Prove a function is a norm?












0












$begingroup$


I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.



Where the properties of a norm are





  1. ${||x||} ge 0$ for all $x$ in${mathbb R}^n$


  2. ${||x||} = 0$ when $x=0$


  3. ${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$

  4. $||x+y|| le ||x||+||y||$


I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?






functions norm







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$endgroup$












  • $begingroup$
    3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
    $endgroup$
    – Mustafa Said
    Jan 25 at 2:40


















0












$begingroup$


I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.



Where the properties of a norm are





  1. ${||x||} ge 0$ for all $x$ in${mathbb R}^n$


  2. ${||x||} = 0$ when $x=0$


  3. ${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$

  4. $||x+y|| le ||x||+||y||$


I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?






functions norm







share|cite|improve this question











$endgroup$












  • $begingroup$
    3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
    $endgroup$
    – Mustafa Said
    Jan 25 at 2:40
















0












0








0





$begingroup$


I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.



Where the properties of a norm are





  1. ${||x||} ge 0$ for all $x$ in${mathbb R}^n$


  2. ${||x||} = 0$ when $x=0$


  3. ${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$

  4. $||x+y|| le ||x||+||y||$


I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?






functions norm







share|cite|improve this question











$endgroup$




I want to show that $${||x||}_{infty} = max{{|x_1|,|x_1|,...,|x_n|}}$$
is a norm.



Where the properties of a norm are





  1. ${||x||} ge 0$ for all $x$ in${mathbb R}^n$


  2. ${||x||} = 0$ when $x=0$


  3. ${||ax||} = |a|{||x||}$ for all $a$ in ${mathbb R}$

  4. $||x+y|| le ||x||+||y||$


I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?






functions norm




functions norm






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share|cite|improve this question













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edited Jan 25 at 5:28









Chinnapparaj R

5,7032928




5,7032928










asked Jan 25 at 2:33









Chemical EngineerChemical Engineer

597




597












  • $begingroup$
    3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
    $endgroup$
    – Mustafa Said
    Jan 25 at 2:40




















  • $begingroup$
    3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
    $endgroup$
    – Mustafa Said
    Jan 25 at 2:40


















$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40






$begingroup$
3. follows since $||ax||=max{|ax_1|, ... , |ax_n|}=a max{|x_1|, ... , |x_n|}$ and 4. follows from the triangle inequality on $mathbb{R}$.
$endgroup$
– Mustafa Said
Jan 25 at 2:40












2 Answers
2






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$begingroup$

For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).






share|cite|improve this answer









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    0












    $begingroup$

    To prove 4., notice that



    $$ ||x+y||_{infty} = |x_l+y_l | $$



    for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






      active

      oldest

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      active

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      active

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      -1












      $begingroup$

      For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
      For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
      For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
      For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).






      share|cite|improve this answer









      $endgroup$


















        -1












        $begingroup$

        For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
        For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
        For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
        For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).






        share|cite|improve this answer









        $endgroup$
















          -1












          -1








          -1





          $begingroup$

          For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
          For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
          For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
          For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).






          share|cite|improve this answer









          $endgroup$



          For 1. $max = |x_i| geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
          For 2. let $||x||=0 rightarrow 0 = max geq |x_i| geq 0 forall x_i rightarrow x=0$. Now let $x=0 rightarrow forall x_i, x_i = 0 rightarrow max = 0 rightarrow ||x||=0$.
          For 3. $||ax|| = max |ax| = |a||x_i| = |a|max |x| = |a|||x||$, where $|x_i|$ is the max element.
          For 4. $||x+y|| = |x_i+y_i| leq |x_i| + |y_i| leq max x + max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 2:50









          lightxbulblightxbulb

          1,115311




          1,115311























              0












              $begingroup$

              To prove 4., notice that



              $$ ||x+y||_{infty} = |x_l+y_l | $$



              for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                To prove 4., notice that



                $$ ||x+y||_{infty} = |x_l+y_l | $$



                for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  To prove 4., notice that



                  $$ ||x+y||_{infty} = |x_l+y_l | $$



                  for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows






                  share|cite|improve this answer









                  $endgroup$



                  To prove 4., notice that



                  $$ ||x+y||_{infty} = |x_l+y_l | $$



                  for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| leq |x| + |y| $ and since for example $|x_l| leq max |x_i| = ||x|| $, the result follows







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 25 at 2:44









                  Jimmy SabaterJimmy Sabater

                  3,013325




                  3,013325






























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