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Consider the continuous and periodic function $f:\mathbb R \rightarrow \mathbb R$ with period $T > 0$ so that $f(x)=f(x+T)$ for any $x$.

Question: Prove that there exists a $c$ such that $f(c)=f(c+ \pi)$.

Note: This is not claiming that the period of $f$ is $\pi$, and is not a restatement of the definition of periodic. Many comments below reflect this incorrect interpretation of the statement of the problem.

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closed as off-topic by Shailesh, choco_addicted, JKnecht, Yves Daoust, Arjang May 3 '16 at 11:10

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  • $\begingroup$ what is pi suppose to be? $\endgroup$ – Arjang May 3 '16 at 9:30
  • $\begingroup$ Pi is the irrational number 3.1415926.... $\endgroup$ – user334433 May 3 '16 at 9:33
  • $\begingroup$ what thought have you put into the question? $\endgroup$ – Siddharth Bhat May 3 '16 at 9:56
  • $\begingroup$ I was thinking that there always exists a line of symmetry. Intermediate value theorem might turn out to be useful. But can't use it properly $\endgroup$ – user334433 May 3 '16 at 10:00
  • $\begingroup$ Is the question complete? What if the function has a period larger than $\pi$? $\endgroup$ – N. F. Taussig May 3 '16 at 10:03
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Periodicity and continuity of $f$ implies boundedness, hence $f$ has a maximum and a minimum within the closure of a period $P=[0, T]$. Say the minimum and maximum are attained at $t_1\in P$ and $t_2\in P$. We have then

$$f(t_1) - f(t_1 + \pi) \leq 0 \quad f(t_2) - f(t_2 + \pi) \geq 0$$

as well as continuity of $f(t) - f(t + \pi)$.

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  • $\begingroup$ I think this is right $\endgroup$ – user334433 May 3 '16 at 11:20
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    $\begingroup$ This is correct. Thanks for the clean argument, Coolwater! I knew your answer was solid, but had no choice but to work up a few examples to convince myself since the question looked like an incorrect restatement of the definition of periodic. Apparently many other users were confused as well! Note that the use of $\pi$ here is irrelevant, thus we can generalize: For any continuous periodic function $f$, and $\forall a$, $\exists c$ such that $f(c)=f(c+a)$. $\endgroup$ – jdods May 3 '16 at 18:04

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