Let $f: [0, 1]\rightarrow R$ be continuous and suppose that $f(0) = f(1)$. Prove there is $x$ such that $f(x)=f(x+\frac{1}{n})$

Question

Let $f: [0, 1]\rightarrow R$ be continuous and suppose that $f(0) = f(1)$. Prove that for each positive integer $n$ there is an $x$ in $[0, 1 - \frac{1}{n}]$ such that $f(x) = f(x + \frac{1}{n}$).

Without loss of generality, take $f(0)=f(1)=0$. I have been able to show that if $f(x)\ge 0$ or $f(x)\le 0$ the statement is true. It actually isn't that difficult to show. The trick is to define $g_n(x) = f(x+\frac{1}{n})-f(x)$, on $[0,1-\frac{1}{n}]$ then notice that $g(0)=f(\frac{1}{n})$ and $g(1-\frac{1}{n}) = -f(1-\frac{1}{n})$. We want to exclude the cases where $f(\frac{1}{n})=0$ and $f(1-\frac{1}{n})=0$ since that will immediately prove our statement. Using our condition $f(x)\ge 0$ and $f(x)\le 0$ and applying intermediate value theorem, we deduce that there's some $c$ such that $g_n(c)=0$.

When we don't have the condition $f(x)\ge 0$ and $f(x)\le 0$, we could have continuous function such that $f(\frac{1}{n})\cdot f(1-\frac{1}{n}) = -|f(\frac{1}{n})\cdot f(1-\frac{1}{n})|$ (this is a long way of saying they have different sign). Notice that the method I described in the paragraph above can be extended to any interval $[a,b]$ if $b-a\ge \dfrac{1}{n}$. The main obstacle I face right now is that $f$ could have at least $n$ zeros such that the distance between any consecutive zeros the distance is at most $\frac{1}{n}$.

However, I have struggled with this problem for a while now and I have made little progress. I would appreciate some hints, but please don't post complete solutions.

Answer 1

Let $g_n(x)=f\left(x+\frac{1}{n}\right)-f(x)$, then $$\sum_{k=0}^{n-1}g_n\left(\frac{k}{n}\right)=f(1)-f(0)=0$$ Thus there exists $k$ such that $g_n\left(\frac{k}{n}\right) g_n\left(\frac{k+1}{n}\right)\leqslant 0$. Since $f$ is continuous, $g_n$ is continuous and there exists $x\in\left[\frac{k}{n},\frac{k+1}{n}\right]$ such that $g_n(x)=0$ using intermediate value theorem. Since $g_n$ is defined on $\left[0,1-\frac{1}{n}\right]$, $x\in \left[0,1-\frac{1}{n}\right]$.

Answer 2

Suppose on the contrary that no such $x$ exists. Then the function $g(x) = f(x) - f(x + 1/n)$ is continuous; WLOG, take $g$ to be strictly positive over $[0, \frac{n - 1}{n}]$. Then we have

\begin{equation} \begin{split} g(0) + g(1/n) + ... + g(\frac{n - 1}{n}) &= f(0) - f(1/n) + f(1/n) - f(2/n) + ... + f(\frac{n - 1}{n}) - f(1) \\ &= f(0) - f(1) \\ &= 0 \end{split} \end{equation}

But this contradicts the fact that $g$ is strictly positive.