Timeline for Duals of ideals of operators between Banach spaces
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2017 at 17:34 | comment | added | M.González | OK: $\mathcal{C}^{dual}\subsetneq \mathcal{DP}\cap\mathcal{R}$. | |
| Dec 29, 2017 at 14:45 | comment | added | user114263 | However, the formal identity from $\ell_1$ to $\ell_2$ lies in $\mathfrak{DP}\cap \mathfrak{R}$ and not in $\mathfrak{V}^\text{dual}$ since $I_{\ell_1}\in \mathfrak{DP}$ and $I_{\ell_2}\in \mathfrak{R}$. But the adjoint of the formal identity maps the weakly null $\ell_2$ basis to a normalized sequence in $\ell_\infty$, and is therefore not completely continuous. | |
| Dec 29, 2017 at 14:43 | comment | added | user114263 | Then $QA:X\to F/G$ is also a surjection ($A$ maps $E$ onto $F$, which is mapped onto $F/G$ by $Q|_F=q$). Then $QA$, and therefore $A^*Q^*$, is not compact. Since $Q^*$ is weakly compact, $A^*$ isn't completely continuous, or else $A^*Q^*$ would be compact. Therefore if $A^*$ is completely continuous, $A\in \mathfrak{R}$. | |
| Dec 29, 2017 at 14:42 | comment | added | user114263 | If $A:X\to Y$ is not in $\mathfrak{R}$, there is a subspace $E$ of $X$ such that $E, F:=A(E)$ are isomorphic to $\ell_1$ and $A|_E:E\to F$ is an isomorphism. There exists a subspace $G$ of $F$ such that $E/G$ is isomorphic to $\ell_2$ and the quotient map $q:F\to F/G$ is in $\Pi_2$. By the $\Pi_2$ lifting property, there exists a surjection $Q:Y\to F/G$. This is weakly compact, since $F/G$ is reflexive. | |
| Dec 29, 2017 at 14:38 | comment | added | user114263 | One can show that $\mathfrak{V}^\text{dual}\subset \mathfrak{DP}\cap \mathfrak{R}$, but the reverse inclusion does not hold. A characterization of being in $\mathfrak{DP}$ (analogous to the spatial case) is: $A:X\to Y \in \mathfrak{DP}$ if and only if for any weakly null $(x_i)$ in $X$ and any weakly null $(y^*_i)$ in $Y^*$, $\lim_i y^*_i(Ax_i)=0$. If $A^*$ is completely continuous, $(x_i)$ is weakly null, and $(y^*_i)$ is weakly null, then $(A^*y^*_i)$ is norm null and $(x_i)$ is bounded, so $\lim_i A^*y^*_i(x_i)=\lim_i y^*_i(Ax_i)=0$, and $A$ lies in $\mathfrak{DP}$. | |
| Dec 28, 2017 at 8:35 | comment | added | M.González | It appears as 4.16 Corollary in the book [Joe Diestel, Hans Jarchow, Andrew Tonge. Absolutely Summing Operators. Cambridge Univ. Press, 1995]. | |
| Dec 27, 2017 at 18:45 | comment | added | user114263 | I was unaware of the fact that a space having an $\ell_1$ subspace must have an $\ell_2$ quotient. Do you have a reference or proof of this fact? I don't have access to the Diestel survey. | |
| Dec 26, 2017 at 14:58 | comment | added | M.González | Suppose that $E^*$ is Schur. Then $E^*$ has the DPP, hence $E$ has the DPP. Moreover, if a space contains a copy of $\ell_1$ then it has a quoptient isomorphic to $\ell_2$, hence the dual contains a copy of $\ell_2$, thus the dual is not Schur. | |
| Dec 26, 2017 at 14:56 | comment | added | M.González | Since $E$ has the Dunford-Pettis property, $T$ is completely continuous. But $E$ contains no copies of $\ell_1$ implies that every bounded sequence in $E$ has a weakly Cauchy subsequence. Then $T$ is compact. Hence $(f_n)$ converges in norm to $0$. | |
| Dec 26, 2017 at 14:53 | comment | added | M.González | The operator $T:E\to c_0$ defined by $Tx =(f_n(x))$ is weakly compact. | |
| Dec 26, 2017 at 14:52 | comment | added | M.González | Suppose that E has the DPP and contains no copies of $\ell_1$. Let $(f_n)$ be a weakly null sequence in $E^*$. We have to show that $(f_n)$ converges in norm to $0$. | |
| Dec 26, 2017 at 14:50 | comment | added | M.González | A reference for that fact is J. Diestel, "A survey or results related to the Dunford–Pettis property" , Contemp. Math. , 2 , Amer. Math. Soc. (1980) pp. 15–60. The proof is not difficult. | |
| Dec 26, 2017 at 13:46 | comment | added | user114263 | I can see two of the three requisite implications to show that $\mathfrak{DP}\cap \mathfrak{R}= \mathfrak{V}^\text{dual}$. Do you have a reference for the fact: A Banach space $E$ has the DPP and contains no copies of $\ell_1$ if and only if $E^*$ has the Schur property? | |
| Dec 25, 2017 at 11:41 | comment | added | M.González | Do you know if $\mathcal{DP}\cap\mathcal{R} = \mathcal{B}^{dual}$? Here $\mathcal{R}$ are the operators that factor through a Banach space containing no copies of $\ell_1$. | |
| Dec 25, 2017 at 11:38 | comment | added | M.González | A Banach space $E$ has the Dunford-Pettis property and contains no copies of $\ell_1$ if and only if the dual space $E^*$ has the Schur property. | |
| Dec 23, 2017 at 22:50 | comment | added | user114263 | For every countable, non-zero ordinal $\xi$, the notion of $\xi$-Banach-Saks has been defined by Argyros, Merkourakis, and Tsarpalias, and $1$-Banach-Saks coincides with the Banach-Saks property. It was shown by Beanland and Causey that for every $0<\xi<\omega_1$, the $\xi$-analogue of the result you mention holds. | |
| Dec 23, 2017 at 22:42 | comment | added | user114263 | Thank you for your answer, but this doesn't address the question, which is asking specifically about the two ideals $\mathfrak{V}$ and $\mathfrak{DP}$. | |
| Dec 23, 2017 at 11:12 | history | answered | M.González | CC BY-SA 3.0 |