Let $n\in \mathbb{N}$. $f$ is a real valued function, defined and continuous on $[0,n]$ and $f(0)=f(n)$. Prove that there exists $x_1,x_2\in [0,n]$, so that $x_1-x_2=1$ and $f(x_2)=f(x_1)$.
Hints on how to approach this problem would be appreciated.
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Have you considered using induction? – Nick Sep 17 '14 at 18:12
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@user176744: What about if $n=0$? – Lehs Sep 17 '14 at 18:14
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I'm guessing that they don't define $0$ as a natural number – Oria Gruber Sep 17 '14 at 18:14
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question is posed to me as $n\in \mathbb{N}$ not $n\in \mathbb{N}_0$. – user176744 Sep 17 '14 at 18:15
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@user176744 In some countries, it's usual to have $\Bbb N={0, 1, \dots}$. That's why this question about $0\in\Bbb N$ arises quite often on MSE. – Jean-Claude Arbaut Sep 18 '14 at 11:34
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Hint: Assuming you meant continuous functions from $[0,n]$ to $\mathbb{R}$, consider a new function $$ g(x) = f(1+x)-f(x), \forall x\in[0,n-1]$$.
Try and prove that $g(x)$ is continuous and crosses zero. So, $\exists x\in[0,n-1]$ such that $f(1+x)=f(x)$.
Shash
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Looks obscure at first sight, but it's indeed a valuable hint :-) – Jean-Claude Arbaut Sep 18 '14 at 11:41
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I think your question is incomplete; consider the map $$f:\ [0,2]\ \longrightarrow\ \Bbb{R}^2:\ x\ \longmapsto (\cos(\pi x),\sin(\pi x)).$$ It is clear that $f$ is continuous and that $f(0)=f(2)$, but for $x_1,x_2\in[0,2]$ we have $f(x_1)=f(x_2)$ if and only if $x_1,x_2\in\{0,2\}$. In particular we cannot have $x_1-x_2=1$.
Servaes
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I edited the question, as it's most probably what the OP meant. – Jean-Claude Arbaut Sep 18 '14 at 11:36
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Just to expand on Shash's answer:
Notice that due to Rolle's theorem, we know there is $\xi \in (0,n)$ such that $f'(\xi)=0$.
Suppose wlog that $f(\xi)$ is local maximum, and $\xi >1$.
In that case, $g(0)=f(1)-f(0)$ is larger than zero, because $1$ is on the way $\xi$ and the function is ascending.
That also means that $g(n-1)=f(n)-f(n-1)$ is negative due to a similar argument.
$g$ is continuous, $g(0)*g(n-1) <0$, this means (think bisection method) that there is $\psi \in (0,n-1)$ such that $g(\psi)=0$. this means that $f(1+\psi)=f(\psi)$, and we are done.
Disclaimer: this argument succeeds if $f(\xi)$ is local minimum / maximum. In that case that it isn't (meaning $f''(x)=0$) then I'm not 100% sure that what you are trying to prove holds).
Disclaimer 2: The assumption that $\xi >1$ is important and non trivial. even if $f(\xi)$ is local extremum, if $\xi <1$ then my proof won't work.
Oria Gruber
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Well I already assumed so many assumptions, might just as well add one more to the list... – Oria Gruber Sep 17 '14 at 19:34