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If I want to prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$, I need to show that:

$\exists\epsilon>0$ $\forall\delta>0$ $\exists{x,y}\in\mathbb{R}\ : |x-y|<\delta$ and $|\sin(x^2) - \sin(y^2)|\geq\epsilon$.

So let's take $\epsilon = 1$. Then I want $|\sin(x^2)-\sin(y^2)|\ge1$. That's the case if $\sin(x^2)$=0 and $\sin(y^2)=\pm1$. Thus $x^2=n\pi$ and $y^2=n\pi + \frac{1}{2}\pi$. Now I'm stuck on expressing x and y, which I want to express in $\delta$, to ensure that $|x-y|<\delta$.

Thanks in advance for any help.

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3 Answers 3

up vote 8 down vote accepted

You have chosen $x^2=n\pi$ and $y^2 = n\pi+\frac{\pi}{2}$, so you can take $x=\sqrt{n \pi}$ and $y=\sqrt{n \pi + \frac{\pi}{2}}$.

Then,

$|x-y|=\sqrt{n \pi + \frac{\pi}{2}}-\sqrt{n \pi}=\frac{n\pi + \frac{\pi}{2}-n\pi}{\sqrt{n \pi + \frac{\pi}{2}}+\sqrt{n \pi}}=\frac{\frac{\pi}{2}}{\sqrt{n \pi + \frac{\pi}{2}}+\sqrt{n \pi}}<\frac{2}{2\sqrt{n \pi}}<\frac{1}{\sqrt{n}}$

If $n > \frac{1}{\delta^2}$ then $|x-y|<\delta$ but $|f(x)-f(y)|\geq 1$. The values $x,y$ are very close but $f(x)$ and $f(y)$ are far apart. Intuitively, for $\epsilon = 1$ there is no $\delta$ that allows you to know $f(x)$ within precision $\epsilon$ if you know $x$ within precision $\delta$. The oscillations in $\sin(x^2)$ get faster and faster, on arbitrarily small intervals the function changes its value from 0 to 1.

Note that if you change the function to $\sin x$ the proof will fail, because taking $x=n\pi$ and $y=n\pi+\frac{\pi}{2}$ does not make $|x-y|<\delta$ for small $\delta$ (indeed, $y-x$ is constant). In fact, $\sin x$ is uniformly continuous.

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Hint: If $x^2=n\pi$ and $\delta>0$, note that $(x+\delta)^2>x^2+2\delta x$. To make this $>n\pi +\frac12\pi$, it suffices to have $2\delta x > \frac12 \pi$. This gives you a condition on $n$.

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Why would I want $(x+\delta)^2 \gt n\pi+\frac{1}{2}\pi$ ? I'm still not feeling it. –  Jeroen Oct 22 '12 at 22:27

if f(x)=sin(x^2) then derivative of f(x) is not bounded on real line, but it is bounded on any bounded subset of R, so f(x) is not uniformly continuous on the domain R but uniformly continuous on any bounded subset of R.

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1  
This even gives a reason that $f$ is Lipschitz on each bounded subset of $\mathbb R$. For uniform continuity on bounded intervals, it suffices that the function is continuous. This doesn't answer the question. –  Jonas Meyer Sep 17 '13 at 13:41

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