real analysis - If $f:[0,1]toBbb R$ is continuous and $f(0)=f(1)$ show that $forall ninBbb N,exists x_n,y_nin[0,1]:|x_n-y_n|=1/n$ and $f(x_n)=f(y_n)$

If $f:[0,1]\to\Bbb R$ is continuous and $f(0)=f(1)$ show that $\forall n\in\Bbb N,\exists x_n,y_n\in[0,1]:|x_n-y_n|=1/n$ and $f(x_n)=f(y_n)$

It is supposed that I must use the intermediate value theorem, not something more advanced (like any kind of derivative or so).

The case for $n=2$ is the unique that I can prove, it is easy because if we create a function $h(x)=f(x+1/2)-f(x)$ (what is continuous because is a sum of two continuous functions) then we can see that $h(0) h(1/2)<0$ and cause the intermediate value theorem then exists a point $c$ such that $h(c)=0$ and $f(c+1/2)=f(c)$.

But for the general case $1/n$ Im lost... I was trying to see anything, with graphs, but seems too complicate. I feel that one possible solution could come from some recursive mechanic involving $f(x)$ and $h_n(x)$, but it seems too complicate and I dont think this will be the real solution.

Can you help me please, leaving some hint or so? Thank you very much.

Answer

The idea is still to use the intermediate value theorem. For fixed $n$, setting $g(x):=f(x+\frac1n)-f(x)$, we need to show that $g$ has a zero in the interval $[0,\frac{n-1}{n}]$. To do this, we consider the value of $g$ at $0,\frac{1}{n},\dots\frac{n-1}{n}$. Note that
$$\sum_{k=0}^{n-1}g(\frac{k}{n})=\sum_{k=0}^{n-1}f(\frac{k+1}{n})-f(\frac{k}{n})=f(1)-f(0)=0.$$
Now if there exists $0\leq k\leq n-1$ such that $g(\frac{k}{n})=0$, then we are done. Otherwise, $g(\frac{k}{n}$) are not zero for $0\leq k\leq n-1$, then there must exist $0\leq k_1\leq n-1$ and $0\leq k_2\leq n-1$ such that $g(\frac{k_1}{n})$ and $g(\frac{k_2}{n})$ has opposite sign, thus we see from the intermediate value theorem that $g$ must have a zero between $\frac{k_1}{n}$ and $\frac{k_2}{n}$.