# Thread: Problem with periodic function

1. ## Problem with periodic function

Mój problem jest następujący:My problem is as follows

Let $f(x)=xh(x)$, where $h(x)$ is periodic function with period 1. Prove or give a counterexample to the following statement: Function $f$ is increasing if and only if $h$ is constant (and the constant is positive). We assume that $h$ is defined on the whole R.

Thank you for help

2. the "If" part is trivial.
for the " only if" part: since h is periodic function with period 1, we only need to focus on the closed interval [0,1].
To prove h is positive constant, it is surffice to prove that h is nondecreasing in [0,1] which implies h is constant in [0,1].
this can be Proved by contradiction!
suppose there is two points $x_1 < x_2$ in [0,1] such that $h(x_1)>h(x_2)$.
then $k+x_2>k+x_1$,where $k$ is arbitrary positive integer. since f is increasing, we have
$f(k+x_2)-f(k+x_1)=k(h(x_2)-h(x_1))+x_2h(x_2)-x_1h(x_1)\geq 0$ (*)
hold for any arbitrary positive integer $k$.
since $h(x_1)>h(x_2)$, let k approches infinity, then
$k(h(x_2)-h(x_1))+x_2h(x_2)-x_1h(x_1)$ tend to negative infinity, which is contradict to (*).
thus h is nondecreasing in [0,1]. combined with h(0)=h(1) gives that h is constant in [0,1], therefore constant in all real line.
since f is increasing, the constant is definitely positive, of course.