Fixed point of continuous function Let $f:[0,5] \rightarrow \mathbb R$ be continuous where $f(0) = f(5)$. Then $\exists c \in [0,4]: f(c) = f(c+1)$.
My first idea was to show that $h:[0,4] \rightarrow \mathbb R: x \mapsto f(x)-f(x+1)$ has a zero.
What is the easiest way to prove this ?
 A: If $f(0)=f(1)$, then you're finished. Assume $f(0)<f(1)$. The map $g(x)=f(x+1)-f(x)$ is continuous and $g(0)>0$. If $g(y)\le0$ for some $y\in[0,4]$, then you can use the intermediate value theorem to conclude the existence of a $c\in[0,4]$ such that $g(c)=0$. So assume that $g(y)>0$ for all $y\in[0,4]$. Now, $f(5)-f(0)=\sum_{i=0}^4 (f(i+1)-f(i))=\sum_{i=0}^4 g(i)>0$ which contradicts $f(5)=f(0)$.
This worked so nicely because the domain is of length an integer times the difference between the points with the same image. Call the length of the domain $l$, and the (shorter) difference $k$, so here $l=5, k=1$.
There is a counterexample where $1<\frac lk<2$. Consider $\sin(x)$ on $[0,2\pi]$, so $l=2\pi$. The zeros of $\sin$ are at $0, \pi, 2\pi$. Between $0$ and $\pi$ the function is positive, between $\pi$ and $2\pi$ it is negative. Now, if $k=\frac 32\pi$, then $\sin(c)$ will always be $\ge0$ and $\sin(c+k)$ will always be $\le0$. But here we have $\frac lk=\frac 43.$
A: If $f(1)=f(0)$ we are done otherwise WLG suppose $f(1)>f(0)$. Repeat the same reasoning: if $f(2)=f(1)$ we are done otherwise there's two cases:


*

*if $f(2)<f(1)$ then the function $g(x)=f(x+1)-f(x)$ is continuous and change signs so there's $c\in(0,1)$ such that $g(c)=0$

*the case $f(2)>f(1)$: we repeat this reasoning until we assume that $f(5)>f(4)$ (the intermediate other cases are now clear) then
$$0=f(5)-f(0)=(f(5)-f(4))+(f(4)-f(3))+\cdots+(f(1)-f(0))>0$$
which's a contradiction.

A: You shouldn't be looking for a fixed point: a fixed point of your $h$ would satisfy $f(x) - f(x+1) = x$, whereas you are looking for $f(x) - f(x+1) = 0$; i.e. a zero of $h$. This feels to me like it should be an application of the intermediate value theorem.
