Prove that the dual space of $R^n$ is $R^n$












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I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.



(Apologies for not being able to type the whole of it)



enter image description here



I got stuck after the line where we took the supremum over all x of norm 1.




How did we arrive at the step of equality in the Cauchy-Schwarz inequality?




Any good explanation would be helpful.






functional-analysis proof-explanation







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.



    (Apologies for not being able to type the whole of it)



    enter image description here



    I got stuck after the line where we took the supremum over all x of norm 1.




    How did we arrive at the step of equality in the Cauchy-Schwarz inequality?




    Any good explanation would be helpful.






    functional-analysis proof-explanation







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.



      (Apologies for not being able to type the whole of it)



      enter image description here



      I got stuck after the line where we took the supremum over all x of norm 1.




      How did we arrive at the step of equality in the Cauchy-Schwarz inequality?




      Any good explanation would be helpful.






      functional-analysis proof-explanation







      share|cite|improve this question











      $endgroup$




      I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.



      (Apologies for not being able to type the whole of it)



      enter image description here



      I got stuck after the line where we took the supremum over all x of norm 1.




      How did we arrive at the step of equality in the Cauchy-Schwarz inequality?




      Any good explanation would be helpful.






      functional-analysis proof-explanation




      functional-analysis proof-explanation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 27 at 10:16







      Shatabdi Sinha

















      asked Jan 27 at 9:57









      Shatabdi SinhaShatabdi Sinha

      18612




      18612






















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          $begingroup$

          Just calculate it directly;
          $$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
          since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.






          share|cite|improve this answer









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            $begingroup$

            Just calculate it directly;
            $$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
            since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Just calculate it directly;
              $$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
              since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Just calculate it directly;
                $$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
                since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.






                share|cite|improve this answer









                $endgroup$



                Just calculate it directly;
                $$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
                since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 10:30









                jmerryjmerry

                16.1k1633




                16.1k1633






























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