Recall that the James $p$-space $J\_{p}(1<p<\infty)$ is the (real) Banach space of all sequences $(a\_{n})\_{n}$ of real numbers such that $\lim\_{n\rightarrow \infty}a\_{n}=0$ and 
$$\|(a\_{n})\_{n}\|\_{pv}=\sup\{(\sum\_{j=1}^{m}|a\_{i\_{j-1}}-a\_{i\_{j}}|^{p})^{\frac{1}{p}}:1\leq i\_{0}<i\_{1}<\cdots<i\_{m}, m\in \mathbb{N}\}<\infty.$$


The sequence $(e\_{n})\_{n}$ of standard unit vectors forms a monotone shrinking basis for $J\_{p}$ in norm $\|\cdot\|\_{pv}$. It is known that $J\_{p}$ is non-reflexive and is codimension of $1$ in $J^{\*\*}\_{p}$, but every infinite-dimensional closed subspace of $J\_{p}$ contains a subspace isomorphic to $l\_{p}$.


Q1: Let $T:J\_{p}\rightarrow X$ be an operator. If $T$ is strictly singular, is $T$ weakly compact?


Q2: Let $T:J\_{p}\rightarrow X$ be an operator. If $T$ is non-weakly compact, does $T$ fix a copy of $l\_{p}$?


Thank you!



Non-weakly compact operators on the James $p$-space $J_{p}(1<p<\infty)$
The answer to both questions is no.


Let $I:J\_p\to c\_0$ denote the formal identity defined by
\begin{equation\*}Ie\_n=f\_n,\;\;\;n\in\mathbb{N},\end{equation\*}
where $(e\_n)\_{n=1}^\infty$ is the canonical basis for $J\_p$ and $(f\_n)\_{n=1}^\infty$ is the canonical basis for $c\_0$. It is well-known that $x\_n=\sum\_{k=1}^ne\_k$ forms a normalized spreading basis for $J\_p$. Note that
\begin{equation\*}Ix\_n=s\_n,\;\;\;n\in\mathbb{N},\end{equation\*}
where $(s\_n)\_{n=1}^\infty$ is the summing basis for $c\_0$. It follows that $I$ is not weakly compact. On the other hand, it is clear that $I$ is strictly singular since $J\_p$ is $\ell\_p$-saturated whereas $c\_0$ is $c\_0$-saturated. And for the latter reason, $I$ fails to fix a copy of $\ell\_p$.


