$f:[0,2\pi]\to\mathbb{R}, $ f continuous. $f(0)=f(2\pi)$ Show that there exists an $x_0\in(0,\pi)$ such that $f(x_0)=f(x_0+\pi)$
continuous and periodic function such that $f(0)=f(2\pi)$.
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functions
continuity
periodic-functions
1 Answers
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Hint: Consider $g:[0, \pi] \to \mathbb{R}$, defined as $g(t) = f(t) - f(t+\pi)$ and use the intermediate value theorem.
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0Perhaps add a hint to look at the signs of $g(0)$ and $g(\pi)$ and observe that $g(0)=-g(\pi)$. – 2017-02-21