Let $f\colon \mathbb R \to \mathbb R$ be a continuous $2\pi$-periodic function, i.e. for every $t \in \mathbb R,$ we have $f(t) = f(t+ 2\pi)$. Which of the following statements are true?
a. There exists $t_0 \in \mathbb R$ such that $f(t_0) = f(−t_0).$
b. There exists $t_0 \in \mathbb R$ such that $f(t_0) = f(t_0+\pi/2).$
c. There exists $t_0 ∈ \mathbb R$ such that $f(t_0) = f(t_0 +\pi/4).$
My Try:-
(a) Let $f(-\pi+2\pi)=f(\pi)=f(-\pi)$
(b) let $g(t)=f(t+\pi/2)-f(t)$
$g(0)=f(\pi/2)-f(0)$
$g(3\pi/2)=f(2\pi)-f(3\pi/2)$
I am wishing to apply Intermediate Value Theorem, But not able to apply.
(c) $g(t)=f(t+\pi/4)-f(t)$
Please help me to apply IVT in a correct way.
Here also I wish to apply Intermediate value theorem.
$$0=f(0)-f(2\pi)=(f(0)-f(\pi/2))+(f(\pi/2)-f(\pi)$$ $$+(f(\pi))-f(3\pi/2))+(f(3\pi/2)-f(2\pi)).$$ Hence at least one of the terms here has to be positive and at least one negative. It follows that the continuous function $f(x)-f(x+\pi/2)$ takes both positive and negative values. Hence it must vanish at some point. This proves b) and c) is similar.