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Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous with $f(0)=f(1)$. If $h \in (0,\frac{1}{2})$ is not of the form $\frac{1}{n}$, there does not necessarily exist $|x-y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=\frac{2}{5}$

I'm thinking I need to use some kind of modification to a sin function to get this to work. Not sure how to come up with an explicit formula, though. I could just draw a picture, but I want to find some explicit formula so I can show that it's true.

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If you accept a piecewise function as an explicit formula, then the function in the figure below works.

enter image description here

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A modified $\sin$ function could work, but I think it would be easier to reason using a piecewise linear function. Try defining a function like the one below. The $x$-intercepts are specifically chosen to be $0, \frac16, \frac13,\frac12,\frac23,\frac56,1$. What has to be true about $|x-y|$ if $f(x)=f(y)$? (If you really want to use $\sin$, then modify your $\sin$ function so that it has $x$-intercepts at the same places as this function. That should also work.)

enter image description here

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  • $\begingroup$ can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)? $\endgroup$ – Alain Jun 22 '17 at 5:51
  • $\begingroup$ You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work. $\endgroup$ – kccu Jun 24 '17 at 17:56

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