Duals of ideals of operators between Banach spaces

Question

Given an operator ideal $\mathfrak{I}$, $\mathfrak{I}^\text{dual}$ is the class of all operators $A:X\to Y$ between Banach spaces $X$ and $Y$ such that $A^*\in \mathfrak{I}$. Given an operator ideal $\mathfrak{I}$, it is often of interest to know when there is another ideal $\mathfrak{J}$ of independent interest such that either $\mathfrak{I}^\text{dual}=\mathfrak{J}$ or $\mathfrak{I}=\mathfrak{J}^\text{dual}$.

Natural examples are the compact and weakly compact operators, which are their own dual ideals. Another interesting example is that an operator $A:X\to Y$ is Asplund (or decomposing) if and only if $A^*$ has the Radon-Nikodym property. Therefore if $\mathfrak{P}$ denotes the ideal of Radon-Nikodym operators and $\mathfrak{D}$ denotes the ideal of Asplund operators, $\mathfrak{P}^\text{dual}=\mathfrak{D}$.

Let $\mathfrak{V}$ denote the class of completely continuous operators and let $\mathfrak{DP}$ denote the ideal of operators $B:X\to Y$ such that for any Banach space $Z$ and any weakly compact operator $A:Y\to Z$, $AB$ is completely continuous.

Is it known whether $\mathfrak{V}^\text{dual}$ or $\mathfrak{DP}^\text{dual}$ is another ideal of independent interest? Or whether there exists another ideal $\mathfrak{I}$ of independent interest such that either $\mathfrak{I}^\text{dual}= \mathfrak{V}$ or $\mathfrak{I}^\text{dual}= \mathfrak{DP}$?

Answer 1

Let $\mathcal{K}$, $\mathcal{W}$ and $\mathcal{C}$ denote the compact, weakly compact and completely continuous operators, respectively, and let $\mathcal{W}^{-1}\circ \mathcal{K}$ denote the operators $T:X\to Y$ such that $BT\in\mathcal{K}$ for each $Z$ and each $B\in\mathcal{W}(Y,Z)$.

CLAIM: $\mathcal{C}^{dual} = \mathcal{W}^{-1}\circ \mathcal{K}$.

Suppose $T\in \mathcal{C}^{dual}(X,Y)$. If $B\in\mathcal{W}(Y,Z)$, then $(BT)^*=T^*B^* \in \mathcal{K}$ because $T^*\in \mathcal{C}$ and $B^*\in\mathcal{W}$, hence $BT\in \mathcal{K}$, and we conclude $T\in\mathcal{W}^{-1}\circ \mathcal{K}$.

Suppose $T\in (\mathcal{W}^{-1}\circ \mathcal{K})(X,Y)$, and let $(g_n)$ be a weakly null sequence in $Y^*$. The operator $B:Y\to c_0$ defined by $Bx =(g_n(x))$ is weakly compact. Then $BT\in \mathcal{K}$. Since the unit vector basis $(e_n)$ is weak$^*$-null in $\ell_1=c_0^*$, $(T^*g_n)= (T^*B^*e_n)$ is norm null. Thus $T^*\in \mathcal{C}$.

Answer 2

An operator $T:X\to Y$ is Banach-Saks if for every bounded sequence $(x_n)$ in $X$ there is a subsequence $(Tx_{n_k})$ such that the Cesàro means $N^{-1}(\sum_{k=1}^N Tx_k)$ from a norm-convergent subsequence. The class of Banach-Saks operators $\mathcal{BS}$ is an operator ideal which is contained in the operator ideal of weakly compact operators. In particular, if the identity $I_E$ of a Banach space $E$ is in $\mathcal{BS}$ then $E$ is reflexive.

In the paper [Ann. Acad. Sci. Fennicae Math. 18 (1993), 3-11], it is proved that $T:X\to Y$ is $\mathcal{BS}$ if and only if the intermediate space $E$ in the Davis-Figiel-Johnson-Pelczynski factorization of $T$ has the Banach-Saks property (i.e. $I_E\in \mathcal{BS}$), $T^* \in \mathcal{BS}$ if and only if $I_{E^*}\in \mathcal{BS}$, and $T^{**} \in \mathcal{BS}$ if and only if $I_{E^{**}}\in \mathcal{BS}$. Since $I_E\in \mathcal{BS}$ does not imply $I_{E^*}\in \mathcal{BS}$, we get:

$\mathcal{BS}^{dual}\neq \mathcal{BS}$, and $\mathcal{BS}^{dual\, dual}= \mathcal{BS}$.