Question about Kadets Klee norm Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Bounded operator from a Hilbert space to $ell^1$ is compactCharacterisation of norm convergenceIf $C$ is convex , weakly-closed and norm-bounded $Longrightarrow$ $C$ is weakly-compactQuestion about compact operatorsShorter proof for $T$ compact and $x_n to x$ weaky then $Tx_n to Tx$ stronglyOn the weak closure of a sequence of projectionsLet $(x_n)$ be a bounded sequence such that $(T^* Tx_n)$ is Cauchy. Then $T^*x_n$ is also Cauchy.If every subsequence of $(x_n)$ has a subsequence converging weakly to $x$ then $x_n$ converges weakly to $x$.Discuss the strong and weak convergence of $B=_L^2le1$A weakly convergent sequence in a compact set, is strongly convegnet

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Question about Kadets Klee norm



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Bounded operator from a Hilbert space to $ell^1$ is compactCharacterisation of norm convergenceIf $C$ is convex , weakly-closed and norm-bounded $Longrightarrow$ $C$ is weakly-compactQuestion about compact operatorsShorter proof for $T$ compact and $x_n to x$ weaky then $Tx_n to Tx$ stronglyOn the weak closure of a sequence of projectionsLet $(x_n)$ be a bounded sequence such that $(T^* Tx_n)$ is Cauchy. Then $T^*x_n$ is also Cauchy.If every subsequence of $(x_n)$ has a subsequence converging weakly to $x$ then $x_n$ converges weakly to $x$.Discuss the strong and weak convergence of $B=uin W_0^1,2(0,1):$A weakly convergent sequence in a compact set, is strongly convegnet










1












$begingroup$


I'm reading a book that claims the following: If $C$ is a closed bounded convex subset of the sphere in a space with the Kadets Klee property, then $C$ is weakly compact.



I'm not sure why this is true. If $(x_n)_n$ is any sequence in $C$. If it has a weakly converging subsequence converging to $x$, then $xin C$ since $C$ is also weakly closed. Then $||x_n_k||to ||x||$ since the sequence and $x$ are in the sphere, and this would imply norm convergence. But I don't see why $(x_n)_n$ has a weakly converging subsequence to begin with.










share|cite|improve this question









$endgroup$











  • $begingroup$
    what is Kadets Klee property?
    $endgroup$
    – supinf
    Mar 27 at 16:46










  • $begingroup$
    If you converge weakly and the norms converge, then the sequence converges in norm
    $endgroup$
    – user124910
    Mar 27 at 19:03















1












$begingroup$


I'm reading a book that claims the following: If $C$ is a closed bounded convex subset of the sphere in a space with the Kadets Klee property, then $C$ is weakly compact.



I'm not sure why this is true. If $(x_n)_n$ is any sequence in $C$. If it has a weakly converging subsequence converging to $x$, then $xin C$ since $C$ is also weakly closed. Then $||x_n_k||to ||x||$ since the sequence and $x$ are in the sphere, and this would imply norm convergence. But I don't see why $(x_n)_n$ has a weakly converging subsequence to begin with.










share|cite|improve this question









$endgroup$











  • $begingroup$
    what is Kadets Klee property?
    $endgroup$
    – supinf
    Mar 27 at 16:46










  • $begingroup$
    If you converge weakly and the norms converge, then the sequence converges in norm
    $endgroup$
    – user124910
    Mar 27 at 19:03













1












1








1


1



$begingroup$


I'm reading a book that claims the following: If $C$ is a closed bounded convex subset of the sphere in a space with the Kadets Klee property, then $C$ is weakly compact.



I'm not sure why this is true. If $(x_n)_n$ is any sequence in $C$. If it has a weakly converging subsequence converging to $x$, then $xin C$ since $C$ is also weakly closed. Then $||x_n_k||to ||x||$ since the sequence and $x$ are in the sphere, and this would imply norm convergence. But I don't see why $(x_n)_n$ has a weakly converging subsequence to begin with.










share|cite|improve this question









$endgroup$




I'm reading a book that claims the following: If $C$ is a closed bounded convex subset of the sphere in a space with the Kadets Klee property, then $C$ is weakly compact.



I'm not sure why this is true. If $(x_n)_n$ is any sequence in $C$. If it has a weakly converging subsequence converging to $x$, then $xin C$ since $C$ is also weakly closed. Then $||x_n_k||to ||x||$ since the sequence and $x$ are in the sphere, and this would imply norm convergence. But I don't see why $(x_n)_n$ has a weakly converging subsequence to begin with.







functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 27 at 16:43









user124910user124910

1,262817




1,262817











  • $begingroup$
    what is Kadets Klee property?
    $endgroup$
    – supinf
    Mar 27 at 16:46










  • $begingroup$
    If you converge weakly and the norms converge, then the sequence converges in norm
    $endgroup$
    – user124910
    Mar 27 at 19:03
















  • $begingroup$
    what is Kadets Klee property?
    $endgroup$
    – supinf
    Mar 27 at 16:46










  • $begingroup$
    If you converge weakly and the norms converge, then the sequence converges in norm
    $endgroup$
    – user124910
    Mar 27 at 19:03















$begingroup$
what is Kadets Klee property?
$endgroup$
– supinf
Mar 27 at 16:46




$begingroup$
what is Kadets Klee property?
$endgroup$
– supinf
Mar 27 at 16:46












$begingroup$
If you converge weakly and the norms converge, then the sequence converges in norm
$endgroup$
– user124910
Mar 27 at 19:03




$begingroup$
If you converge weakly and the norms converge, then the sequence converges in norm
$endgroup$
– user124910
Mar 27 at 19:03










0






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