$f:[0,1] \to \Bbb R$ is continuous function s.t. $f(0)=f(1)$. Prove that $\forall n\geq 1$ $\exists x \in [ 0 ,1] $ s.t. $f(x)=f(x+\frac{1}{n})$

Question

Problem: Let $ f: [ 0,1 ] \to \Bbb R $ be continuous function s.t. $ f(0) = f(1) $. Prove that for all natural $ n \geq 1 $ there exists $ x \in [ 0 ,1] $ s.t. $ f(x) = f(x + \frac{1}{n} ) $

Attempt: For $ n = 1 $ we'll choose $ x =0 $ and then we'll have $ f(0) = f(1) $. Let $ n\geq 2 $. If $ f $ is constant then we're finished. Assume $ f $ is not constant. Define $ g(x) = f(x) - f(x+\frac{1}{n}) $, $\forall x \in [0,1-\frac{1}{n}] $, note this function is continuous by the continuity of $ f $ on the given interval. [ I don't know how to proceed from here ]

Do you have any ideas as to how to proceed? I don't have any intuition as to what I should do next.
Thanks for the help in advance!

Answer 1

Let us define $0 = x_0<x_1<....<x_n = 1$ and $x_i= \frac{i}{n}$ for $i = 0,...,n$. We define also the function $g(x) = f(x+\frac{1}{n})-f(x).$

If there exists $x_i$ (for $0 \le i \le n-1$) such that $g(x_i) = 0$ then the problem is solved.

If not, calculate the sum of $g(x_i)$, we have $$\sum_{i=0}^{n-1}g(x_i) =\sum_{i=0}^{n-1}\left(f(x_{i+1}) -f(x_{i}) \right) = f(1) -f(0) = 0 \tag{1}$$

As $g(x_i) \ne 0$ then $g(x_i)<0$ or $g(x_i)>0$. From $(1)$ we deduce that there must exist 2 values $x_i <x_j$ such that $g(x_i)g(x_j)<0$ (if not, the LHS of $(1)$ is strictly positive or negative).

As $g(x)$ is a continuous function, by the intermediate value theorem, there exists $x'$ satisfying $x_i<x' <x_j$ and $g(x') = 0$. Then $f(x') = f(x'+\frac{1}{n})$.

Q.E.D