Explanation:
In order to prove that affirmation, we define the function g over the interval [0, 1/2] with the formula [tex] g(x) = f(x+1/2)-f(x) . [/tex]
If we evaluate g at the endpoints we have
g(0) = f(1/2)-f(0) = f(1/2) - f(1) (because f(0) = f(1))
g(1/2) = f(1) - f(1/2) = -g(0)
Since g(1/2) = -g(0), we have one chance out of three
We will prove that g has a zero on [0,1/2]. If g(0) = 0, then it is trivial. If g(0) ≠ 0, then we are in one of the first two cases, and therefore g(0) * g(1/2) < 0. Since f is continuous, so is g. Bolzano's Theorem assures that there exists c in (0,1/2) such that g(c) = 0. This proves that g has at least one zero on [0,1/2].
Let c be a 0 of g, then we have
[tex] 0 = g(c) = f(c+1/2)-f(c) [/tex]
Hence, f(c+1/2) = f(c) as we wanted.
