Recall that an operator $T:X\rightarrow Y$ is called absolutely summing if there exists a constant $C>0$ such that
$$\sum\_{i=1}^{n}\|Tx\_{i}\|\leq C \sup\_{x^{\*}\in B\_{X^{\*}}}\sum\_{i=1}^{n}|\langle x^{\*},x\_{i}\rangle|,$$ for all finite families $(x\_{i})\_{i=1}^{n}$ in $X$. The least $C$ for which the above inequality always holds is denoted by $\pi(T)$. The set of all absolutely summing operators from $X$ to $Y$ is denoted by $\Pi(X,Y)$.


An operator $T:X\rightarrow Y$ is said to be nuclear provided that there are sequences $(x^{\*}\_{n})\_{n}$ in $X^{\*}$ and $(y\_{n})\_{n}$ in $Y$ such that $\sum\limits\_{n}\|x^{\*}\_{n}\|\|y\_{n}\|<\infty$ and $Tx=\sum\limits\_{n}x^{\*}\_{n}(x)y\_{n}$ for $x\in X$. The nuclear norm $\nu(T)$ is defined to be the infimum of $\sum\limits\_{n}\|x^{\*}\_{n}\|\|y\_{n}\|<\infty$ over all possible such representations. The space of all nuclear operators from $X$ to $Y$ is denoted by $\mathcal{N}(X,Y)$. It is known that every nuclear operator is absolutely summing.


Question 1. For $1<p,q<\infty$, when is $\Pi(l\_{p},l\_{q})$ reflexive ?


Question 2. For $1<p,q<\infty$, $\Pi(l\_{p},l\_{q})=\mathcal{N}(l\_{p},l\_{q})$ ?


Question 3. For $1<p<q<\infty$, is the inclusion map $i\_{p,q}$ from $l\_{p}$ to $l\_{q}$ absolutely summing ?


Thank you !



Absolutely summing operators from $l_{p}$ to $l_{q}$
As for Question 3, note that an absolutely summing operator is completely continuous, i.e., maps weakly null sequences to null sequences, a property not shared by $i\_{p,q}$ if $p>1$. Therefore this question has a negative answer.


