What is the derivative of total variation?

Let f(x) be a function of bounded variation on closed, bounded interval [a, b].

Let g(x) = V(f, [a, x])

where $\displaystyle V(f, [a, x]) = \sum_{k=1}^n |f(x_{i}) - f(x_{i-1}) |$ given some partition $\displaystyle P = \left \{a, x_{1}, x_{2}, ...., x_{n-1}, x \right \}$ of [a, x].

How can I calculate the derivate of g(x)?

I know of differentiation term-by-term but not sure if this would apply here.

Thanks

Re: What is the derivative of total variation?

I would start with the definition of the derivative and see what you get:

$\displaystyle g'(x) = \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h}$

Re: What is the derivative of total variation?

This is what I get

$\displaystyle g'(x) = \lim_{h \rightarrow 0} \frac{V(f_{[a, x+h]}) - V(f_{[a, x]})}{h} = \lim_{h \rightarrow 0} \frac{V(f_{[x, x+h]})}{h} = \lim_{h \rightarrow 0} \frac{\sum_{i=1}^m|f(x_{i}) - f(x_{i-1})|}{h}$

If I consider the supremum of g' over all partitions of [a, b], is it true that $\displaystyle |g'| \leq f'$?

For example, if my partition is just two points $\displaystyle P = \big\{x, x+h\big\}$ then I get

$\displaystyle g'(x) = \lim_{h \rightarrow 0} \frac{|f(x+h) - f(x)|}{h} = |f'(x)|$

Re: What is the derivative of total variation?

No, quite the contrary. Instead, you just showed that if $g'(x)$ exists, then $g'(x) \ge |f'(x)|$ since you want the supremum over all possible partitions (one of which is the partition with just two points).

Re: What is the derivative of total variation?

Yes, sorry, that's what I meant to write.

Thanks Slip.