Continuity with $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. Is there $x \in \mathbb{R}$ with $f(x)=f(x+\sqrt{2})$?

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be continous with $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. Show there exists $x \in \mathbb{R}$ with $f(x)=f(x+\sqrt{2})$.

My question is, do we even need $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$ to show this, or is is sufficient, that f is bounded? What happens if $f$ is monotonous?

I posted a sketch for the proof of the original/task problem below.

Answer 1

The function $\arctan x$ is bounded and continuous, but the $x$ the problem asks for does not exist, so the periodicity assumption is indeed important.

Answer 2

This is how I would approach the proof for the original problem/task:

First consider $f:[0,1]\to \mathbb{R}$. Since $f$ is continuos, we can apply the extreme value theorem. Thus, there exists $a:= \min_{x\in [0,1]}$ and $b:=\max_{x\in [0,1]}$.

Now we can use $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. As a result there cannot exists $\xi \in \mathbb{R}$ with $f(\xi)<f(a)$ or $f(\xi)>f(a)$. Has anyone an idea how this can be shown rigorously? As a consequence $a$ is a global minimum and $b$ is a global maximum.

Let $g: \mathbb{R} \to \mathbb{R},\; g(x)=f(x+\sqrt{2})-f(x) $.

Then we have $g(a)=f(a+\sqrt{2})-f(a)\geq 0$, and $g(b)=f(b+\sqrt{2})-f(b)\leq 0$. Because $a$ is a global minimum and $b$ is a global maximum of $f$.

If $g(a)=0$ or $g(b)=0$ we are done. Else, by the intermediate value theorem, there exists $x \in (a, b)$, such that $g(x)=0$.

Why are we done? Because $g(x)=0 \iff f(x+\sqrt{2})- f(x)=0 \iff f(x)=f(x+\sqrt{2})$.

Answer 3

You can note that if you have $\xi\in\mathbb{R}$, you have $f(\xi)=f(\xi-\lfloor \xi\rfloor) $ and you have $\xi-\lfloor\xi\rfloor \in [0,1]$, so you get that the maximum and the minimum are global