TITLE: If $f(a)=f(b)$ then $\forall\alpha$, $f(x+\alpha)=f(x)$ for some $x$
QUESTION [6 upvotes]: Given that $f$ is continuous in $[a,b]$ with $f(a)=f(b)$. Prove or disprove: given any $\alpha\in[0,{b-a\over2}]$, there exists some $x\in[a,b]$ such that $f(x+\alpha)=f(x)$.

If this were true then I'd suppose we prove by contradiction. Suppose $f(x)\ne f(x+\alpha_0)$ for some $\alpha_0\in[0,{b-a\over2}]$ for all $x$, then since $f$ is continuous, suppose w.l.o.g. $f(x+\alpha_0)-f(x)>0,\forall x\in[a,b]$. How should I proceed?

REPLY [5 votes]: This is not true. Take $a = 0$, $b = 5$, and $\alpha = 2$.
Consider the piecewise linear function $f : [0, 5] \to \mathbb{R}$ which is linear on $[i, i + 1]$ for $i = 0, 1, 2, 3, 4$, such that $f(0) = 0$, $f(1) = 2$, $f(2) = -1$, $f(3) = 1$, $f(4) = -2$, and $f(5) = 0$.
Consider the function $g : [0, 3] \to \mathbb{R}$ defined by $g(x) = f(x + 2) - f(x)$. Note that $g$ is linear on $[i, i + 1]$ for $i = 0, 1, 2$. Note also that $g(0), g(1), g(2), g(3) = -1$. Therefore, $g$ is strictly negative - in fact, we always have $g(x) = -1$ for all $x \in [0, 3]$. So for all $x \in [0, 3]$, we have $f(x + 2) = f(x) - 1 < f(x)$.