Operator on $l^1(N)$ space

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Let $A = (A_n,m)$ with $A_n,m in mathbbC, n,m in mathbbN$ be a matrix. Assume $||A|| = sup_m sum_n|A_n,m| < infty$. Show that for $T:l^1(mathbbN) to l^1(mathbbN), (Tf)(n) = sum_mA_n,mf(m)$ we have $||T|| = ||A||$.



Showing $||T|| leq ||A||$ was easy. But how can I show $||T|| geq ||A||$?




functional-analysis norm




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    Let $A = (A_n,m)$ with $A_n,m in mathbbC, n,m in mathbbN$ be a matrix. Assume $||A|| = sup_m sum_n|A_n,m| < infty$. Show that for $T:l^1(mathbbN) to l^1(mathbbN), (Tf)(n) = sum_mA_n,mf(m)$ we have $||T|| = ||A||$.



    Showing $||T|| leq ||A||$ was easy. But how can I show $||T|| geq ||A||$?




    functional-analysis norm




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $A = (A_n,m)$ with $A_n,m in mathbbC, n,m in mathbbN$ be a matrix. Assume $||A|| = sup_m sum_n|A_n,m| < infty$. Show that for $T:l^1(mathbbN) to l^1(mathbbN), (Tf)(n) = sum_mA_n,mf(m)$ we have $||T|| = ||A||$.



      Showing $||T|| leq ||A||$ was easy. But how can I show $||T|| geq ||A||$?




      functional-analysis norm




      share|cite|improve this question













      Let $A = (A_n,m)$ with $A_n,m in mathbbC, n,m in mathbbN$ be a matrix. Assume $||A|| = sup_m sum_n|A_n,m| < infty$. Show that for $T:l^1(mathbbN) to l^1(mathbbN), (Tf)(n) = sum_mA_n,mf(m)$ we have $||T|| = ||A||$.



      Showing $||T|| leq ||A||$ was easy. But how can I show $||T|| geq ||A||$?





      functional-analysis norm





      share|cite|improve this question












      share|cite|improve this question




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      edited Jul 30 at 8:25









      Bob

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      asked Jul 30 at 8:02









      AM88

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          Let $e_k$ be the sequence with $1$ at the $k-th$ place and $0$ everywhere else. Then $||Te_k||=sum_n |T(e_k)(n)|= sum _n |A_n,k|$ so $||T|| geq sum _n |A_n,k|$ for all $k$. Take sup over $k$.






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            Let $e_k$ be the sequence with $1$ at the $k-th$ place and $0$ everywhere else. Then $||Te_k||=sum_n |T(e_k)(n)|= sum _n |A_n,k|$ so $||T|| geq sum _n |A_n,k|$ for all $k$. Take sup over $k$.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Let $e_k$ be the sequence with $1$ at the $k-th$ place and $0$ everywhere else. Then $||Te_k||=sum_n |T(e_k)(n)|= sum _n |A_n,k|$ so $||T|| geq sum _n |A_n,k|$ for all $k$. Take sup over $k$.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Let $e_k$ be the sequence with $1$ at the $k-th$ place and $0$ everywhere else. Then $||Te_k||=sum_n |T(e_k)(n)|= sum _n |A_n,k|$ so $||T|| geq sum _n |A_n,k|$ for all $k$. Take sup over $k$.






                share|cite|improve this answer













                Let $e_k$ be the sequence with $1$ at the $k-th$ place and $0$ everywhere else. Then $||Te_k||=sum_n |T(e_k)(n)|= sum _n |A_n,k|$ so $||T|| geq sum _n |A_n,k|$ for all $k$. Take sup over $k$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 30 at 8:10









                Kavi Rama Murthy

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