Prove that $\exists x_1,x_2 \in \mathbb{R}$ with $|x_1-x_2|=\frac{T}{2}$ such that $f(x_1)=f(x_2)$

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be continuous periodic function with T>0.

Prove that $\exists x_1,x_2 \in \mathbb{R}$ with $|x_1-x_2|=\frac{T}{2}$ such that $f(x_1)=f(x_2)$.

I don't even know where to start... I tried to find any connection between $x_1,x_2,T$ and then use that the function is periodic, but no avail. I know that continuous periodic function is bounded (does it say that f is differentiable?), but that didn't help me too...

Please help, thank you!

Answer 1

Take $g(x)=f(x+T/2)-f(x)$, it's sufficient to show $g(x)$ has zero point. According to periodicity of function $f$, we know $$g(T/2)=f(T)-f(T/2)=f(0)-f(T/2)=-g(0)$$ Moreover, $f$ is continuous implies $g$ is also continuous. Then there exists $x_1\in[0,T/2]$ such that $0=g(x_1)=f(x_1+T/2)-f(x_1)=f(x_2)-f(x_1)$. The reason follows from $g(T/2)\cdot g(0)=-g(0)^2\leq0$.