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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!

07216
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  • Quick comment: by periodic do you mean that for a finiteness constant interval the function repeats? Because if that is the case sin(x^2) by itself fails that form of periodicity which when added to a period functions produces another non periodic – Sidharth Ghoshal May 22 '13 at 03:56
  • As opposed to dynamic interval length: like the function from 0 to 1 is the same as 1 to 3/2 – Sidharth Ghoshal May 22 '13 at 03:56
  • I know it can be solved by proving $\sin{x^2}$ isn't periodic,but I want to fellow the whole problem. – 07216 May 22 '13 at 04:09

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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.

André Nicolas
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