Any ideas on the problem below?

Question

suppose f:[0,1]->R is contineous how can it be proves that for x between 0 and 0.5 there is an x such that f(x)=f(x+1/2). IF it is known that f(0)=f(1)

Answer 1

Let g(s) = f(s) - f(s + 1/2), s in [0, 1/2]. Then, g is continuous (because so is f) and g(0) = f(0) - f(1/2) and g(1/2) = f(1/2)  - f(1) =  f(1/2)  - f(0) = - g(0).

If g(0) =0, then f(0) = f(1/2), so that x =0 satisfies the desired condition. Otherwise, g(0) and g(1/2) have opposite signs. Since g is continuous, there is an x in (0, 1/2) such that  g (x) = f(x) - f(x + 1/2) = 0 →  f(x) = f(x + 1/2), as desired.

Hence, there's x ∈ [0, 1/2] with f(x) = f(x + 1/2).

Answer 2

It seems some info is not given. The fuction should be symmetric about x = 1/2, then only it is possible. Only continuity and f(0) = f(1)doesnot suffice.