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

$$f(x)= f(x+\pi)$$

Hello my friend,

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 $x \in [0, 2\pi]$?