Let $f: [0,1] \to \mathbb R$ be a continuous function with $f(0)= f(1)$. Then for any $c \in [0,1/2]$ , $f(c) = f(c+1/2)$.

Question

Let $f: [0,1] \to \mathbb R$ be a continuous function and $f(0)= f(1)$. Then prove that for any $c \in [0,1/2]$ , $f(c) = f(c+1/2)$.

According to me there can be many cases as the question doesn't give information. I can think of a case where in $f(x)$ is constant function. How many other cases are possible? I want a rigorous proof with reasoning.

Answer 1

Consider the function $g:\big[0,\frac{1}{2}\big]\to \Bbb R$ defined by $$g(x)= f(x)- f\bigg(x+\frac{1}{2}\bigg),x\in \bigg[0,\frac{1}{2}\bigg].$$ Then, $g$ is continouos and $g(0)=f(0)-f\big(\frac{1}{2}\big)$ and $g\big(\frac{1}{2}\big)=f\big(\frac{1}{2}\big)-f(1)=f\big(\frac{1}{2}\big)-f(0)$. So $g(0)\geq 0\implies g\big(\frac{1}{2}\big)\leq 0$ and $g(0)\leq 0\implies g\big(\frac{1}{2}\big)\geq 0$. Now using Intermediate Value Property we have $c\in \big[0,\frac{1}{2}\big]$ such that, $g(c)=0$. So $f(c)=f\big(c+\frac{1}{2}\big)$.