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 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|,|x-y|} 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: 6,834 Thanks: 733 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*)|,|(x-y)+(x*-y*)|} I know from absolute value properties that: |(x+y)+(x*+y*)|=<|(x+y)|+|(x*+y*)| and |(x-y)+(x*-y*)|=<|(x-y)|+|(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 can´t it be that ||(x,y)||=|(x-y)| in the even though |(x+y)+(x*+y*)|>|(x-y)+(x*-y*)|. In other words, can I be certain that if |(x+y)+(x*+y*)|>|(x-y)+(x*-y*)| then |(x+y)|>|(x-y)| 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: 6,834 Thanks: 733 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*)||=|(x-y)+(x*-y*)|=<|(x-y)|+|(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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