weighted norm modification

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Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
$$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?



I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?










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    up vote
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    Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
    $$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
    Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?



    I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?










    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
      $$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
      Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?



      I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?










      share|cite|improve this question















      Fix constants $ w_1,...,w_n > 0 $, and for $ x,y in mathbb{R}^n $, define:
      $$ big < x,y big >_w = sum_{k=1}^n w_ix_iy_i text{.} $$
      Verify that this yields an inner product on $ mathbb{R}^n $. How would we need to modify this definition for it to yield an inner product on $ mathbb{C}^n $? What about $ l^2(mathbb{N}) $?



      I only have a question about the inner product on $ l^2 $. Would I need to add conditions like $ w_n $ converges as well as $ x_n = y_n $ and $ x_n $ converges in $ l^2$ to make it work?







      functional-analysis lp-spaces inner-product-space






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      edited Dec 3 at 9:53









      Davide Giraudo

      124k16150258




      124k16150258










      asked Dec 2 at 19:18









      Chase Sariaslani

      705




      705






















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          For $mathbb{C}^n$:



          $$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$



          where $overline{y_i}$ denotes the complex conjugate of $y_i$.



          For $mathcal{l}^2(mathbb{N})$:



          $$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$



          To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.






          share|cite|improve this answer





















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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            0
            down vote



            accepted










            For $mathbb{C}^n$:



            $$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$



            where $overline{y_i}$ denotes the complex conjugate of $y_i$.



            For $mathcal{l}^2(mathbb{N})$:



            $$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$



            To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              For $mathbb{C}^n$:



              $$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$



              where $overline{y_i}$ denotes the complex conjugate of $y_i$.



              For $mathcal{l}^2(mathbb{N})$:



              $$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$



              To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                For $mathbb{C}^n$:



                $$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$



                where $overline{y_i}$ denotes the complex conjugate of $y_i$.



                For $mathcal{l}^2(mathbb{N})$:



                $$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$



                To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.






                share|cite|improve this answer












                For $mathbb{C}^n$:



                $$big < x,y big >_w = sum_{i=1}^n w_ix_ioverline{y}_i,$$



                where $overline{y_i}$ denotes the complex conjugate of $y_i$.



                For $mathcal{l}^2(mathbb{N})$:



                $$big < x,y big >_w = sum_{i=1}^{+infty} w_ix_iy_i.$$



                To ensure convergence, a necessary condition is that $w_i to 0$ when $i to +infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 at 19:25









                the_candyman

                8,67822044




                8,67822044






























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