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

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?

• 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. Commented May 9, 2018 at 3:02
• 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 $\delta$.
– Dave
Commented May 9, 2018 at 3:03
• Sorry, I want to write uniform continuity...... Commented May 9, 2018 at 3:04
• I've corrected it. Commented May 9, 2018 at 3:04

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

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

• Thank you a lot ! Commented May 9, 2018 at 3:51
• @R.Sherlock. You're welcome.
– RRL
Commented May 9, 2018 at 3:52
• May I ask another question? What mistake I made in my wrong solution? Commented May 9, 2018 at 3:54
• @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 $\epsilon$.
– RRL
Commented May 9, 2018 at 4:10
• Oh, I notice that my $\delta$ depends on x and y , thanks again! Commented May 9, 2018 at 4:14