f is a continuous function on $[0,1]$, f' exists on all $(0,1)$ and f' is monotonically increasing on $(0,1)$. Do not assume $f'$ is bounded. Show $f$ is of bounded variation on $[0,1]$.
Need to show that for all partitions $P$ on $[0,1]$ the sum given by $\sum_{P} |f(x_k)-f(x_{k-1})|$ is bounded by some number independent of the partition chosen - or alternatively - that the supremum of such sums over all partitions is bounded.
Things we know:
~By the Mean Value Theorem, we can find a $c_k \in (x_{k-1},x_k)$ for all $k$ such that $f(x_k)-f(x_{k-1}) = f'(c_k)(x_k-x_{k-1})$. Then we have $|f(x_k)-f(x_{k-1})| = |f'(c_k)|*|(x_k-x_{k-1})|$. We don't know that f' is bounded though...
~We also know that f is continuous which means $\forall \epsilon > 0, \exists \delta > 0$ such that $|x-y| < \delta \Rightarrow |f(x)-f(y)| < \epsilon$. If $P$ is a partition where the lengths of the subintervals are less than $\delta$, then for all $k$, we have $|x_k-x_{k-1}|$ and therefore $|f(x_k)-f(x_{k-1})| < \epsilon$.
~$f'$ is strictly increasing which says $y > x \Rightarrow f'(y) > f'(x)$.
How do I get that $f$ is of bounded variation?
Since $f'$ is increasing, we have that it changes sign at most once. If $f$ is strictly increasing or strictly decreasing, the problem is trivial, so assume $f'$ has a zero at $x_0$. Then the variation is $f(0)-f(x_0)+f(1)-f(x_0)=f(0)+f(1)-2f(x_0)$.
The idea is just splitting the interval into pieces where $f$ is monotonic.