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July 1st, 2015, 01:14 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, 04:17 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,160 Thanks: 866 
On what set?

July 2nd, 2015, 07:31 AM  #3 
Newbie Joined: Mar 2014 Posts: 16 Thanks: 2 
On R^2

July 2nd, 2015, 09:22 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,160 Thanks: 866 
"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)? 
July 3rd, 2015, 02:44 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 02:48 PM. 

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continuous, function, show, uniformly 
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