Find all positive real numbers a so that for any continuous function satisfying some conditions, there exists some $x$ with $f(x)=f(x+a)$

Question

Find all positive real numbers a so that for any continuous function on $[0,1]$ with $f(0)=f(1)=0$, there exists some $x$ with $f(x)=f(x+a)$.

Note that any $a=1/n$ for $n$ a positive integer works. Indeed, we can let $S = \{|f(m/n)| : m\in\mathbb{Z},0\leq m < n\}$ and choose $m$ so that $|f(m/n)|$ is maximum. Note that if $f$ is identically zero then the claim trivially holds, so suppose $f$ is not identically zero. Then $0 < m < n$ since $f(0)=f(1)=0$. WLOG, suppose $f(m/n)$ is positive. Note that $f(x)-f(x+1/n)$ is defined for x in $[(m-1)/n,m/n]$ from the above observation. Then $f(x)-f(x+1/n)$ is nonpositive at $x=(m-1)/n$ and nonnegative at $x=m/n$, so by the IVT it has a zero in $[(m-1)/n, m/n]$. But I'm not sure how to show that if $a$ is not of the above form, then there is some continuous function on $[0,1]$ with $f(0)=f(1)=0$ so that there is no $x$ with $f(x)=f(x+a)$. The mean value theorem and the intermediate value theorem might be useful. It could be useful to construct some piecewise linear function that's slightly modified from a function that satisfies $f(x)=f(x+a)$ for some x, $f(0)=f(1)=0$.

Answer 1

Suppose $0 < a < 1$. Let $$ f(x) = \cos(2\pi x/a) + x(1 - \cos(2\pi/a)) - 1. $$ It is easily verified that $f(0) = f(1) = 0$. Now assume that there is some $x$ such that $0 \le x < x + a \le 1$ and $f(x) = f(x+a)$. This simplifies to $\cos(2\pi/a) = 1$, from which it follows that $a = 1/n$ for some positive integer $n$.