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This question already has an answer here:

Prove that the function $f:\Bbb R \rightarrow \Bbb R ;x\mapsto\sin(x^{2})$ is not periodic.

Let's assume the opposite, i.e that $f$ is periodic. Then for all $\tau \in\Bbb R$, $\sin(\tau +x)^{2}=\sin(x^{2})$.

How do I continue from here?

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marked as duplicate by Martin Sleziak, Michael Albanese, Namaste, E. Joseph, Daniel W. Farlow Nov 20 '16 at 0:15

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    $\begingroup$ In particular $f(\tau) = f(0) = 0$. That limits the possible values of $\tau$. Then strategically choose $y$ to see that $f(y+\tau) \neq f(y)$ for all the possibilities you got from $f(\tau) = 0$. $\endgroup$ – Daniel Fischer Nov 19 '16 at 15:07
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    $\begingroup$ That is not what it means for $f$ to be periodic. $f$ is periodic if there exists some value $\tau \in \mathbb{R}$ such that $f(\tau + x)=f(x)$ for all $x \in \mathbb{R}$, i.e., $\sin((\tau+x)^2)=\sin(x^2)$. $\endgroup$ – kccu Nov 19 '16 at 15:07
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    $\begingroup$ Have a look at this math.stackexchange.com/questions/2020495/… $\endgroup$ – David Quinn Nov 19 '16 at 15:08
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    $\begingroup$ Can we replace the squaring function with a more general class of function? Noting that sin is periodic it seems that the squaring function "messes up" the periodicity. $\endgroup$ – Jacob Wakem Nov 19 '16 at 15:35
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    $\begingroup$ Conjecture: Given f is periodic and not constant f o s is not periodic where s is an increasingly increasing function. $\endgroup$ – Jacob Wakem Nov 19 '16 at 15:57
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As per your approach, assume that $\sin(x^2)$ is periodic .

Then since every continuous periodic function is uniformly continuous so will be $\sin (x^2)$.

But $\sin (x^2)$ is not uniformly continuous.

Take $x_n=\sqrt{2n\pi+\frac{\pi}{2}}$ and $y_n=\sqrt {2n\pi}$

$|x_n-y_n|$ can be made sufficiently small but $|f(x_n)-f(y_n)|=1$ can't be made so.

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  • $\begingroup$ Nice observation! +1 $\endgroup$ – user384138 Nov 19 '16 at 15:41
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If $\sin(x^2)$ were periodic, then its derivative $2x\cos (x^2)$ would also be periodic. But a continuous periodic function is bounded, and $2x\cos (x^2)$ is not. The conclusion follows.

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You have $\sin (x+\tau)^2 = \sin x^2$, then $$\cos\left(\frac{(x+\tau)^2-x^2}{2}\right)\sin\left(\frac{(x+\tau)^2-x^2}{2}\right) = 0$$

Then, you can find the condition of $\tau$.

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  • $\begingroup$ Then I get $1)$ $2x\tau+\tau^{2}-k\pi=0$ or $2)$ $2x\tau+\tau^{2}-2k\pi=0$ What can I do after that? $\endgroup$ – lmc Nov 19 '16 at 16:08
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Hint:

The function $f(x)$ has period $T$ if $f(x)=f(x+T)$ for some constant $T$, independent from $x$. In your case this means: $$ \sin x^2=\sin(x+T)^2 $$ but we have: $$ x^2+2k\pi=(x+T)^2 \iff T^2+2xT-2k\pi=0 $$ and the solutions of this equation in $T$ are dependent from $x$.

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  • $\begingroup$ Why does it have to be independent from x? $\endgroup$ – lmc Nov 19 '16 at 16:00
  • $\begingroup$ Because if $T$ is a function of $x$ it is not a constant. $\endgroup$ – Emilio Novati Nov 19 '16 at 16:02
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    $\begingroup$ How do you get from $\sin x^2=\sin(x+T)^2$ to $x^2+2k\pi=(x+T)^2$ (presumably with $k\in\mathbb{Z}$)? In general, $\sin u=\sin v$ does not imply $u+2k\pi=v$ (e.g., $u=\pi/3$, $v=2\pi/3$). $\endgroup$ – Barry Cipra Nov 19 '16 at 18:27
  • $\begingroup$ I suppose that the OP knows how to solve an equation of the form $\sin X=\sin \alpha$ and how find the two set of solutions: $X=\alpha+2k\pi$ and $X=\pi- \alpha +2k \pi$ with $ k \in \mathbb{Z}$. My answer is simply an hint suggesting the fact that, in this case, we need a solution in $T$ and that this solution cannot be independent of $x$. For this it sufficient to look at a set of solutions. $\endgroup$ – Emilio Novati Nov 20 '16 at 9:39
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There should exist $T>0$ such that, if $z$ is a zero of $\sin(x^2)$, then also $z+T$ is a zero of $\sin(x^2)$.

Since $0$ is a zero, also $T$ should be, which means $$ T^2=h\pi $$ for some integer $h$.

Since $\sqrt\pi$ is a zero, also $\sqrt{\pi}+T$ should be, which means $$ (\sqrt{\pi}+T)^2=(k+1)\pi $$ for some integer $k$.

Also $\sqrt{2\pi}$ is a zero, so we get $$ (\sqrt{2\pi}+T)^2=(l+2)\pi $$ for some integer $l$. Therefore $$ \begin{cases} T^2=h\pi \\[4px] T^2=k\pi-2T\sqrt{\pi} \\[4px] T^2=l\pi-2T\sqrt{2\pi} \end{cases} $$ From the first two equations we get $2T=(k-h)\sqrt{\pi}$ and, substituting in the third equation, $$ h\pi=l\pi-(k-h)\sqrt{2}\pi $$ so $$ \sqrt{2}=\frac{l-h}{k-h} $$ would be rational.

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The function $f(x)=\sin(x^2)$ has a finite number of zeroes in any finite interval. If it were periodic with period $T$, then there would be a uniform upper bound on the number of zeroes in any interval of the form $(x,x+T)$. But the spacing between consecutive zeroes of $\sin(x^2)$ tends to $0$ as $x\to\infty$. Hence there is no upper bound on the number of zeroes in intervals of the form $(x,x+T)$, hence $f(x)=\sin(x^2)$ is not periodic.

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