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 July 1st, 2015, 02:14 PM #1 Newbie   Joined: Mar 2014 Posts: 16 Thanks: 2 Show a Function is not Uniformly Continuous Show that the function L(x, y) = xy is not uniformly continuous.
 July 2nd, 2015, 05:17 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 On what set?
 July 2nd, 2015, 08:31 AM #3 Newbie   Joined: Mar 2014 Posts: 16 Thanks: 2 On R^2
 July 2nd, 2015, 10:22 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 "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 $\epsilon$ a value of $\delta$ 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)? Thanks from dpsmith
 July 3rd, 2015, 03:44 PM #5 Newbie   Joined: Mar 2014 Posts: 16 Thanks: 2 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 03:48 PM.

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