limit of sequence of bounded operators

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Suppose $phi_n:Arightarrow B(H_n)$ is a sequence of nonzero representations, where $A$ is a nonunital $C^*$-algebra,$H_n$ is a Hilbert space, and $P_n$ is a sequence of projections on $H_n$. Does there exist $a_0$ such that $P_nphi_n(a_0)$ is norm convergent?




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    Suppose $phi_n:Arightarrow B(H_n)$ is a sequence of nonzero representations, where $A$ is a nonunital $C^*$-algebra,$H_n$ is a Hilbert space, and $P_n$ is a sequence of projections on $H_n$. Does there exist $a_0$ such that $P_nphi_n(a_0)$ is norm convergent?




    operator-theory operator-algebras c-star-algebras




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      Suppose $phi_n:Arightarrow B(H_n)$ is a sequence of nonzero representations, where $A$ is a nonunital $C^*$-algebra,$H_n$ is a Hilbert space, and $P_n$ is a sequence of projections on $H_n$. Does there exist $a_0$ such that $P_nphi_n(a_0)$ is norm convergent?




      operator-theory operator-algebras c-star-algebras




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      Suppose $phi_n:Arightarrow B(H_n)$ is a sequence of nonzero representations, where $A$ is a nonunital $C^*$-algebra,$H_n$ is a Hilbert space, and $P_n$ is a sequence of projections on $H_n$. Does there exist $a_0$ such that $P_nphi_n(a_0)$ is norm convergent?





      operator-theory operator-algebras c-star-algebras





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      edited Aug 1 at 4:39









      Martin Argerami

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      asked Jul 31 at 14:54









      mathrookie

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          This makes no sense as written. The elements $P_nphi_n(a_0)$ are in different Hilbert spaces, if the $H_n$ are different.



          If the Hilbert space is fixed, the answer is no unless the sequence is zero. Take $A=c_0$, $phi_n(a)=a(1),Iin B(ell^2(mathbb N))$. Take $P_n$ to a sequence of pairwise orthogonal projections.



          Then, if $a(1)=0$, we have $P_nphi_n(a_0)=0$ for all $n$. But if $a(1)ne0$, then $P_nphi(a_0)=a(1),P_n$ is a sequence with no convergent subsequences.






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            This makes no sense as written. The elements $P_nphi_n(a_0)$ are in different Hilbert spaces, if the $H_n$ are different.



            If the Hilbert space is fixed, the answer is no unless the sequence is zero. Take $A=c_0$, $phi_n(a)=a(1),Iin B(ell^2(mathbb N))$. Take $P_n$ to a sequence of pairwise orthogonal projections.



            Then, if $a(1)=0$, we have $P_nphi_n(a_0)=0$ for all $n$. But if $a(1)ne0$, then $P_nphi(a_0)=a(1),P_n$ is a sequence with no convergent subsequences.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              This makes no sense as written. The elements $P_nphi_n(a_0)$ are in different Hilbert spaces, if the $H_n$ are different.



              If the Hilbert space is fixed, the answer is no unless the sequence is zero. Take $A=c_0$, $phi_n(a)=a(1),Iin B(ell^2(mathbb N))$. Take $P_n$ to a sequence of pairwise orthogonal projections.



              Then, if $a(1)=0$, we have $P_nphi_n(a_0)=0$ for all $n$. But if $a(1)ne0$, then $P_nphi(a_0)=a(1),P_n$ is a sequence with no convergent subsequences.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                This makes no sense as written. The elements $P_nphi_n(a_0)$ are in different Hilbert spaces, if the $H_n$ are different.



                If the Hilbert space is fixed, the answer is no unless the sequence is zero. Take $A=c_0$, $phi_n(a)=a(1),Iin B(ell^2(mathbb N))$. Take $P_n$ to a sequence of pairwise orthogonal projections.



                Then, if $a(1)=0$, we have $P_nphi_n(a_0)=0$ for all $n$. But if $a(1)ne0$, then $P_nphi(a_0)=a(1),P_n$ is a sequence with no convergent subsequences.






                share|cite|improve this answer













                This makes no sense as written. The elements $P_nphi_n(a_0)$ are in different Hilbert spaces, if the $H_n$ are different.



                If the Hilbert space is fixed, the answer is no unless the sequence is zero. Take $A=c_0$, $phi_n(a)=a(1),Iin B(ell^2(mathbb N))$. Take $P_n$ to a sequence of pairwise orthogonal projections.



                Then, if $a(1)=0$, we have $P_nphi_n(a_0)=0$ for all $n$. But if $a(1)ne0$, then $P_nphi(a_0)=a(1),P_n$ is a sequence with no convergent subsequences.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Aug 1 at 4:39









                Martin Argerami

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