A problem about periodic functions Suppose that $f(x):\mathbb{R}\rightarrow \mathbb{R}$ is a periodic function with a minimal positive period $T$. Can $g(x)=f(x^2)$ be periodic?
I know it is impossible if the condition $\forall x \in (0,T),f(x)\not=f(0)$ is added. But Is there a counterexample for the general case? Thanks for any help in advance.
 A: This question is well above my league, I will nevertheless attempt a solution for learning purposes, apologising to anybody's whose time I waste in the process.
$f$ is a periodic function, with minimal period $T$.
Without loss of generality, we can rescale the $x$-axis so to have period $T=1$. Let us still call the re-scaled function $f$.
By definition, $$f(0) = f(1) = \cdots =  f(j)$$ for any $j \in \mathbb{N}$.
As $g(x) = f(x^2)$, this implies
$$g(0) = g(1) = g(\sqrt{2}) = \cdots = g(\sqrt{j}) $$ for any $j \in \mathbb{N}$.
If $g$ were periodic of period $T'$, the number of occurences on which  $g$ attained the value $g(0)$ between  any $x$ and $x+ T'$ would be constant.
If the number of such occurrances were finite, we would reach a contraddiction.
Indeed it would not be possible  as there is an increasing number of values of the type $\sqrt{j}$, $j \in \mathbb{N}$, in intervals $(x, x+ T')$ as $x$ increases.
The case whereby the function $g$ attains the value $g(0)$ infinitely often over a period is being considered, following the observations made in the comments by M. Winter.
