This proof is almost done except for the step of showing that the function's derivative is 0 a.e.
Let I={[pn,qn]} denote the set of all closed intervals in R with rational endpoints pi,qi∈Q.
Let ϕn denote a variant of the Cantor Ternary Function (i.e., Devil's Staircase Function) extended to all R which satisfies ϕn(x)=0 for all x<pi, ϕn(x)=1/2n for all x>qi and is non-decreasing, continuous, and of zero derivative a.e. on [pi,qi] so that ϕn satisfies
- ϕ′n(x)=0 a.e. on R
- ϕn non-decreasing, continuous on R
- ϕn is bounded above by 1/2n
- ϕn(pn)<ϕn(qn) (trivial but important later)
Now let ∑ϕn=ϕ.
Claim
- ∑ϕn→ϕ uniformly
- ϕ is strictly increasing and continuous on R yet also satisfies ϕ′(x)=0 a.e. on R.
Attempted Proof
First since ∀n∈N, we have that |ϕn|≤1/2n with ∑1/2n<∞, it follows that ∑ϕn→ϕ uniformly (a fact from Real Analysis).
Consider that the finite sum of k continuous functions is itself a continuous function. Since all of the ϕn are continuous functions, we therefore have that fk=∑kn=1ϕn is also a continuous function. Moreover, since (1) implies that fk→ϕ unformly, we now have that ϕ is also continuous (via another Real Analysis fact that a uniformly convergent sequence of continuous functions converges to another continuous function).
To show ϕ is strictly increasing, consider x,y∈R s.t. x<y. Since ∃k∈N s.t. x<pk<qk<y, we have that ϕk(x)≤ϕk(pk)<ϕk(qk)≤ϕk(y) implies that ϕk(x)<ϕk(y) so that for N≥k we have ∑Nn=1ϕn(x)<∑Nn=1ϕn(y) and more generally ϕ(x)<ϕ(y). This shows that indeed ϕ is strictly increasing on R.
To show that ϕ′(x)=0 a.e. on R, we will show that ϕ′=∑ϕ′n=∑0n, where each 0n is the derivative function of ϕn which is 0 almost everywhere.
But I'm not sure how to complete (4)?
Answer
From your answer I see that it is ok to use standard facts from real analysis, so let's use them!
Fubini's theorem on differentiation. Assume (fn)n∈N is a sequence of non-decreasing functions on [a,b], and the series ∑∞n=1fn(x) converges for all x∈[a,b], then
(∞∑n=1fn(x))′=∞∑n=1f′n(x)
a.e. on [a,b].
Now we turn to the original problem. Consider arbitrary interbal [a,b]. By assumption ϕ′n=0 a.e. on R and a fortiori on [a,b]. By Fubini's theorem on differentiation we get
ϕ′(x)=(∞∑n=1ϕn(x))′=∞∑n=1ϕ′n(x)=∞∑n=10=0
a.e. on [a,b]. Since [a,b] is an arbitrary interval, then ϕ′=0 a.e. on R.
No comments:
Post a Comment