Banach algebra $l^p$ is not isomorphic to $C^{*}$ algebra












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Consider commutative Banach algebra $l^p$, $p in [1,infty)$ with multiplication by coordinates. I know, that $Delta (l^{p})={e_n : n in mathbb{N}}$ - set of canonical functionals. We know that $widehat{x}(e_n)=x_n$ and $widehat{x}: l^{p} to C_0(Delta (l^{p}))$. Because there exists $xin l^{p}$ such that $x_ineq x_j$ for $ineq j$, and Gelfand transformation is continuous, GT must be discrete. I would like to show, that GT is or is not surjective, but I do not have any idea. Many thanks










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    $begingroup$


    Consider commutative Banach algebra $l^p$, $p in [1,infty)$ with multiplication by coordinates. I know, that $Delta (l^{p})={e_n : n in mathbb{N}}$ - set of canonical functionals. We know that $widehat{x}(e_n)=x_n$ and $widehat{x}: l^{p} to C_0(Delta (l^{p}))$. Because there exists $xin l^{p}$ such that $x_ineq x_j$ for $ineq j$, and Gelfand transformation is continuous, GT must be discrete. I would like to show, that GT is or is not surjective, but I do not have any idea. Many thanks










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Consider commutative Banach algebra $l^p$, $p in [1,infty)$ with multiplication by coordinates. I know, that $Delta (l^{p})={e_n : n in mathbb{N}}$ - set of canonical functionals. We know that $widehat{x}(e_n)=x_n$ and $widehat{x}: l^{p} to C_0(Delta (l^{p}))$. Because there exists $xin l^{p}$ such that $x_ineq x_j$ for $ineq j$, and Gelfand transformation is continuous, GT must be discrete. I would like to show, that GT is or is not surjective, but I do not have any idea. Many thanks










      share|cite|improve this question











      $endgroup$




      Consider commutative Banach algebra $l^p$, $p in [1,infty)$ with multiplication by coordinates. I know, that $Delta (l^{p})={e_n : n in mathbb{N}}$ - set of canonical functionals. We know that $widehat{x}(e_n)=x_n$ and $widehat{x}: l^{p} to C_0(Delta (l^{p}))$. Because there exists $xin l^{p}$ such that $x_ineq x_j$ for $ineq j$, and Gelfand transformation is continuous, GT must be discrete. I would like to show, that GT is or is not surjective, but I do not have any idea. Many thanks







      abstract-algebra lp-spaces group-isomorphism banach-algebras gelfand-representation






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      edited Jan 7 at 23:16









      Martin Sleziak

      44.7k9117272




      44.7k9117272










      asked Jan 7 at 16:25









      LeoLeeLeoLee

      112




      112






















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