I'm trying to find out if
$$
f(x,y)=\cos\left(\sqrt{xy}\right)
$$
is uniformly continuous on the set $\{(x, y)\in\mathbb{R}^2 : x\geq0, y\geq 0\}$. The theorems I have available to use for this are for showing that the function is uniformly continuous (and merely sufficient rather than necessary), but I happen to know that it is not uniformly continuous in this case. But I can't show why.
I've tried working with the definition to get:
$$
|(x, y)-(x_0, y_0)|=\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta
$$
but then I need to show that this does not imply that
$$
|f(x, y)-f(x_0, y_0)|=|\cos\sqrt{xy}-\cos\sqrt{x_0y_0}|<\epsilon.
$$
But how would I do that? Some hints would be appreciated. Thanks.
Best Answer
Look at special points. Say points with the same $x$-coordinate and a small difference in the $y$-coordinate,
$$f(x,y+h) - f(x,y).$$
Looking at $\sqrt{xy}$, you should be able to see for what choice of $y$ you get the largest difference.
You need one $\varepsilon > 0$ so that for every $\delta > 0$ there are points less than $\delta$ apart so that $\lvert f(x_1,y_1) - f(x_2,y_2)\rvert \geqslant \varepsilon$.