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This question already has an answer here:

Suppose that the function $f:[0,1] \to \mathbb{R}$ is continuous on $[0,1]$ and $f(0)=f(1)$. Prove that for each natural number $n$, there exists $x_n \in \mathbb{R}$ such that $0 \leq x_n \leq 1-\frac{1}{n}$ and $f(x_n)=f(x_n+\frac{1}{n})$.

Though I don't know how the proof would look like, I have a strong feeling that it has something to do with the Intermediate Value Theorem, judging by the continuity of $f$ and the existence of such a $x_n$. So I guess I'm supposed to define a function $g(x)=f(x)-f(x+\frac{1}{n})$ on $[0,\frac{1}{n}]$ and try to claim that $g(x_n)=0$ for some $x_n \in [0,\frac{1}{n}]$. Unfortunately I don't know how to proceed further from here, maybe because I haven't made use of the fact that $f(0)=f(1)$.

Any hint and suggestion is much appreciated. Thank you!

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marked as duplicate by Aryabhata, Thomas Andrews, Amzoti, drawar, Pedro Tamaroff Apr 24 '13 at 17:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Oh my bad, sorry for asking a duplicated question, I did a full search and couldn't find any similar problems so I decided to post it here. Anyway, thanks to all involved, for taking your time answering my question. One more vote and it should be closed. $\endgroup$ – drawar Apr 24 '13 at 16:27
  • $\begingroup$ drawar, it is hard to find that, you don't have to apologize. $\endgroup$ – Aryabhata Apr 24 '13 at 17:26
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Consider $g_n(x)=g(x+\frac{1}{n})-g(x)$ for $x\in [0,\frac{n-1}{n}]$.

Now $0=g(1)-g(0) = g_n(0) + g_n(\frac{1}{n}) + \dots g_n(\frac{n-1}n)$. If $g_n(\frac{k}k)=0$ for any $k$ you are done. On the other hand, if not, since their sum is $0$, at least one of the values $g_n(k/n)$ must be positive and at least one must be negative.

Now use the intermediate value theorem.

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  • $\begingroup$ Very nice, thank you! $\endgroup$ – drawar Apr 24 '13 at 15:59
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This problem is tricky, by which I mean there's a proof that doesn't use the usual toolkit (IVT, for instance).

Suppose not. Then there exists $n \geq 1$ for which, for every $x \in [0,1-1/n]$, $f(x) \neq f(x + 1/n)$. The left and right hand side are continuous functions, and so $\neq$ is either a $<$ or a $>$ identically on the domain $[0,1-1/n]$. Choose $>$ without loss.

Plugging in $x = k/n$, we have $$ f(k/n) > f((k+1)/n) $$ Applying in succession, we obtain $$ f(0) > f(1/n) > \cdots > f((n-1)/n) > f(1) $$ which is a contradiction.

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  • $\begingroup$ The IVT comes in, for example, in the assertion that $\neq$ is either a $<$ or a $>$. I guess I meant that the IVT comes up in a tricky way. $\endgroup$ – A Blumenthal Apr 24 '13 at 15:33
  • $\begingroup$ I think I see your point here. Did you mean if there exist $x_1$ and $x_2$ s.t $f(x_1) > f(x_1+\frac{1}{n})$ and $f(x_2) < f(x_2+\frac{1}{n})$ then there is an $x$ between $x_1$ and $x_2$ s.t $f(x) = f(x+\frac{1}{n})$, which is a contradiction? $\endgroup$ – drawar Apr 24 '13 at 16:12
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    $\begingroup$ @drawar Yeah, that's one way to see it. Be careful where the contradiction hypothesis comes in, though. $\endgroup$ – A Blumenthal Apr 24 '13 at 16:24

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