# Let $f:\mathbb{I} \to \mathbb{R}$ continuous function such that $f(0)=f(1)$.

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

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?

• 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 $\mathbb{I}$. Nov 16, 2018 at 2:00
• No, only this is questions and it is about of continuous function Nov 16, 2018 at 2:00
• Any hunch? What was your 1st idea when see this? What have you learned?
– xbh
Nov 16, 2018 at 2:01
• I think that i should work with succesiones for the $\frac{1}{n} + x$ Nov 16, 2018 at 2:03
• Is that induction? Nov 16, 2018 at 2:05

Suppose there is no such $$x$$. Then either $$f(x+\frac 1 n ) >f(x)$$ for all $$x$$ or $$f(x+\frac 1 n ) for all $$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) which is a contradiction.
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.
• Continuity implies what you conjecture, plus compactness of $\mathbb{I}$, I think. Nov 16, 2018 at 2:07
• @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. Nov 16, 2018 at 2:13
• 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? Nov 16, 2018 at 2:23
• 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 $x \in [0, 1-\alpha]$. Nov 17, 2018 at 2:14