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f continuous and {f(x)} = f({x}) implies f(x) or f(x)-x periodic

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote an unsolved problem from another forum posted on January 8th, 2013.

I don't know how to solve this problem:
Let f be a continuous real function such that [tex]\{f(x)\} = f(\{x\})[/tex] for each x ([tex]\{x\}[/tex] is the fractional part of number [tex]x[/tex]).

Prove that then [tex]f[/tex] or [tex]f(x)-x[/tex] is a periodic function.

Could you help me?
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Consider the function [tex]g:\mathbb{R}\to\mathbb{R}[/tex], [tex]g(x)=f(x+1)-f(x)[/tex]. Clearly, [tex]g[/tex] is continuous. As [tex]\{x+n\}=\{x\}[/tex] for all integer [tex]n[/tex]:

[tex]\{f(x+1)\}=\{f(x)\}=f(\{x\})=\{f(x)\}[/tex]

As [tex]\{f(x+1)\}=\{f(x)\}[/tex], [tex]f(x+1)-f(x)[/tex] must be integer. Being [tex]\mathbb{R}[/tex] connected, there exists [tex]m\in \mathbb{Z}[/tex] such that [tex]f(x+1)-f(x)=m[/tex] for all [tex] x\in \mathbb R[/tex] (Why?). This implies that [tex]h(x)=f(x)-mx[/tex] is periodic (Why?).

If we prove that [tex]m=0[/tex] or [tex]m=1[/tex] then, [tex]h(x)=f(x)[/tex] is periodic or [tex]h(x)=f(x)-x[/tex] is periodic. We have

[tex]f(0)=f(\{0\})=\{f(0)\}\in[0,1)[/tex]

But [tex]f(1)=f(0)+m\in[m,m+1)[/tex]. On the other hand, if [tex]x\in[0,1)[/tex] we have [tex]f(x)=f(\{x\})=\{f(x)\}\in[0,1)[/tex]. Using the continuity of [tex]f[/tex]:

[tex]f(1)=\lim_{x\to 1^-}f(x)\in[0,1][/tex]

That is, [tex]f(1)\in [m,m+1)\cap [0,1][/tex]. This intersecion is not empty if and only [tex]m=0[/tex] or [tex]m=1[/tex]
 

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
Hi Fernando,
I saw this problem earlier and puzzled over it for some time. My only progress was proof that f is periodic if f(1)<1.
I'm probably just dense, but I don't see why there is an integer m such that for all x, f(x+1)-f(x) =m.
(For a given x, f(x+1)-f(x)=floor(f(x+1))-floor(f(x)), certainly an integer. By why is there one integer for any x?)
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I'm probably just dense, but I don't see why there is an integer m such that for all x, f(x+1)-f(x) =m.
(For a given x, f(x+1)-f(x)=floor(f(x+1))-floor(f(x)), certainly an integer. By why is there one integer for any x?)
Well, according to a well-known theorem, a real continuous function on a connected set asumes as a value each number between any two of its values.

In our case, $g(x)=f(x+1)-f(x)$ is continuous and $\mathbb{R}$ is connected. If $g(x_1)=m$ and $g(x_2)=n$ with $m,n$ distinct integers, $g$ would assume non integers values between $m$ and $n$. Contradiction.
 

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
Thanks Rinaldo. I was just being dense. Good proof.