Does bounded variation and continuous means total variation continuous

Question

$F$ is of bounded variation and continuous. Is it true that total variation is continuous ?

In case, $F$ is absolutely continuous it is trivial to see. But for the above case how to proceed ?

Answer 1

Yes, this is true. Let $F:[a,b]\to\mathbb R$ be as in your post, and denote by $V(x)$ the variation of $F$ on $[a,x]$. It is a standard fact that both $V+F$ and $V-F$ are nondecreasing functions on $[a,b]$.

Suppose $V$ is discontinuous from the left at $c\in (a,b]$ (discontinuity from the right can be ruled out in the same way; or just flip the function). This means $V$ has an upward jump at that point: $\delta:= V(c)-V(c-)>0$. Define $$\tilde V (x) = \begin{cases} V(x), \quad & x<c \\ V(x)-\delta,\quad & x\ge c\end{cases}$$ Observe that $\tilde V+F$ and $\tilde V-F$ are still nondecreasing, since both $V\pm F$ had the same upward jump, the removal of which preserves the nondecreasing property.

The representation $$F = \frac12(\tilde V+F) - \frac12(\tilde V-F)$$ implies that the variation of $F$ on $[a,b]$ is at most the sum of variations of the two functions, i.e., $$\frac{1}{2} (\tilde V(b) +F(b)-F(a)) + \frac{1}{2} (\tilde V(b) - F(b)+ F(a)) $$ But the latter simplifies to $\tilde V(b)$, which is strictly less than $V(b)$ — contradicting the fact that $V(b)$ is the variation of $F$ on $[a,b]$.