if |f| is periodic then f is periodic

Question

Decide whether the following statement about a function f: R -> R is true. If |f| is periodic, then f is periodic. Give a proof or counterexample.

Answer 1

That's not true. To get a counter example you just have to have a nonperiodic function $\sigma$ that takes values $\pm1$ and a periodic $\phi$. Then $f(x)=\sigma(x)\phi(x)$ is non-periodic but $|f(x)| = \phi(x)$ is periodic.

For example let

$$\sigma(x) = \begin{cases}+1& \mbox{if } x\ge0\\ -1 & \mbox{otherwise} \end{cases}$$

the function $\phi$ can be any periodic, trivial example would be $\phi(x)=1$ which makes the above example the counter example, but you could also select $\phi(x)=\sin x$ which makes the counter example continuous.