Wednesday, May 25, 2016

real analysis - Existence of a Strictly Increasing, Continuous Function whose Derivative is 0 a.e. on mathbbR



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,qiQ.



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





  1. ϕn(x)=0 a.e. on R

  2. ϕn non-decreasing, continuous on R

  3. ϕn is bounded above by 1/2n

  4. ϕn(pn)<ϕn(qn) (trivial but important later)



Now let ϕn=ϕ.



Claim





  1. ϕnϕ uniformly

  2. ϕ is strictly increasing and continuous on R yet also satisfies ϕ(x)=0 a.e. on R.



Attempted Proof




  1. First since nN, we have that |ϕn|1/2n with 1/2n<, it follows that ϕnϕ uniformly (a fact from Real Analysis).



  2. 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 fkϕ 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).


  3. To show ϕ is strictly increasing, consider x,yR s.t. x<y. Since kN 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 Nk 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.


  4. To show that ϕ(x)=0 a.e. on R, we will show that ϕ=ϕn=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)nN 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=1fn(x)
a.e. on [a,b].




Now we turn to the original problem. Consider arbitrary interbal [a,b]. By assumption ϕn=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=1ϕn(x)=n=10=0
a.e. on [a,b]. Since [a,b] is an arbitrary interval, then ϕ=0 a.e. on R.


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