February 26th, 2016, 01:50 PM  #1 
Newbie Joined: Feb 2016 From: Montevideo Posts: 3 Thanks: 0  Vectorial norm help
Hello, I am new to the forums. I was not sure if this question should be asked in this section, but as the exercise is from my calculus course I have posted it here. Here is the question: Does the function (x,y)=max{x+y,xy} define a function in R2? I am having trouble in proving (or disproving) the triangle inequality. Thank you in advance for your help. 
February 26th, 2016, 05:22 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689 
What have you tried so far?

February 26th, 2016, 08:51 PM  #3 
Newbie Joined: Feb 2016 From: Montevideo Posts: 3 Thanks: 0 
Correction: It should say define a NORM in R2* What I have tried: I want to prove that (x,y)+(x*,y*)=<(x,y)+(x*,y*) I know that: (x,y)+(x*,y*)=max{(x+y)+(x*+y*),(xy)+(x*y*)} I know from absolute value properties that: (x+y)+(x*+y*)=<(x+y)+(x*+y*) and (xy)+(x*y*)=<(xy)+(x*y*) Now, the problem I have is the following: Lets say that (x,y)+(x*,y*)=(x+y)+(x*+y*)=<(x+y)+(x*+y* ) What I would need to prove is that: (x,y)=(x+y) and (x*,y*)=(x*+y*) My doubt is that why canĀ“t it be that (x,y)=(xy) in the even though (x+y)+(x*+y*)>(xy)+(x*y*). In other words, can I be certain that if (x+y)+(x*+y*)>(xy)+(x*y*) then (x+y)>(xy) and (x*+y*)>(x*y*) and the same for the other case? If I can prove that then the problem is done. 
February 27th, 2016, 01:23 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689 
I am confused about your doubt. There are 2 cases: (x,y)+(x*,y*)=(x+y)+(x*+y*)=<(x+y)+(x*+y* )=<x,y+x*,y* (x,y)+(x*,y*)=(xy)+(x*y*)=<(xy)+(x*y*)=<x,y+x*,y* 
February 27th, 2016, 02:24 PM  #5 
Newbie Joined: Feb 2016 From: Montevideo Posts: 3 Thanks: 0 
Actually, maybe you did not understand me but you wrote exactly what I needed. I was writing a reply and then it clicked, maybe I was confusing myself, looking for a problem when there was not one (classic math). You may not know how you helped, but you did. I am truly grateful Cheers! 

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