# Find a Continuous Function [duplicate]

Possible Duplicate:
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 $f(x) \geq f(0)$

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## marked as duplicate by Hagen von Eitzen, Martin Sleziak, Noah Snyder, no identity, Ｊ. Ｍ.Oct 10 '12 at 17:04

You might wanna do some reading about thwe Universal Chord Theorem. –  Pedro Tamaroff Oct 8 '12 at 16:45

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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@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 8 '12 at 15:57
@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 8 '12 at 15:59
@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 8 '12 at 16:17
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.