Universal Chord Theorem

Interestingly,

The numbers of the form $r = \displaystyle \frac{1}{n} \ \ n \ge 1$ are the only positve numbers such that for any continuous function $\displaystyle f:[0,1] \to \mathbb{R}$ such that $\displaystyle f(0) = f(1)$, there is some point $\displaystyle c \in [0,1-r]$ such that $\displaystyle f(c) = f(c+r)$.

For any other positive $r$ we can find such a continuous function for which there is no $c$ such that $f(c) = f(c+r)$.

For a proof that $\displaystyle r = \frac{1}{n}$ satisifies this property, let $\displaystyle g(x) = f(x) - f(x+ \frac{1}{n})$, for $\displaystyle x \in [0, 1-\frac{1}{n}].$

Then we have that $\displaystyle \sum_{k=0}^{n-1} \ g\left(\frac{k}{n}\right) = 0$.

Thus, if none of $\displaystyle g\left(\frac{k}{n}\right)$ are $\displaystyle 0$, then $\displaystyle \exists i,j \in [0, 1, ..., n -1] \ni \displaystyle g\left(\frac{i}{n}\right) \gt 0$ and $\displaystyle g\left(\frac{j}{n}\right) \lt 0$.

For any positive $\displaystyle r$, consider the following example, due to Paul Levy.

$\displaystyle f(x) = \sin^2\left(\frac{\pi x}{r}\right) - x \ \sin^2\left(\frac{\pi}{r}\right)$. Clearly, $f$ is continuous and $f(0)=0=f(1).$

If $\displaystyle f(x) = f(x+r)$, then, $\displaystyle r\ \sin^2\left(\frac{\pi}{r}\right) = 0$ and hence, $\displaystyle r = \frac{1}{m}$ for some integer $\displaystyle m$.

Apparently this is called the Universal Chord Theorem (due to Paul Levy!).

You want to use the intermediate value theorem, but not applied to $f$ directly. Rather, let $g(x)=f(x)-f(x+1/2)$ for $x\in[0,1/2]$. You want to show that $g(a)=0$ for some $a$. But $g(0)=f(0)-f(1/2)=f(1)-f(1/2)=-(f(1/2)-f(1))=-g(1/2)$. This gives us the result: $g$ is continuous and changes sign, so it must have a zero.

Hint: consider $g(x)$, defined on $[0,1/2]$ by $g(x)=f(x+1/2)-f(x)$