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I've been reading through Zorich's "Analysis I" book recently, and I came across this nice little exercise.

Let $f: [0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$. Show that

  • for any $n\in \mathbb N$ there exists a horizontal closed interval of length $\frac 1n$ with endpoints on the graph of this function;

  • if the number $\ell$ is not of the form $\frac 1n$ there exists a function of this form on whose graph one cannot inscribe a horizontal chord of length $\ell$.

The first part can be proven like this: Consider $g: [0,(n-1)/n] \to \mathbb R$ given by $g(x) = f(x) - f(x+1/n)$. Then

$$\sum_{k=0}^{n-1} g(k/n) = 0$$

and therefore either all of these points are zero or there exists both a point where $g$ is positive and a point where $g$ is negative. By continuity, there must then also be a point where $g = 0$. So we are done.

Now, the second statement seemed rather counterintuitive, and I have given it some time now, but don't see a counterexample for $\ell < 1/2$.

(For $\ell > 1/2$ the function $f(x) = \sin(2\pi x)$ will do.)

Can anyone help me out?

Cheers,

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I think in the example for $l > 1/2$, you meant to say $f(x) = \sin(2\pi x)$ right? –  Shuhao Cao May 3 '11 at 19:52
    
@MathChief: Ah, yes. Of course you're right. Thanks! –  Sam May 3 '11 at 20:47
    
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1 Answer 1

up vote 4 down vote accepted

Try $f(x) = \sin^2(\pi x/\ell) - x \sin^2(\pi/\ell)$

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Great example. Thanks –  Sam May 3 '11 at 21:09
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