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Is $f(x,y)=\sin(xy)$ is uniformly continuous in $\mathbb{R}^2$?

My solution is:

By the definition, we have to prove that $$ \forall \epsilon > 0 , \forall (x_1,y_1),(x_2,y_2) \in \mathbb{R}^2 ,\exists \delta $$ such that $$ |(x_1,y_1)-(x_2,y_2)| < \delta $$ implies $$ |\sin(x_1y_1) - \sin(x_2y_2)| < \epsilon $$ Then I take $x_2 = x_1 + \delta$,$y_2 = y_1 + \delta$

then $$|\sin(x_1y_1) - \sin(x_2y_2)| = |\sin(x_1y_1) - \sin(x_1y_1 + \delta(x_1+y_1 + \delta^2))|$$ Since $\sin(z)$ is continuous on $\mathbb{R}$ then $\forall x_1,y_1$ we have $$ \sin(x_1y_1 + \delta(x_1+y_1 + \delta^2)) \to \sin(x_1y_1) $$ when $\delta \to 0$. Then we can conclude that $\sin(xy)$ is uniform continuity on $\mathbb{R}^2$. My question is: (1) Am I right (2) If wrong, how to solve this problem?

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  • $\begingroup$ I think you might have some definitions mixed up - it looks like you want to talk about uniform continuity, but you actually wrote out the definition for continuity (i.e. the usual kind) - you would need to switch the order of the $\exists \delta$ and the preceding term to get uniform continuity. $\endgroup$ Commented May 9, 2018 at 3:02
  • $\begingroup$ Just a note about the title/question: do you mean "uniformly continuous" rather than "uniform convergence"? If so, the definition of uniform continuity has the points $(x,y)$ independent of $\delta$. $\endgroup$
    – Dave
    Commented May 9, 2018 at 3:03
  • $\begingroup$ Sorry, I want to write uniform continuity...... $\endgroup$
    – xxyshz
    Commented May 9, 2018 at 3:04
  • $\begingroup$ I've corrected it. $\endgroup$
    – xxyshz
    Commented May 9, 2018 at 3:04

1 Answer 1

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It's not uniformly continuous on $\mathbb{R}^2$.

You can pick $x_1 = y_1$ and $x_2 = y_2$, force $|(x_1,y_1) - (x_2,y_2)| = \sqrt{2}|x_1 - x_2|$ to be arbitrarily small and have $|\sin (x_1y_1) - \sin (x_2y_2)| \geqslant 1$ since $\sin x^2$ is not uniformly continuous. Take $x_1 = \sqrt{n\pi + \pi/2}$ and $x_2 = \sqrt{n \pi}$, for example, as $n \to \infty$.

Note that

$$\sqrt{n\pi + \pi/2} - \sqrt{n \pi} = \frac{\pi/2}{\sqrt{n\pi + \pi/2} + \sqrt{n \pi} },$$

and the RHS tends to $0$ as $n \to \infty,$ but

$$|\sin x_1^2 - \sin x_2^2| = | \sin(n\pi + \pi/2) - \sin(n\pi)| = 1.$$

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  • $\begingroup$ Thank you a lot ! $\endgroup$
    – xxyshz
    Commented May 9, 2018 at 3:51
  • $\begingroup$ @R.Sherlock. You're welcome. $\endgroup$
    – RRL
    Commented May 9, 2018 at 3:52
  • $\begingroup$ May I ask another question? What mistake I made in my wrong solution? $\endgroup$
    – xxyshz
    Commented May 9, 2018 at 3:54
  • $\begingroup$ @R.Sherlock: You are trying to prove by a direct $\delta-\epsilon$ approach. Relying on the continuity of the sine function you produce a $\delta$ but have not shown that it depends only on $\epsilon$ and not on $(x_1,y_1)$ and $(x_2,y_2)$. In fact, it can't as I showed. SInce the domain is all of $\mathbb{R}^2$ we can pick points $(x,y)$ where the x and y coordinates become arbitrarily large such that, even though $|(x_1,y_1) - (x_2,y_2)| < \delta$, the same $\delta$ does not control the function difference to be less than $\epsilon$. $\endgroup$
    – RRL
    Commented May 9, 2018 at 4:10
  • $\begingroup$ Oh, I notice that my $\delta$ depends on x and y , thanks again! $\endgroup$
    – xxyshz
    Commented May 9, 2018 at 4:14

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