Re: bounded variation




david petry wrote:
> If a function f is differentiable, then the total variation on an
> interval (a,b) could be defined as integral (t=a..b) |f'(t)| dt.
> (i.e. the integral of the norm of the derivative). The function has
> bounded variation if that is finite.

I have a question about the integral you mentioned. Is this a Riemann
integral or a Lebesgue integral? Does it matter? What I mean is, for
the equation V(f) = integral( | f ' | ) to hold, is it neccesary that |
f ' | exist AND be Riemann integrable? Example:

Let f (x) = x * Sqrt[x] sin(1 / x) when x=/=0, and f (0) = 0. We want
to know the variation of f on the interval [0,1]. It can easily be
checked that f ' exists on the interval [0,1], and that
| f ' (x) | = | 3 / 2 Sqrt[x] sin (1 / x) - cos(1 / x) / Sqrt[x] | when
x=/=0, and f ' (0) = 0. Thus, f ' is unbounded on [0,1], so it is not
Riemann integrable. However, the Lebesgue integral of f ' on [0,1]
exists and is finite ( | f ' | is bounded above by 1 / Sqrt[x], whose
improper integral and Lebesgue integral converge on [0,1]). Does this
imply that f is of bounded variation on [0,1]?

.



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