I want to mathematically prove the following statement: "if $f(t)$ has a bounded first derivative on $[a,b]$, then it has bounded first variation on $[a,b]$."
- First order variation is defined as: $$ \lim_{\Vert p\Vert \to 0} \sum^{n-1}_{j=0}|\Delta f_j| $$
So far, I got:
Bounded first derivative hypothesis can be written as: $$ \left|\lim_{t_{j+1}\to t_j}\frac{f(t_{j+1})-f(t_j)}{t_{j+1}-t_j}\right|\leq M, $$ where $M$ is a positive real number.
Again, it can be written as $$ \lim_{\Vert p\Vert\to 0}\left|\frac{\Delta f(t_{j})}{\Delta t_{j}}\right|\leq M, $$ where $p$ is a partition of $[a,b]$ and $\Vert p\Vert$ is the mesh of the partition.
Now, the first variation: $$ \lim_{\Vert p\Vert \to 0} \sum^{n-1}_{j=0}|\Delta f_j|=\lim_{\Vert p\Vert \to 0} \sum^{n-1}_{j=0}\left|\frac{\Delta f_j}{\Delta t_j}\right||\Delta t_j|\leq M\cdot\lim_{\Vert p\Vert \to 0} \sum^{n-1}_{j=0}|\Delta t_j| $$
How should I proceed from here? How do I know that the limit in the last expression is bounded/zero? I am looking for a formal proof and any help would be appreciated!