Prove or disprove: If $ f $ is periodic, then $|f|$ is also periodic.

Question

I started with the definition of periodic function and absolute value function. And I do it with discussing different cases of $x$ and $p$. But I got stuck with when $ -p\leq x <0 $ , I want to show $ |f(x+p) |= |f(x)| $.

Answer 1

let $p$ be our periodicity and $h=g\circ f$ with $f$ being periodict, we then have that $$h(n+p)=g\circ f (n+p)=g(f(n+p))=g(f(n))=g\circ f (n) = h(n)$$ so our composition is also periodic, in our case we have $g(x)=|x|$