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July 1st, 2015, 01:14 PM   #1
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Show a Function is not Uniformly Continuous

Show that the function L(x, y) = xy is not uniformly continuous.
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July 2nd, 2015, 04:17 AM   #2
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On what set?
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July 2nd, 2015, 07:31 AM   #3
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On R^2
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July 2nd, 2015, 09:22 AM   #4
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"Continuous" on a R^2 means that, given a point (x_0, y_0) in R^2, given any $\displaystyle \epsilon> 0$, there exist $\displaystyle delta> 0$ such that if $\displaystyle \sqrt{(x- x_0)^2+ (y- y_0)^2}< \delta$ then $\displaystyle |xy- x_0y_0|< epsilon$.

"Uniformly continuous" means that, for a given a value of can be chosen that works for all points in R^2.

Could you prove that xy is continuous on R^2 at, say, (0, 0)? Where would that proof break down if you used a general point rather than (0, 0)?
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July 3rd, 2015, 02:44 PM   #5
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Actually, I think I've solved the problem. Here's what you have to do.

Find an epsilon (I don't know how enter the math symbol) so that, given any delta > 0, you can find two points in the domain within delta of each other, BUT their function values differ by greater than epsilon.

To that end, for L(x, y) = xy, let epsilon = 1. Let delta > 0. Choose N so that sqrt(2)/N < delta. Let M > N/2.

Look at the two points (M, M) and (M + 1/N, M + 1/N). They differ by sqrt(2)/N < delta.

Yet, | L(M + 1/N, M + 1/N) - L(M, M) | = (2M)/N + 1/N^2 > 1 = epsilon.

Voila.

Last edited by dpsmith; July 3rd, 2015 at 02:48 PM.
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