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$\mathbb{I} = [0,1]$

Let $f:\mathbb{I} \to \mathbb{R}$ continuous function such that $f(0)=f(1)$. Prove that for all $n \in \mathbb{N} $ there $ x \in \mathbb{I}$ such that $ x + \frac{1}{n} \in \mathbb{I}$ and $f( x + \frac{1}{n})=f(x)$

Could you help me by giving me an idea of ​​how to do it?

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  • $\begingroup$ No derivatives, but maybe there's a theorem we can use to show that $f(x+1/n)-f(x)$ has at least one zero on $\mathbb{I}$. $\endgroup$
    – Prototank
    Nov 16, 2018 at 2:00
  • $\begingroup$ No, only this is questions and it is about of continuous function $\endgroup$ Nov 16, 2018 at 2:00
  • $\begingroup$ Any hunch? What was your 1st idea when see this? What have you learned? $\endgroup$
    – xbh
    Nov 16, 2018 at 2:01
  • $\begingroup$ I think that i should work with succesiones for the $\frac{1}{n} + x $ $\endgroup$ Nov 16, 2018 at 2:03
  • $\begingroup$ Is that induction? $\endgroup$
    – Prototank
    Nov 16, 2018 at 2:05

2 Answers 2

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Suppose there is no such $x$. Then either $f(x+\frac 1 n ) >f(x)$ for all $x$ or $f(x+\frac 1 n ) <f(x)$ for all $x$ (by IVP applied to the continuous function $f(x+\frac 1 n ) -f(x))$. Assume that $f(x+\frac 1 n ) >f(x)$ for all $x$. (the proof is similar in the other case). Then $f(0)<f(\frac 1 n) <f(\frac 2 n)<\cdots <f(1)$ which is a contradiction.

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Extend $f$ to $\mathbb{R}$ periodically (and so continuously by $f(0) = f(1)$). Let $g(x): = f(x+1/n) - f(x)$. Then we have $$ g(x) + g\left(x+\frac{1}{n}\right) + \cdots + g\left(x + \frac{n-1}{n}\right) = 0 $$ This implies that $g(x)$ can't have always same sign, and so $g$ has a zero by intermediate value theorem.

I'm sure that there's better way to show that $f(x+\alpha) = f(x)$ has root for any $0<\alpha < 1$, since this method only works for $\alpha\in \mathbb{Q}$. But I don't have any idea for this now.

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  • $\begingroup$ Continuity implies what you conjecture, plus compactness of $\mathbb{I}$, I think. $\endgroup$
    – Prototank
    Nov 16, 2018 at 2:07
  • $\begingroup$ @Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $\alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number. $\endgroup$
    – Seewoo Lee
    Nov 16, 2018 at 2:13
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    $\begingroup$ The sequence of $r_n$ converges to $\alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:K\to\mathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=\lbrace (x,y):x,y\geq 0 \ x+y\leq 1\rbrace$. Choose a convergent subsequence in $K$. How does that look? $\endgroup$
    – Prototank
    Nov 16, 2018 at 2:23
  • $\begingroup$ @Prototank That seems good for me. Thanks! $\endgroup$
    – Seewoo Lee
    Nov 16, 2018 at 3:37
  • $\begingroup$ But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $\alpha$ other than some $1/n$ the claim is false (even if $\alpha$ is rational). Let p be a periodic function whose period is such an $\alpha$, ($0 < \alpha < 1$), and $p(0) = 0$, $p(1) = c \neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + \alpha) - f(x) = -c\alpha \neq 0$ for all $x \in [0, 1-\alpha]$. $\endgroup$ Nov 17, 2018 at 2:14

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