Tuesday, 23 February 2016

real analysis - Classification of all functions satisfying f such that f(x)=f(x2).



I know that of functional identity f(a+b)=f(a)+f(b), y=kx is definitely a solution. Other solutions may be constructed by treating the real numbers as a vector field over the rational numbers, which are pathological.



How to construct all pathological functions satisfying the identity f(x)=f(x2)? I know that f must be constant if it's continuous, but what are the other cases?


Answer



Edited 4 Jan to incorporate details I skipped pointed out by user517969.



Let ϕ(x) and ψ(x) be periodic with period log2, let a,bR. Then f(x)={aif x=0ϕ(log(log|x|))0<|x|<1b|x|=1ψ(log(log(|x|))|x|>1
satisfies your functional equation , and can be as pathological (or as smooth) as ϕ and ψ are. Because f(x)=f(x) there is no loss of generality in using the absolute value signs in the innermost logarithm. I think this is the most general form of solution.



No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...