real analysis - Prove that there exists $x_n$ such that $0 leq x_n leq 1-frac{1}{n}$ and $f(x_n)=f(x_n+frac{1}{n})$.

Suppose that the function $f:[0,1] \to \mathbb{R}$ is continuous on
$[0,1]$ and $f(0)=f(1)$. Prove that for each natural number $n$, there
exists $x_n \in \mathbb{R}$ such that $0 \leq x_n \leq 1-\frac{1}{n}$
and $f(x_n)=f(x_n+\frac{1}{n})$.

Though I don't know how the proof would look like, I have a strong feeling that it has something to do with the Intermediate Value Theorem, judging by the continuity of $f$ and the existence of such a $x_n$. So I guess I'm supposed to define a function $g(x)=f(x)-f(x+\frac{1}{n})$ on $[0,\frac{1}{n}]$ and try to claim that $g(x_n)=0$ for some $x_n \in [0,\frac{1}{n}]$. Unfortunately I don't know how to proceed further from here, maybe because I haven't made use of the fact that $f(0)=f(1)$.

Any hint and suggestion is much appreciated. Thank you!

Answer

Consider $g_n(x)=g(x+\frac{1}{n})-g(x)$ for $x\in [0,\frac{n-1}{n}]$.

Now $0=g(1)-g(0) = g_n(0) + g_n(\frac{1}{n}) + \dots g_n(\frac{n-1}n)$. If $g_n(\frac{k}k)=0$ for any $k$ you are done. On the other hand, if not, since their sum is $0$, at least one of the values $g_n(k/n)$ must be positive and at least one must be negative.

Now use the intermediate value theorem.