Exercise 7.5.11

Assume f is integrable on [ a , b ] and has a "jump discontinuity" at c ( a , b ) . This means that both one-sided limits exist as x approaches c from the left and from the right, but that

lim x c f ( x ) lim x c + f ( x ) .

(This phenomenon is discussed in more detail in Section 4.6.)

(a)
Show that, in this case, F ( x ) = a x f is not differentiable at x = c .
(b)
The discussion in Section 5.5 mentions the existence of a continuous monotone function that fails to be differentiable on a dense subset of R . Combine the results of part (a) with Exercise 6.4.10 to show how to construct such a function.

Answers

(a)
For conciseness, define l = lim x c f ( x ) and r = lim x c + f ( x ) as the left and right limits of f ( c ) . Define
g ( x ) = { f ( x ) x < c l x = c f ( x ) r + l x > c

where we’ve essentially shifted f after x > c to line up with x < c , so that g is continuous at c . As a result, G ( x ) = a x g is differentiable at c .

Now if F = a x f was also differentiable at c , then we would also have G F = a x g f differentiable at c . But actually evaluating a x g f leaves us with

a x g f = { 0 x c ( l r ) ( x c ) x > c

which is not differentiable at c .

(b)
Define h and u n similarly as is done in Exercise 6.4.10, except we only enumerate over the rational numbers in [ 0 , 1 ] . Given that h converges uniformly, and each u n is integrable over [ 0 , 1 ] , we have h integrable over [ 0 , 1 ] as well. From part (a) we have that g ( x ) = a x h is not differentiable at all rational points (dense in [ 0 , 1 ] ), while Theorem 7.5.1 (ii) indicates g is continuous. Since h is nonnegative, g must be monotone increasing. We can then extend g over all of R by repeating it and applying appropriate offsets.
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2022-01-27 00:00
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