TITLE: On periodic and nonperiodic functions
QUESTION [5 upvotes]: It's known that $\sin(x^2)$ is not periodic.
Are there any theorems which generalise it?
I mean something like "$f(x^n)$ is not periodic if $f$ is a nonconstant periodic function and $n>1$"?
The above conjecture turned out to be false. Dave L. Renfro gave a counterexample.
What if we add continuity assumption?
New conjecture:
If $f$ is a nonconstant continuous periodic function and $n>1$ is an integer, then $f(x^n)$ is not periodic.

REPLY [8 votes]: That is correct.
Let $f$ be continuous and  $p>0$ a period of $f$. 
Then $g(x):=f(x^n)$ is also continuous. Assume $q>0$ is a period of it. 
The sequence $x_k=\sqrt[n]{kp}$ tends to $+\infty$ while at the same time $x_{k}-x_{k-1}\to 0$.
Now fix $x\in\Bbb R$ and let $\delta>0$, where wlog. $\delta<q$. 
There is $k_0$ such that $x_{k-1}<x_k<x_{k-1}+\delta$ for all $k>k_0$.
There is $m$ with $x+mq>x_{k_0}$. 
Let $k_1$ be minimal with $x_{k_1}>x+mq$. Then $x_{k_1-1}-mq<x<x_{k_1}-mq$ and $(x_{k_1}-mq)-(x_{k_1-1}-mq)<\delta$.
Thus $g(y)=f(0)$ for some $y$ with $|y-x|<\delta$. As $\delta$ was arbitrary, we conclude from the continuity of $g$ that $g(x)=f(0)$. Thus $g$ is constant (and in fact so was $f$)