Proof of Cantor Set: Consider g'(x), f_n(c), and f'(x)

[kathrynag]

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 post, but I didn't see any other way to post this since all of these are directly related to each other. I figured out 1 already, but kept it there since g(x) will be used in future posts.

For 2 a) Is the limit 0?
b) Not quite sure

 

[lippyka]

how to answer this. For 3 a) Yes, the derivative of f(x) exists for all c not in C because f(x) is continuous everywhere outside of C, so its derivative must exist.b) Yes, |f(x)|<=(x-c)^2 for all x in [0,1] because as x tends to c, f(x) tends to 0, which implies that |f(x)|<=(x-c)^2. This in turn implies that f'(c)=0 since the derivative of a function at a point is the limit of (change in y)/(change in x) as x tends to c, and the change in y is 0 when x=c.c) The reason f'(x) fails to be continuous on C is because the derivative of a function at a point is the limit of (change in y)/(change in x) as x tends to c, and in this case, the change in y is 0 when x=c which implies the limit is 0, which means that the derivative is not continuous at c.4. The reason f'(x) is Riemann integrable on [0,1] is because it is continuous on [0,1] except at the points in the Cantor set, but since the Cantor set has measure zero, it does not affect the integration of the function. 5. The limit of the sum of the lengths of the intervals making up each C_n is 1/2. This is because the length of each interval at each stage is 1/3^n, and the number of intervals is 2^(n-1). So the sum of the lengths of the intervals making up each C_n is 2^(n-1)*(1/3^n). As n tends to infinity, this sum converges to 1/2.