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fa.functionalanalysis
2d

revised 
On the relationship between the factorizations of an operator $T$ and its second adjoint $T^{**}$
Added explanation of the "of course" statement. 
2d

answered  A question on unconditionally $p$summable sequences 
Oct
7 
comment 
Is there an easier proof to show that the closed convex hull of a normalized weakly null sequence is weakly compact?
True, Rauni, but the proof of the factorization theorem gives an injective $J$. Also, if you have such a factorization with a non injective $J$, you can mod out the kernel of $J$ to get a factorization through another reflexive space for which the mapping from the reflexive space is injective. 
Oct
6 
awarded  Enlightened 
Oct
6 
awarded  Nice Answer 
Oct
3 
comment 
On compact operators with domain $c_{0}$
Yes. If $T:Y\to X^*$ is even just weakly compact, then $T^{**}$ maps $Y^{**}$ into $X^*$ and extends $T$. In a first course in functional analysis one should learn that an bounded linear operator $S:V\to U$ is weakly compact if and only if $S^{**}V^{**} \subset U^{**}$. So this question is arguably more appropriate for another site. (I say ``arguably" because such a basic fact is probably hard to find in text books on real analysis.) 
Oct
2 
comment 
Is there an easier proof to show that the closed convex hull of a normalized weakly null sequence is weakly compact?
Yes, DFJP, Rauni. 
Oct
2 
comment 
Quotient of a Banach algebra
You are right, Yemon. Sorry. 
Oct
1 
answered  Quotient of a Banach algebra 
Sep
29 
comment 
An operator factoring through a Banach space containing no copy of $l_{1}$
Of course. The identity operator on $C[0,1]$. 
Sep
28 
comment 
A question on characterizing a Banach space containing no copy of $l_{1}$
There is a non compact operator from $X$ into $\ell_1$ iff and only if $\ell_1$ is isomorphic to a complemented subspace of $X$. For that you need one additional comment; namely, that every subspace of $\ell_1$ contains a small complemented subspace that is isomorphic to $\ell_1$. 
Sep
27 
comment 
What is the name for a Banach space property closed under ultraproducts?
Because any hereditary property possessed by all finite dimensional spaces and ultraproducts is possessed by all Banach spaces. Incidentally, James' original definition of superQ was that every space that is finitely representable in such a space must have Q. In particular, the property must be hereditary. A space $Y$ is finitely representable in a space $X$ iff $Y$ embeds isometrically into some ultrapower of $X$. 
Sep
27 
answered  Approximation via finite rank CameronMartin projections 
Sep
27 
comment 
Approximation via finite rank CameronMartin projections
For an example, take for $W$ any separable Banach space that fails the approximation property. 
Sep
27 
comment 
What is the name for a Banach space property closed under ultraproducts?
Your last sentence is not correct. Finite dimensional spaces have cotype $q$ for all $q$, and every Banach space embeds isometrically into an ultraproduct of finite dimensional spaces. That explains why there is no name for properties that are preserved under arbitrary ultraproducts. Finite inequalities are approximately preserved under ultraproducts; see papers by Ward Henson to see how this elementary observation leads to a model theoretic characterization. 
Sep
27 
answered  A question on compact operators with domain $l_{p}$ 
Sep
23 
comment 
How to characterize a Banach space $X$ such that any operator from $X$ to $l_{p}$ is compact?
As far as I know the only characterization is basically trivial: There is a non compact operator from $\ell_p$ to $X$ iff there is a normalized basic sequence $(x_n)$ in $X$ and a $C$ s.t. for all finite sequence $(a_n)$ of scalars we have $\ \sum_n a_n x_n \^p \le C \sum_n a_n^p$. 
Sep
22 
comment 
Is $T^{**}$ unconditionally $p$summing whenever $T$ is unconditionally $p$summing?
Corrected; thanks. 
Sep
22 
revised 
Is $T^{**}$ unconditionally $p$summing whenever $T$ is unconditionally $p$summing?
Corrected typo. 
Sep
21 
comment 
Is $T^{**}$ unconditionally $p$summing whenever $T$ is unconditionally $p$summing?
Yes. If $Y$ does not contain an isomorphic copy of $c_0$, then every operator from $c_0$ to $Y$ is compact. But if $X^*$ contains a copy of $c_0$, then $X$ contains a complemented copy of $\ell_1$ (an old result of Pelczynski that is in standard textbooks; probably AlbiacKalton has this). 