A Hahn-Banach separation theorem argument, claryfying the details











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I have a question regarding the proof of proposition 6.1. in https://arxiv.org/pdf/1509.01870.pdf, how exactly Hahn-Banach separation theorem has been used.



Proposition 6.1: A discrete group is $C^*$-simple iff for every bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ and every $ain C_r^*(G)$ $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|=0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$.



Proof: By a result stated in the paper, $G$ is not $C^*$-simple if and only if there is a bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ such that $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$. By the Hahn-Banach separation theorem, this is equivalent to the existence of $ain C_r^*(G)$ such that $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|>0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$. (QED)



Note that ${lambda_galambda_{g^{-1}}mid gin G}$ is the G-orbit of $a$ in $C_r^*(G)$.
It's clear to me that if $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$, that the functionals must have strictly positive disctance somewhere. But how precisly is that Hahn-Banach's separation theorem? I see it is applied to the convex disjoint sets A={the weak* closed convex hull of the orbit $Gphi$} and $B={phi(1)tau_lambda}$ right? But then there must be a functional which seperated $A$ and $B$. And here it doesn't look to me as an application of Hahn-Banach.










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    I have a question regarding the proof of proposition 6.1. in https://arxiv.org/pdf/1509.01870.pdf, how exactly Hahn-Banach separation theorem has been used.



    Proposition 6.1: A discrete group is $C^*$-simple iff for every bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ and every $ain C_r^*(G)$ $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|=0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$.



    Proof: By a result stated in the paper, $G$ is not $C^*$-simple if and only if there is a bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ such that $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$. By the Hahn-Banach separation theorem, this is equivalent to the existence of $ain C_r^*(G)$ such that $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|>0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$. (QED)



    Note that ${lambda_galambda_{g^{-1}}mid gin G}$ is the G-orbit of $a$ in $C_r^*(G)$.
    It's clear to me that if $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$, that the functionals must have strictly positive disctance somewhere. But how precisly is that Hahn-Banach's separation theorem? I see it is applied to the convex disjoint sets A={the weak* closed convex hull of the orbit $Gphi$} and $B={phi(1)tau_lambda}$ right? But then there must be a functional which seperated $A$ and $B$. And here it doesn't look to me as an application of Hahn-Banach.










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      down vote

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      up vote
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      I have a question regarding the proof of proposition 6.1. in https://arxiv.org/pdf/1509.01870.pdf, how exactly Hahn-Banach separation theorem has been used.



      Proposition 6.1: A discrete group is $C^*$-simple iff for every bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ and every $ain C_r^*(G)$ $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|=0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$.



      Proof: By a result stated in the paper, $G$ is not $C^*$-simple if and only if there is a bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ such that $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$. By the Hahn-Banach separation theorem, this is equivalent to the existence of $ain C_r^*(G)$ such that $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|>0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$. (QED)



      Note that ${lambda_galambda_{g^{-1}}mid gin G}$ is the G-orbit of $a$ in $C_r^*(G)$.
      It's clear to me that if $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$, that the functionals must have strictly positive disctance somewhere. But how precisly is that Hahn-Banach's separation theorem? I see it is applied to the convex disjoint sets A={the weak* closed convex hull of the orbit $Gphi$} and $B={phi(1)tau_lambda}$ right? But then there must be a functional which seperated $A$ and $B$. And here it doesn't look to me as an application of Hahn-Banach.










      share|cite|improve this question













      I have a question regarding the proof of proposition 6.1. in https://arxiv.org/pdf/1509.01870.pdf, how exactly Hahn-Banach separation theorem has been used.



      Proposition 6.1: A discrete group is $C^*$-simple iff for every bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ and every $ain C_r^*(G)$ $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|=0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$.



