Consider g(x)=x^2sin(1/x) if x>0 and 0 if x<=0

1. a) Find g'(0)

b) Compute g'(x) for x not 0

c)Explain why, for every delta>0, g'(x) attains every value between 1 and -1 as ranges over the set (-delta,delta). Conclude that g'(x) is not continuous at x=0.

Next, we want to explore g with the Cantor set.

We have f_1(x)=0 if x is in [0,1/3]

=g(x-1/3) if x is to right of 1/3

=g(-x+2/3) if x is to left of 2/3

=0 if x is in [2/3,1]

Now f_2(x)=1/3f_1(3x) for x in [0,1/3]

=f_1(x) if x is in [1/3,2/3]

=1/3f_1(3x-2) if x is in [2/3,1]

2. a) if c is in C(Cantor set) what is lim f_n(c)?

b) Why does lim f_n(x) exist for x not in C?

Now set f(x)=lim f(x)

3.a) Explain why f'(x) exists for all c not in C

b) If c in C, argue that |f(x)|<=(x-c)^2 for all x in [0,1]. Show how this implies f'(c)=0

c) Give a careful argument for why f'(x) fails to be continuous on C.

4. Why is f'(x) Riemann integrable on [0,1]

The reason the Cantor set has measure zero is that, at each stage, 2^(n-1) open intervals of length 1/3^n are removed from C_(n-1). The resulting sum:

sum(2^(n-1)(1/3^n) converges to 1, which means that C1, C2, C3,.... have total lengths tending to zero. Now let's remove intervals of lengths 1/3^(n+1)

5.Show that under these circumstances, the sum of the lengths of the intervals making up each C_n, no longer tends to zero as n tends to infinity. What is this limit?

I know this is way too many questions for one thread, but these are all directly related to each other.

I had already figured out 1, but left it there since it is used in the next problem.

2a) I think the answer is 0.

b) not quite sure

Any hints would help!