Prove that $\sin^2{x}+\sin{x^2}$ isn't periodic by using uniform continuity Before the problem is a proof that says a periodic function whose domain is $\mathbb{R}$ is uniformly continuous.So actually the problem is to prove $\sin^2{x}+\sin{x^2}$ isn't uniformly continuous.I hope to fellow the problem.Thanks for the zealous!
 A: We proceed from the definition of uniform continuity. (One can get the result more simply by considering the derivative of $\sin(x^2)$.)
It is enough to show that $\sin(x^2)$ is not uniformly continuous. For if $\sin^2 x+\sin(x^2)$ were uniformly continuous, then since $\sin^2 x$ is uniformly continuous, the difference  $\sin^2 x+\sin(x^2)-\sin^2 x$ would be uniformly continuous. 
Let $\epsilon=\frac{1}{2}$. We show that there does not exist a $\delta\gt 0$ such that for all $x$, if $|x-y|\lt \delta$, then $|\sin(x^2)-\sin(y^2)|\lt \epsilon$.
Let $\delta$ be a fixed positive quantity. Let $x=\sqrt{2n\pi}$, where $n$ is a large integer to be chosen later. Then $\sin(x^2)=0$. Let $n$ be such that $2\delta\sqrt{2n\pi}\gt \frac{\pi}{2}$. Note that 
$$(x+\delta)^2=x^2+2x\delta+\delta^2\gt x^2+2x\delta\gt 2n\pi +2\delta\sqrt{2n\pi}\gt 2n\pi+\frac{\pi}{2}.$$  
It follows that there is a $y$ such that $x\lt y\lt x+\delta$ and $\sin(y^2)=1$. In particular, $|\sin(x^2)-\sin(y^2)|\gt \frac{1}{2}$. This completes the proof. 
