Norms and Parallelogram Identity












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I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.



The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2



I am quite confused here any help would be great.



Thanks






linear-algebra normed-spaces inner-product-space







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    0












    $begingroup$


    I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.



    The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2



    I am quite confused here any help would be great.



    Thanks






    linear-algebra normed-spaces inner-product-space







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.



      The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2



      I am quite confused here any help would be great.



      Thanks






      linear-algebra normed-spaces inner-product-space







      share|cite|improve this question











      $endgroup$




      I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.



      The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2



      I am quite confused here any help would be great.



      Thanks






      linear-algebra normed-spaces inner-product-space




      linear-algebra normed-spaces inner-product-space






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 8 at 21:46







      jake walsh

















      asked Jan 8 at 21:30









      jake walshjake walsh

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          $begingroup$

          In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:



          Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
          $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
          However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.



          Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
          $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
          However,
          $$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
          Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.






          share|cite|improve this answer









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            1












            $begingroup$

            In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:



            Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
            $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
            However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.



            Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
            $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
            However,
            $$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
            Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:



              Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
              $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
              However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.



              Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
              $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
              However,
              $$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
              Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:



                Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
                $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
                However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.



                Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
                $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
                However,
                $$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
                Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.






                share|cite|improve this answer









                $endgroup$



                In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:



                Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
                $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
                However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.



                Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
                $$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
                However,
                $$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
                Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 23:50









                HarambeHarambe

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