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I can prove that it is continuous, because as long as $$\left|\sqrt{(x-x')^2+(y-y')^2}\right|<\min{\left\{\frac{\epsilon}{2(1+|x|+|y|)},1\right\}}$$, we have $|xy-x'y'|<\epsilon$.

But how do I prove that it is not uniformly continuous?

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Look at the slope in the $y$ direction at $(x = a, 0)$. This is just $a$. Therefore, as $x \to \infty$, the function is steeper and steeper. Now, you can use this to produce a counterexample directly for the definition of uniform continuity (just take any $\epsilon$ and show that no $\delta$ value will work by taking sufficiently big $x$). – Marek Sep 7 '12 at 7:22
I think you need to be careful to define the domain of your function if you are introducing "uniform continuity" - a continuous function on a closed and bounded (=compact) set is uniformly continuous. It would be good to state that you are considering this as a function from $$\mathbb R^2 \to \mathbb R$$ (if this indeed the case). – Mark Bennet Sep 7 '12 at 7:33
up vote 4 down vote accepted

You want to show that there is some $\epsilon>0$ for which, for any $\delta>0$, there is some $x,y,x',y'$ with $$\sqrt{(x-x')^2+(y-y')^2}<\delta \;\text{ yet }\; |xy-xy'|\geq \epsilon.$$ Let $\epsilon=1/2$ and note that $|xy-xy'|\geq |x||y-y'|$ so for any $\delta>0$, if $|x|>1/\delta$ we have $$|y-(y+\delta/2)|=\delta/2<\delta\; \text{ yet }\;|xy-x(y+\delta/2)|=|x|\delta/2>1/2$$ hence $xy$ is not uniformly continuous.

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But isn't it "$...\text{ yet }\; |xy-x'y'|\geq \epsilon.$" instead of "$...|xy-xy'|...$"? – Voldemort Sep 7 '12 at 7:30
@Voldemort We can always let $x'=x$, which is what I do. – Alex Becker Sep 7 '12 at 7:31

HINT: Notice that the $\delta$ that you used to prove continuity depends heavily on $x$ and $y$: when either of them is large in absolute value, your $\delta$ is very small compared with $\epsilon$. Try to make that observation more rigorous: show that that there is an $\epsilon>0$ such that for any $\delta>0$, if you take $|xy|$ large enough, you can find a point $\langle x',y'\rangle$ such that $\sqrt{(x-x')^2+(y-y')^2}<\delta$, but $|xy-x'y'|\ge\epsilon$. That will show that there is no $\delta$ that ‘works’ for that $\epsilon$ independently of the choice of point $\langle x,y\rangle$. (Don’t work too hard to pick this ‘bad’ $\epsilon$; most anything will work!)

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