Let $f$ be continuous on $[0, 1]$ with $f(0) = f(1)$. Prove that there exists $c ∈ \left[0,\frac{1}{2}\right]$ such that $f(c) = f(c+\frac{1}{2})$.

Question

Let $f$ be continuous on $[0, 1]$ with $f(0) = f(1)$. Prove that there exists $c ∈ \left[0,\frac{1}{2}\right]$ such that $f(c) = f\left(c+\frac{1}{2}\right)$.

So, I know I'm supposed to use the Intermediate Value Theorem, and I can see generally how it's gonna be used, but I'm kind of confused how to? 0 and 1 are not opposite signs, so IVT can't be applied, can it? And then for the IVT corollary, it states $f(0)=f(1)$, so we can't use that? Will it be one where we assume to the contrary or...?

Define $g$ with domain $[0,\frac 12], g(x) = f(x) - f(x+\frac 12).$ If this has a sign change you can apply the IVT. If not, you are done. — user317176, Apr 21 '21 at 22:17
@red: This question is one of the most popular here at MSE. It's been posted years ago. You need to learn how to find a duplicate post here. That's a skill you want to work on. I am sure some expert here could post it.... — , Apr 22 '21 at 01:47

score 3 · Accepted Answer · answered Apr 21 '21 at 22:15

3

Hint Define $g: [0, \frac{1}{2}] \to \mathbb R$ as $g(x)=f(x+\frac{1}{2})-f(x)$.

What is $g(0)+g(\frac{1}{2})=?$ How does this help?

answered Apr 21 '21 at 22:15

N. S.

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Let $f$ be continuous on $[0, 1]$ with $f(0) = f(1)$. Prove that there exists $c ∈ \left[0,\frac{1}{2}\right]$ such that $f(c) = f(c+\frac{1}{2})$.

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