Absolutely summing operators from $l_{p}$ to $l_{q}$

Question

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 !

Answer 1

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.

Answer 2

Question 2 has a negative answer.

Hint: Show that if question 2 has a positive answer, then every Hilbert-Schmidt operator must be of trace class.

Answer 3

Question 1 has a negative answer.

In 1977, D. R. Lewis proved that $\Pi_{p}(X,Y)$($1\leq p<\infty$) is reflexive if both $X$ and $Y$ are reflexive and either $X$ or $Y$ has the approximation property.

But what about $\Pi(l_{1},l_{p})$ for $1<p<\infty$ ?