Show that there is a real number $ξ ∈ [1, 2]$ with $f(ξ) = f(ξ − 1).$

Question

Suppose that $f$ is a continuous function on $[0, 2]$ such that $f(0) = f(2)$.

Show that there is a real number $\xi \in [1, 2]$ with $f(\xi) = f(\xi − 1).$

Answer 1

This is the special case $n=2$ of the Theorem of the Horizontal Chord: Let $a<b$ and let $f:[a,b]\to \Bbb R$ be continuous with $f(a)=f(b).$ If $n\in \Bbb N$ then there exists $x\in [a, b-\frac {b-a}{n}]$ such that $f(x)=f(x+\frac {b-a}{n}).$

Proof: Trivial for $n=1,$ with $x=a.$ For $n>1,$ for $x\in [a, b-\frac {b-a}{n}]=dom(g)$ let $g(x)=f\left(x+\frac {b-a}{n}\right)-f(x).$ We have $$\sum_{j=0}^{n-1} g\left(a+j\cdot \frac {b-a}{n}\right)=f(b)-f(a)=0.$$

For example with $[a,b]=[0,1]$ and $n=4$ we have $$\sum_{j=0}^3 g(j/4)=$$ $$=(f(1/4)-f(0)+(f(2/4)-f(1/4)+(f(3/4)-f(2/4))+(f(4/4)-f(3/4)=$$ $$=f(1)-f(0)=0.$$

So the members of $\{g(a+j\cdot \frac {b-a}{a}): 0\leq j\leq n-1\}$ cannot all be positive nor all negative. So the function $g$ cannot be always positive nor always negative. Since $g$ is continuous and its domain is an interval, therefore $0=g(x)=f(x+\frac {b-a}{n})-f(x)$ for some $x\in dom(g)=[a,b-\frac {b-a}{n}].$

Remark. If $0<r<1$ and $r$ is not the reciprocal of a natural number then there exists a continuous $f:[0,1]\to \Bbb R$ with $f(0)=f(1),$ such that $f(x)\ne f(x+r)$ for all $x\in [0,1-r].$

Answer 2

$$f(\xi)=f(\xi-1)\iff f(\xi)-f(\xi-1)=0$$

$$k(\xi)=f(\xi)-f(\xi-1),\,\,k(2)=f(2)-f(1)=-(f(1)-f(2))=-(f(1)-f(0))=-k(1)$$

Hence by the intermediate value theorem exist a number $\xi_0\in[1,2]$ such that $f(\xi_0)-f(\xi_0-1)=0$