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Exercise 7.5.11
Assume is integrable on and has a "jump discontinuity" at . This means that both one-sided limits exist as approaches from the left and from the right, but that
(This phenomenon is discussed in more detail in Section 4.6.)
- (a)
- Show that, in this case, is not differentiable at .
- (b)
- The discussion in Section mentions the existence of a continuous monotone function that fails to be differentiable on a dense subset of . Combine the results of part (a) with Exercise 6.4.10 to show how to construct such a function.
Answers
- (a)
-
For conciseness, define
and
as the left and right limits of
. Define
where we’ve essentially shifted after to line up with , so that is continuous at . As a result, is differentiable at .
Now if was also differentiable at , then we would also have differentiable at . But actually evaluating leaves us with
which is not differentiable at .
- (b)
- Define and similarly as is done in Exercise 6.4.10, except we only enumerate over the rational numbers in . Given that converges uniformly, and each is integrable over , we have integrable over as well. From part (a) we have that is not differentiable at all rational points (dense in ), while Theorem 7.5.1 (ii) indicates is continuous. Since is nonnegative, must be monotone increasing. We can then extend over all of by repeating it and applying appropriate offsets.