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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.

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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 $\lvert f(x_1,y_1) - f(x_2,y_2)\rvert \geqslant \varepsilon$.

$f(2/\delta,0) - f(2/\delta, \delta/2) = 1 - \cos 1$, and $\lVert (2/\delta,0) - (2/\delta,\delta/2)\rVert = \delta/2$.

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  • $\begingroup$ So, please tell me if I'm misunderstanding: irrespective of the choice of $\lambda (>0)$, we can find an $\epsilon$ such that $|f(x_1, y_1)-f(x_2, y_2)|\geq \epsilon$. For example, $\epsilon=0.25$. But, shouldn't this be for all $(x, y)$ in the domain? I can see that the for all $y$ part is satisfied since we can set $y=0$ which is the maximum. But what about $x$? Hope it makes sense... "Maximum" in the sense that is when the greatest difference is achieved. $\endgroup$ – hejseb Jan 31 '14 at 13:52
  • $\begingroup$ No actually, it's only for some $(x, y)$ --- it's of course uniform continuity that should be for all points in the domain. $\endgroup$ – hejseb Jan 31 '14 at 13:57
  • $\begingroup$ We need to find one $\varepsilon > 0$, so that for all $\delta > 0$ we find points with $\lVert a-b\rVert < \delta$, but $\lvert f(a) - f(b)\rvert \geqslant \varepsilon$ to show that $f$ is not uniformly continuous. In short: Yes to your second comment. $\endgroup$ – Daniel Fischer Jan 31 '14 at 14:06

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