Prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$.

Question

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.

Answer 1

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.

Answer 2

Why dont you try to prove it by contradiction???? Suppose it is uniformly continuous...

Answer 3

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 $n$.