- #1

ThomasTheTan

- 6

- 0

## Homework Statement

Given: f is integrable on [a,b] and has a jump discontinuity at c in (a,b) (meaning that both one-sided limits exist as x approaches c from the right and from the left but that Lim(x->c-)[f] is not equal to Lim(x->c+)[f]).

Show that F(x) = Integrate[f,{a,x}] is not differentiable at x = c (That is, F = the definite integral of f from a to x).

## Homework Equations

N/A

## The Attempt at a Solution

So we know that F is continuous by FTC - the only way that I can think of checking that F is not differentiable is... by showing that the derivative doesn't exist (using the limit). However, the limit reduces to a form that's rather unusable (it becomes Lim(x->c)[Integrate[f,{c,x}]/(x-c)], and I don't know how to deal with the Integral in the numerator).

Instead, I've been looking at those one-sided limits (which we know exist). If we knew that f was continuous apart from at this jump discontinuity, this problem would be simple. That fact is not given, however. My thinking is that, since the one-sided limits exist, F must be "one-sided continuous" kind of - that is, for the right hand limit, given epsilon > 0, there exists delta > 0 such that 0<x-c<delta implies |f(x)-L|<epsilon by definition (if L = the limit). Therefore, f is continuous slightly to the right of c. However, it's only continuous for an arbitrarily small segment (which changes with epsilon) - therefore, I can't really use this to make F'(x) = f(x) when 0<x-c<delta, can I?

Any help is appreciated.