Proving that continuous function $f:[0,2] \to \mathbb{R}$ such as $f(0)=f(2)$, $\exists x \in [0,1]:f(x)=f(x+1)$

Question

I am having trouble formalizing the answer to this question. Any help would be appreciated.

Is it true that for every continuous function $f:[0,2] \to \mathbb{R}$ such as $f(0)=f(2)$, $\exists x \in [0,1]:f(x)=f(x+1)$?

If the function is periodic of degree one this seems true to me. I tried proving by contradiction, supposing that $\forall x \in [0,1], f(x) \neq f(x+1)$, but I couldn't get any further.

Answer 1

You might try to define a continous function $g : [0,1] \to \mathbb{R}$ as $$ g(x) = f(x+1) - f(x)$$ and then use intermediate value theorem.