      Proof: By a result stated in the paper, $G$ is not $C^*$-simple if and only if there is a bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ such that $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$. By the Hahn-Banach separation theorem, this is equivalent to the existence of $ain C_r^*(G)$ such that $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|>0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$. (QED)



      Note that ${lambda_galambda_{g^{-1}}mid gin G}$ is the G-orbit of $a$ in $C_r^*(G)$.
      It's clear to me that if $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$, that the functionals must have strictly positive disctance somewhere. But how precisly is that Hahn-Banach's separation theorem? I see it is applied to the convex disjoint sets A={the weak* closed convex hull of the orbit $Gphi$} and $B={phi(1)tau_lambda}$ right? But then there must be a functional which seperated $A$ and $B$. And here it doesn't look to me as an application of Hahn-Banach.







      functional-analysis operator-algebras c-star-algebras






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          Let $A := C^*_r(G)$. Equip $A^*$, the topological dual of $A$, with the weak*-topology. Let us denote, as you already did,
          $$
          K := overline{mathrm{co}}^{w*}(Gphi), qquad F := {phi(1)tau_lambda}.
          $$



          Then, as you noted, by Hahn-Banach you find a continuous functional
          $$
          varphi in (A^*,text{weak*})^*,
          $$

          which separates $K$ and $F$. Now, the crucial point is that $varphi$ is given by some point evaluation, where we use that $A^*$ is equipped with the weak*-topology.



          That gives you the required $a$, which does the job. Furthermore, one has to note that the weak* closure of a convex set in a normed space equals its norm-closure, which again invokes Hahn-Banach.






          share|cite|improve this answer





















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            up vote
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            down vote



            accepted










            Let $A := C^*_r(G)$. Equip $A^*$, the topological dual of $A$, with the weak*-topology. Let us denote, as you already did,
            $$
            K := overline{mathrm{co}}^{w*}(Gphi), qquad F := {phi(1)tau_lambda}.
            $$



            Then, as you noted, by Hahn-Banach you find a continuous functional
            $$
            varphi in (A^*,text{weak*})^*,
            $$

            which separates $K$ and $F$. Now, the crucial point is that $varphi$ is given by some point evaluation, where we use that $A^*$ is equipped with the weak*-topology.



            That gives you the required $a$, which does the job. Furthermore, one has to note that the weak* closure of a convex set in a normed space equals its norm-closure, which again invokes Hahn-Banach.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              Let $A := C^*_r(G)$. Equip $A^*$, the topological dual of $A$, with the weak*-topology. Let us denote, as you already did,
              $$
              K := overline{mathrm{co}}^{w*}(Gphi), qquad F := {phi(1)tau_lambda}.
              $$



              Then, as you noted, by Hahn-Banach you find a continuous functional
              $$
              varphi in (A^*,text{weak*})^*,
              $$

              which separates $K$ and $F$. Now, the crucial point is that $varphi$ is given by some point evaluation, where we use that $A^*$ is equipped with the weak*-topology.



              That gives you the required $a$, which does the job. Furthermore, one has to note that the weak* closure of a convex set in a normed space equals its norm-closure, which again invokes Hahn-Banach.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Let $A := C^*_r(G)$. Equip $A^*$, the topological dual of $A$, with the weak*-topology. Let us denote, as you already did,
                $$
                K := overline{mathrm{co}}^{w*}(Gphi), qquad F := {phi(1)tau_lambda}.
                $$



                Then, as you noted, by Hahn-Banach you find a continuous functional
                $$
                varphi in (A^*,text{weak*})^*,
                $$

                which separates $K$ and $F$. Now, the crucial point is that $varphi$ is given by some point evaluation, where we use that $A^*$ is equipped with the weak*-topology.



                That gives you the required $a$, which does the job. Furthermore, one has to note that the weak* closure of a convex set in a normed space equals its norm-closure, which again invokes Hahn-Banach.






                share|cite|improve this answer












                Let $A := C^*_r(G)$. Equip $A^*$, the topological dual of $A$, with the weak*-topology. Let us denote, as you already did,
                $$
                K := overline{mathrm{co}}^{w*}(Gphi), qquad F := {phi(1)tau_lambda}.
                $$



                Then, as you noted, by Hahn-Banach you find a continuous functional
                $$
                varphi in (A^*,text{weak*})^*,
                $$

                which separates $K$ and $F$. Now, the crucial point is that $varphi$ is given by some point evaluation, where we use that $A^*$ is equipped with the weak*-topology.



                That gives you the required $a$, which does the job. Furthermore, one has to note that the weak* closure of a convex set in a normed space equals its norm-closure, which again invokes Hahn-Banach.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                André S.

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