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If $f:[0,1]\to\Bbb R$ is continuous and $f(0)=f(1)$ show that $\forall n\in\Bbb N,\exists x_n,y_n\in[0,1]:|x_n-y_n|=1/n$ and $f(x_n)=f(y_n)$

It is supposed that I must use the intermediate value theorem, not something more advanced (like any kind of derivative or so).

The case for $n=2$ is the unique that I can prove, it is easy because if we create a function $h(x)=f(x+1/2)-f(x)$ (what is continuous because is a sum of two continuous functions) then we can see that $h(0) h(1/2)<0$ and cause the intermediate value theorem then exists a point $c$ such that $h(c)=0$ and $f(c+1/2)=f(c)$.

But for the general case $1/n$ Im lost... I was trying to see anything, with graphs, but seems too complicate. I feel that one possible solution could come from some recursive mechanic involving $f(x)$ and $h_n(x)$, but it seems too complicate and I dont think this will be the real solution.

Can you help me please, leaving some hint or so? Thank you very much.

Masacroso
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    I think that you mean $h(0)h(1/2)<0$, since $h(1)=f(3/2)-f(1)$ and $f(3/2)$ is undefined unless you extend $f$. – Xiang Yu Jun 03 '16 at 05:54
  • I would try to look around $\max f(x)$, and if $\max f(x)=f(0)$, around $\min f(x)$ – H. Potter Jun 03 '16 at 05:55
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    http://math.stackexchange.com/questions/371571/prove-that-there-exists-x-n-such-that-0-leq-x-n-leq-1-frac1n-and-fx?rq=1 – Marco Cantarini Jun 03 '16 at 06:05
  • Yes @XiangYu, I little typo. – Masacroso Jun 03 '16 at 06:05
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    The blind imitation of the $n=2$ proof is to introduce $h_n(x)=f(x+1/n)-f(x)$ and to note that $h_n(0)+h_n(1/n)+\cdots+h_n(1-1/n)=f(1)-f(0)=0$ hence the numbers $h_n(k/n)$ are neither all positive nor all negative. Can you finish? – Did Jun 03 '16 at 06:07
  • Yes thank you very much @Did. Unfortunately the question is a duplicate but it is hard to search for this question :( – Masacroso Jun 03 '16 at 06:09
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    And a tribute to Paul Lévy is due: http://www.jstor.org/stable/2317756?seq=1#page_scan_tab_contents – Did Jun 03 '16 at 06:09

1 Answers1

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The idea is still to use the intermediate value theorem. For fixed $n$, setting $g(x):=f(x+\frac1n)-f(x)$, we need to show that $g$ has a zero in the interval $[0,\frac{n-1}{n}]$. To do this, we consider the value of $g$ at $0,\frac{1}{n},\dots\frac{n-1}{n}$. Note that $$\sum_{k=0}^{n-1}g(\frac{k}{n})=\sum_{k=0}^{n-1}f(\frac{k+1}{n})-f(\frac{k}{n})=f(1)-f(0)=0.$$ Now if there exists $0\leq k\leq n-1$ such that $g(\frac{k}{n})=0$, then we are done. Otherwise, $g(\frac{k}{n}$) are not zero for $0\leq k\leq n-1$, then there must exist $0\leq k_1\leq n-1$ and $0\leq k_2\leq n-1$ such that $g(\frac{k_1}{n})$ and $g(\frac{k_2}{n})$ has opposite sign, thus we see from the intermediate value theorem that $g$ must have a zero between $\frac{k_1}{n}$ and $\frac{k_2}{n}$.

Xiang Yu
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