Nicholas Hunter

2022-11-23

Suppose $f:\mathbb{R}\to \mathbb{R}$ is continuous and periodic with period 2a for some $a>0$; that is, $f\left(x\right)=f\left(x+2a\right)$ for all $x\in R$. Show there is some $c\in \left[0,a\right]$ such that $f\left(c\right)=f\left(c+a\right)$.
The only way I see to do this question is to apply the intermediate value theorem, but I just don't know how to apply it to this question. I know that $f\left(x\right)=f\left(x+2a\right)$. Therefore, if i can show that $f\left(c+a\right)=f\left(c+2a\right)$ or $f\left(c+a\right)-f\left(c+2a\right)=0$, I'd be done. But I just don't know where to go from there. I tried substituting $x=a$, getting $f\left(2a\right)-f\left(3a\right)$, but I can't show that's less than, equal to, or greater than 0. Any ideas?

Kalmukujobvg

Expert

Hint: Define a function $g\left(x\right)=f\left(x+a\right)-f\left(x\right)$. If $g\left(0\right)=0$, then you're done; if not, note that $g\left(a\right)=-g\left(0\right)$.