Show that $|.|_{1/2}$ is not a norm. [duplicate]












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This question already has an answer here:




  • $l^p$ not norm, $p<1$

    2 answers




Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?



thanks






optimization convex-analysis norm nonlinear-optimization regularization







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marked as duplicate by Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.























    0












    $begingroup$



    This question already has an answer here:




    • $l^p$ not norm, $p<1$

      2 answers




    Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?



    thanks






    optimization convex-analysis norm nonlinear-optimization regularization







    share|cite|improve this question









    $endgroup$



    marked as duplicate by Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.





















      0












      0








      0





      $begingroup$



      This question already has an answer here:




      • $l^p$ not norm, $p<1$

        2 answers




      Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?



      thanks






      optimization convex-analysis norm nonlinear-optimization regularization







      share|cite|improve this question









      $endgroup$





      This question already has an answer here:




      • $l^p$ not norm, $p<1$

        2 answers




      Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?



      thanks





      This question already has an answer here:




      • $l^p$ not norm, $p<1$

        2 answers







      optimization convex-analysis norm nonlinear-optimization regularization




      optimization convex-analysis norm nonlinear-optimization regularization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 29 '18 at 1:39









      HumayooHumayoo

      1013




      1013




      marked as duplicate by Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          1 Answer
          1






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          2












          $begingroup$

          It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.






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          • 1




            $begingroup$
            Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
            $endgroup$
            – DanielWainfleet
            Dec 29 '18 at 6:55


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
            $endgroup$
            – DanielWainfleet
            Dec 29 '18 at 6:55
















          2












          $begingroup$

          It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
            $endgroup$
            – DanielWainfleet
            Dec 29 '18 at 6:55














          2












          2








          2





          $begingroup$

          It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.






          share|cite|improve this answer











          $endgroup$



          It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 3:53

























          answered Dec 29 '18 at 2:02









          FortoxFortox

          6718




          6718








          • 1




            $begingroup$
            Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
            $endgroup$
            – DanielWainfleet
            Dec 29 '18 at 6:55














          • 1




            $begingroup$
            Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
            $endgroup$
            – DanielWainfleet
            Dec 29 '18 at 6:55








          1




          1




          $begingroup$
          Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
          $endgroup$
          – DanielWainfleet
          Dec 29 '18 at 6:55




          $begingroup$
          Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
          $endgroup$
          – DanielWainfleet
          Dec 29 '18 at 6:55



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