Can we apply IVT to solve this?

Question

Let $f:[0,2\pi] \to \mathbb R$ be continuous and periodic (meaning that $f(2\pi) = f(0)$). Show that there exists an $x ∈ [0, π]$ such that $f(x) = f(x+\pi)$. Any hints?

Answer 1

HINT : Let $g : [0,\pi]$ be defined by $g(x) = f(x+\pi) - f(x)$.

• Is $g$ continuous? (Must be)

• What is the relationship between $g(0)$ and $g(\pi)$?

• If $g(x) = 0$ for some $x$, what does that imply about $f$?

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Question : Can we find another constant $c$ not $\pi$ or $2 \pi$ such that $f(x+c) = f(x)$ must occur for some $x$?

Answer 2

If $f(0)=f(\pi)$ then we're done, so suppose otherwise.

Without loss of generality, this gives $f(0)-f(0+\pi)>0$. Then we have $f(\pi)-f(0)<0$, but we know that $f(0)=f(2\pi)=f(\pi+\pi)$, so we get $f(\pi)-f(\pi+\pi)<0$. Do you see how to proceed by considering $f(x)-f(x+\pi)$ with IVT?