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

[Fernando Revilla]

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 \{f(x)\} = f(\{x\}) for each x (\{x\} is the fractional part of number x).

Prove that then f or f(x)-x is a periodic function.

Could you help me?

 

[Fernando Revilla]

Consider the function g:\mathbb{R}\to\mathbb{R}, g(x)=f(x+1)-f(x). Clearly, g is continuous. As \{x+n\}=\{x\} for all integer n:

\{f(x+1)\}=\{f(x)\}=f(\{x\})=\{f(x)\}

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

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

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

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

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

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

 

[johng1]

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]

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.

 

[johng1]

Thanks Rinaldo. I was just being dense. Good proof.