solve the question
Let $f:[0, 2\pi] \rightarrow [0, 2\pi]$ be continuous such that $f(0)=f(2\pi)$. Show that there exists $x \in [0, 2\pi]$ such that
$$f(x)= f(x+\pi)$$
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Analysis

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Re: Analysis
Hello my friend,Asis ghosh wrote:solve the question
Let $f:[0, 2\pi] \rightarrow [0, 2\pi]$ be continuous such that $f(0)=f(2\pi)$. Show that there exists $x \in [0, 2\pi]$ such that
$$f(x)= f(x+\pi)$$
consider the function $g(t)=f(t)  f(t+\pi) \; , \; t \in [0, 2\pi]$. Clearly, $g$ is continuous as a sum of continuous functions. Note that
$$g(0)=f(0)f(\pi) \quad , \quad g(\pi)=f(\pi)f(2\pi)=f(\pi)f(0)$$
Hence
$$g(0) g(\pi) = \left ( f(0) f(\pi) \right )^2 \leq 0$$
Hence by Bolzano's theorem we have that there exists an $x \in (0, \pi) \subset [0, 2\pi]$ such that $g(x)=0$. That is $f(x)=f(x+\pi)$. Can you now complete the proof of the existence of $x \in [0, 2\pi]$?
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