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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 ?

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up vote 3 down vote accepted

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 $\frac lk=\frac 43.$

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This is the proof I would have given, but it uses the fact that the longer interval is of length an integer times the length of the shorter. There ought to be a direct argument not using this accident. – Lubin Aug 7 '13 at 16:12
@Lubin: That's true. Intuitively, it should work for arbitrary lengths. – Stefan Hamcke Aug 7 '13 at 16:45
Can we prove this, too ? Maybe by changing the finite sum into an integral ? – Epsilon Aug 7 '13 at 17:26
@André: I added a counterexample where the interval has not length integer times the difference between the points. I still think it is true if the domain is at least twice as long as the difference. – Stefan Hamcke Aug 7 '13 at 19:19

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.
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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.

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