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