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!

  • 2
    $\begingroup$ I believe the formal identity $I:J_p\to c_0$ is strictly singular but not weakly compact, nor does it fix a copy of $\ell_p$. $\endgroup$
    – Ben W
    Aug 7, 2016 at 1:07

1 Answer 1


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 $\ell_p$.

  • $\begingroup$ Thanks, Ben! The formal identity $I_{p,q}:J_{p}\rightarrow J_{q}$ is non-weakly compact, but is strictly singular, where $1<p<q<\infty$. $\endgroup$ Aug 7, 2016 at 1:31
  • 1
    $\begingroup$ Yes, that's true too. More generally than in my answer, if a Banach space $X$ has a conditional spreading basis $(x_n)_{n=1}^\infty$ which is boundedly complete then the formal identity $I:X\to c_0$ defined by $Ix_n=s_n$ is a bounded linear operator which is strictly singular but not weakly compact. $\endgroup$
    – Ben W
    Aug 7, 2016 at 1:35
  • $\begingroup$ I am wondering whether the formal identity map $I: J_{p}\rightarrow c_{0}$ is bounded. $\endgroup$ Aug 7, 2016 at 14:57
  • $\begingroup$ Yes, since every normalized basis dominates the c0 basis. $\endgroup$
    – Ben W
    Aug 7, 2016 at 15:00

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