Is the total variation function uniform continuous or continuous? I have been doing some excercises on total variation when the following questions came up to my mind:
(1) Let $f$ be continuous on the interval $[0,1]$ and be of bounded variation. Is it true that its total vatiation function $TV(f_{[0,x]})$ is uniformly continuous? i.e. Is it true that for $\forall\space\epsilon>0$, $\exists\space\delta>0$, such that for arbitrary interval $[a,b]$ with $|b-a|<\delta$, we have $TV(f_{[a,b]})<\epsilon$?
(2) Alternatively, if it is not uniformly continuous, can I say that for $\forall\space{}x\in[0,1]\text{ and } \forall\space\epsilon>0$ , $\exists\space\text{ nondegenerate interval }I \text{ such that }x\in{}I\text{ and } TV(f_{I})<\epsilon$?
Thank you!
 A: Since a continuous function on a compact set is uniformly continous, you're actually asking whether the total variation of a continuous function is continuous.
Let $\epsilon\gt0$ and $x_0\in(0,1]$ be given. Since the total variation is non-decreasing, it suffices to find a point $x\lt x_0$ with $TV(f_{[x,x_0]})\lt\epsilon$ to show that the total variation is left-continuous, and thus by symmetry also right-continuous and hence continuous.
Pick some point $x_1\lt x_0$. By  definition there is a partition of $[x_1,x_0]$ such that the sum of absolute differences of function values over the partition is within $\epsilon/2$ of $TV(f_{[x_1,x_0]})$. Since $f$ is continuous, we can find a point $x$ between the last intermediate point of the partition and $x_0$ such that $|f(x)-f(x_0)|\lt\epsilon/2$. Refining the partition with this point doesn't decrease its sum of absolute differences. Now the sum of absolute differences in the partition up to $x$ is within $\epsilon/2+\epsilon/2=\epsilon$ of $TV(f_{[x_1,x_0]})$, and thus so is $TV(f_{[x_1,x]})$; hence $TV(f_{[x,x_0]})\lt\epsilon$ as required.
