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Universal Chord Theorem

I am having a problem with this exercise. Could someone help?

Suppose $a \in (0,1)$ is a real number which is not of the form $\frac{1}{n}$ for any natural number n n. Find a function f which is continuous on $[0, 1]$ and such that $f (0) = f (1)$ but which does not satisfy $f (x) = f (x + a)$ for any x with $x$, $x + a \in [0, 1]$.

I noticed that this condition is satisfied if and only if $f(x) \geq f(0)$

Thank you in advance

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2 Answers2

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Look at $f(x) = \sin(2\pi x)$. For which values of $a$ can you find an $x \in [0,1]$ with $x+a \in [0,1]$ and $f(x) = f(x+a)$? In particular, if you additionally require $a > \frac{1}{2}$, can such an $a$ exist at all?

Once you've answered that you've solved your problem for some values of $a$. Which are those?

A general solution can be found in the answers to Universal Chord Theorem (Link found by the user who asked the question). To quote $$ f(x) = \sin^2\left(\frac{\pi x}{a}\right) - x \ \sin^2\left(\frac{\pi}{a}\right) $$

is a solution. This works because $f(x) = f(x+a)$ implies $a \sin^2\left(\frac{\pi}{a}\right) = 0$ and thus $a=\frac{1}{n}$ for some $n \in \mathbb{N}$. The answers to the linked questions also prove that $a \neq \frac{1}{n}$ for every $n \in \mathbb{N}$ is a necessary condition for a solution to exist.

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fgp
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  • I couldn't find. Could you please help me.. –  Oct 08 '12 at 15:48
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    @user43758 I won't solve it for you, but I'll provide further hints. How far have you got? Haver you answered the first question I posted. If not, I suggest you make a sketch of $\sin(2\pi x)$ for $x \in [0,1]$. Observe what happens to the sign if you go from one point $x$ to a point $x+a$ where $a > 0.5$. – fgp Oct 08 '12 at 15:57
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    @fgp If I understand your answer, you propose for $1/4<a<1/2$ the function $f(x)=\sin(4,\pi,x)$. Let $\delta=(a-1/4)/2$. Then $0<\delta<1/4$, $f(1/4-\delta)=f(1/2+\delta)$ and $(1/2+\delta)-(1/4-\delta)=a$. – Julián Aguirre Oct 08 '12 at 15:59
  • For the first question I should solve $sin(2\pi x)=sin(2\pi(x+a))$ no ?

    I find a=0

     –  Oct 08 '12 at 16:13
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    @user43758 yeah, but you are interested in solutions $a > 0$. For $a=0$, $f(x)=f(x+a)$ quite obviously holds for every functions $f$, no? – fgp Oct 08 '12 at 16:17
  • Concerning your first question, we notice that f changes the sign when a> 1/2. It changes from positive to negative –  Oct 08 '12 at 16:29
  • @JuliánAguirre Ups, you're right. Guess I should have heeded my own advice, and done a sketch before I generalized from $n=1$ (where this works, but only for $a > 0.5$) to the general case. – fgp Oct 08 '12 at 16:30
  • @user43758 Correct. This should solve your problem if you assume $a > 0.5$. I've botched the general case as JuliánAguirre pointed out, you'll have to figure out how to generalize this on your own, or wait for me to fix my answer – fgp Oct 08 '12 at 16:33
  • Oh I understand now. Since the sign of f changes when x=1/2, we have to consider a>1/2 right ? –  Oct 08 '12 at 16:36
  • @user43758 Sort of. Since the sign changes for $a > 0.5$, you can be quite certain that $f(x) \neq f(x+a)$. Thus, $sin(2\pi x)$ is a solution for $a > 0.5$. – fgp Oct 08 '12 at 16:40
  • ok. So now we have to generalise this by consider $sin(2n \pi x)$ and finding for what a do the above properties work right? –  Oct 08 '12 at 16:42
  • @user43758 No, that generalization was wrong, as JuliánAguirre pointed out. Will update my answer if I find a way to fix it. – fgp Oct 08 '12 at 16:45
  • Have you found something? I was looking but didn't find –  Oct 08 '12 at 17:07
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    http://math.stackexchange.com/questions/16374/universal-chord-theorem –  Oct 08 '12 at 17:11
  • Can someone help me with the generalisation please? –  Oct 08 '12 at 17:26
  • @user43758 The link you posted contains the answer. It's at the bottom of the answer by Aryabhata... – fgp Oct 08 '12 at 17:34
  • But I have a problem. It is the real number a such that a is not of the form 1/n .. It the linked problem, we have r=1/m such that m is an integer –  Oct 08 '12 at 17:42
  • What do you think ? –  Oct 08 '12 at 17:51
  • Re-read that question and its answers more carefully. They show both that if $r$ is $1/n$ for some integer $n$, then there always is an $x$ with $f(x) = f(x+r)$ if $f$ is continuous and $f(0)=f(1)$. But if $r$ (your $a$) isn't $1/n$, then there exists an $f$ such that no such $x$ exists (which is what you want). Beware that they call $a$ what you call $x$, and they call $r$ what you call $a$... – fgp Oct 08 '12 at 17:53
  • So the proof that I require is from For any $r \in (0;1)$ to the end right ? –  Oct 08 '12 at 18:02
  • let us continue this discussion in chat –  Oct 08 '12 at 18:06
  • I have a question: Why does $\displaystyle a\ \sin^2\left(\frac{\pi}{a}\right) = 0$ imply that a=1/n ? –  Oct 08 '12 at 18:37
  • @user43758 Draw $\sin(x)$. Where is it zero? Where are hence the zeros of $\sin^2(x)$? Now compare that with $\pi/a$... – fgp Oct 08 '12 at 19:37
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Hint: Since $\frac1a\notin\mathbb N$, there is some $n\in\mathbb N_0$ such that $n<\frac1a<n+1$. Let $r= 1-na$. Then $0<r<a$. Let us try to find a function such that $f(x+a)-f(x)=1$ for all $x$ with $0\le x \le 1-a$. Thus we have by induction $f(x)=k+f(t)$ if $x=ka+t$ with $k\in\mathbb N_0$ and $0\le t<a$ . Then for $f(1)=f(0)$ we need $f(0)=f(1)=f(na+r)=f(r)+n$, i.e. $f(r)=-n$. See if you can build a complete $f$ from this.

Hagen von Eitzen
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