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I'm trying to solve the following problem.

Suppose $f$ is continuous on $[0,12]$ and $f(0) = f(12)$. Define $g(x) = f(x) - f(x+6)$. Prove that there exist $t,s \in [0,12]$, where $s = t + 6$, and $f(s) = f(t)$ and $a,b \in [0,12]$ where $a = b + 3$ such that $f(a) = f(b)$, but that it is not possible to find $q,r$ with $q = r + 4$ and $f(q) = f(r)$.

I was able to figure out the case where $s = t + 6$. I consider $x = 0$, $x = 6$, and prove that $g(0) = - g(6)$, and use the IVT. Either $g(0) = g(6) = 0$ or $g$ crosses the $x$-axis. In the first case, I have $f(0) = f(6)$. In the second, I use the intermediate value theorem to find an appropriate value since $g(0) < 0 < g(6)$ or $g(6) < 0 < g(0)$.

I can't figure out where to start for the case $a = b + 3$ or how to disprove the case $q = r + 4$. I need to somehow consider values of $x$ that are three apart, so my instinct was to consider $x = 9$, but that doesn't lead me anywhere. I need to still work in $0$ and $12$ somewhere, the only values of $x$ I know, to find an equality, but also had no luck. I tried contradiction for the $q = r + 4$ case, but couldn't make any progress.

I'd appreciate any help with these.

Brad G.
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1 Answers1

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The first result can be mimic to obtain the more general following one. For $v \gt 0$ and $f : [u, u+2v] \to \mathbb R$ continuous with $f(u)=f(u+2v)$, it exists $e,e+v \in [u,u+2v]$ with $f(e)=f(e+v)$.

This enables to get the second result when applied to $u=r, v=3$ where $r$ is from the first result.

And as stated in a comment, the third statement is wrong according to the Universal chord theorem.

mathcounterexamples.net
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  • There is no counterexample for the third result. Let $h(x)=f(x)-f(x+4)$ for $x\in [0,8]$. Then $h(0)+h(4)+h(8)=0$ so $h$ cannot be always $>0$ or always $<0$ so $h(x)=0$ for some $x$. These are instances of the Horizontal Chord Theorem: If $f:[a,b]\to \Bbb R$ is continuous with $f(a)=f(b)$ and if $n\in \Bbb N$ then there exists $x\in [a, b-(b-a)/n]$ with $f(x)=f(x+(b-a)/n).$ – DanielWainfleet Jul 16 '21 at 05:49
  • @DanielWainfleet I can follow everything in y our explanation up to $h(0) + h(4) + h(8) = 0$. I can't figure out, from there, how to find $q,r$ with $f(q) = f(r)$ or exactly which value of $h$ you're using. Do you have any tips for this? – Brad G. Jul 16 '21 at 06:40
  • @BradG. For $some$ $r$ we have $0=h(r)=f(r)-f(r+4).$ I dk what you mean by "which value of $h$". I defined $h$ in terms of $f$. – DanielWainfleet Jul 16 '21 at 09:31
  • I understand why $h(0) + h(4) + h(8) = 0$, but I don't know which value of $r$ this corresponds to. My apologies if this is obvious, as I'm sure it is. – Brad G. Jul 16 '21 at 15:21
  • Sorry, I'm still stumped on the second and third cases. I'd appreciate any help on either. – Brad G. Jul 16 '21 at 21:02