If $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $f(0)=f(1)$, then there exists $ x\in [0,1]$ such that $f(x) = f(x+\frac{1}{n})$, where $n$ is any natural number.
Let $g(x)=f(x+\frac{1}{n})-f(x)$ on $[0,1-\frac{1}{n}]$. Now, $f$ is continuous function on closed interval.So, it must attain its bounds. Let $m$ be minima at $c_1 \in [0,1]$ and $M$ be maxima at $c_2 \in [0,1]$.
Now if $c_1,c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $g(x)$ on $[c_1,c_2]$.
If $c_1$ or $c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $[c_1,c_1-\frac{1}{n}]$ or
$[c_2,c_2-\frac{1}{n}]$ respectively. If
$c_1,c_2$ both do not belong to $[0,1-\frac{1}{n}]$, then apply IVT to $g$ on $[c_1-\frac{1}{n},c_2-\frac{1}{n}]$.
Is this proof correct? Thanks.
Best Answer
I don't see what your last paragraph leads to. And you must use the fact that $n\in \Bbb N$ because if $r\in (0,1)\setminus \{1/n: n\in \Bbb N\}$ then there is a continuous $f:[0,1]\to \Bbb R$ with $f(0)=f(1),$ such that $f(x)\ne f(x+r)$ for all $x\in [0,1-r].$
I saw this in American Mathematical Monthly. The title was (approximately?) The Theorem Of The Horizontal Chord. The idea is that $$\sum_{j=0}^{n-1}g(j/n)=\sum_{j=0}^{n-1}f((j+1)/n)-f(j/n)=$$ $$=f(1)-f(0)=0 $$ so the members of $\{g(j/n):j=0,...,n-1\}$ cannot be all positive or all negative, so by the IVT, $g(x)$ must be $0$ for some $x\in [0,1-1/n].$