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.