Antiderivative of periodic function I practise before the real analysis exam and I got stuck on few excercises.
"if $F$ is an antiderivative of a continuous periodic function and $F$ is bounded, then $F$ is periodic"
$~\to~$ one should give a proof or counterexample.
(I'm convinced it's true, but how to prove it?)

Hint(s):
Start with the definition of the antiderivative(s) of $f$:
$$F(x) = \int_0^xf(t)dt + C$$
where $C$ is a constant. Since $f$ is periodic, with period $T>0$, then 
$$F(2T) = \int_0^{2T}f(t)dt + C = \int_0^{T}f(t)dt + \int_T^{2T}f(t)dt + C = 2\int_0^{T}f(t)dt + C$$
$$F(3T) = \int_0^{3T}f(t)dt + C = F(2T) + \int_T^{2T}f(t)dt  = 3\int_0^{T}f(t)dt + C$$
Can you generalize this to find an expression for $F(kT)$ for any integer $k$ (for example by induction)? What does this tell you about $\int_0^{T}f(t)dt$ given the conditions $F$ have to satisfy? 
Finally what is the value of
$$F(x+T) - F(x) = \int_x^{x+T}f(t)dt = \int_x^{T}f(t)dt + \int_T^{x+T}f(t)dt$$
given the result found above?
