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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.

Hoag Seki
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    just apply the definition of being periodic and verify. – Fabrice NEYRET Nov 04 '15 at 10:18
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    @Hetebrij Yes, that's a similar question, but it doesn't ask for a counter example and none is given. – skyking Nov 04 '15 at 10:27
  • "I can think of an example where this wouldn't hold.

    Take 1,-1,1,-1,-1,1,-1,-1,-1,1,-1,-1,-1,-1." It might not be the clearest counterexample, but should get you started, among with the answers, the statement is false.

     – Hetebrij Nov 04 '15 at 10:31

1 Answers1

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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.

skyking
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  • Choosing $\phi(x) = \sin(x)$ wouldn't make the example continuous, it would just change the preimages of $1$. – A.P. Nov 04 '15 at 15:19
  • @A.P. I disagree, where would it be discontinuous? Since it becomes $f(x)=-\sin(x) = \sin-x = \sin|x|$ for negative $x$ and for non-negative it becomes $f(x)=\sin x= \sin|x|$. It's the composition of two continuous functions so it becomes continuous. – skyking Nov 04 '15 at 16:34
  • Oh, probably that's a notation issue. I interpret $f = \sigma \phi$ as $f(x) = \sigma(\phi(x))$. Do you define it as a point-wise product, i.e. as $f(x) = \sigma(x) \phi(x)$? – A.P. Nov 04 '15 at 16:41
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    @A.P. Yes, I clarified that in the text. – skyking Nov 04 '15 at 16:43