The $p$th James space, denoted $J_p$, is just the regular James space using the $p$-norm in place of the 2-norm. See here for a complete definition. To use their notation, let $\mathbb{N}_0$ denote the set of all nonnegative integers, and for let $x=(x(j))_{j=0}^\infty$ denote a scalar sequence. For each $A=\{n_1<\cdots,n_{k+1}\}\subset\mathbb{N}_0$ of cardinality at least two, we define \begin{equation}\nu_p(x,A)=\left(\sum_{j=1}^k|x(n_j)-x(n_{j+1})|^p\right)^{1/p}.\end{equation} For convenience, denote $\nu_p(x,A)=0$ if $\text{card }A\in\{0,1\}$. Now define the $J_p$ norm \begin{equation}\|x\|_{J_p}=\sup\left\{\nu_p(x,A):A\subset\mathbb{N}_0,\text{card }A<\infty\right\}\end{equation} Then we define the $\boldsymbol{p}$th James space as the set of all scalar sequences with finite $J_p$ norm, or, formally, \begin{equation}J_p=\left\{x\in c_0:\|x\|_{J_p}<\infty\right\}.\end{equation}

Question 1. Fix $1<p<2$. Does there exist a normalized (Schauder) basic sequence $(\xi_n)_{n=1}^\infty\subset J_p$ and a constant $C\in[1,\infty)$ such that for all $(a_n)\in c_{00}$ (here, $c_{00}$ denotes the space of all finitely-supported scalar sequences) the inequality \begin{equation}\|\sum_{n=1}^\infty a_n\xi_n\|_{J_p}\leq C\left(\sum_{n=1}^\infty|a_n|^2\right)^{1/2}\end{equation} holds? (In this case, we say that $(\xi_n)_{n=1}^\infty$ is $\boldsymbol{C}$-dominated by the canonical basis of $\ell_2$.)

We could re-cast this question as follows.

Question 2. Fix $1<p<2$ Is there a non-compact operator in the space $\mathcal{L}(\ell_2,J_p)$ (the space of bounded operators from $\ell_2$ into $J_p$?

Question 3. Fix $1<p<2$. Is there a non-compact operator in the space $\mathcal{L}(J_p^*,\ell_2)$?

Note that questions #1-3 are all equivalent.


Seminormalised, weakly-null sequences in $J_p$ have subsequences equivalent to the canonical basis of $\ell_p$. To see this modify Theorem 2.1 here.

Suppose that $T\colon \ell_2\to J_p$ is a non-compact operator ($p<2$). Let $(x_n)$ be a bounded sequence in $\ell_2$ so that $(Tx_n)$ does not have a convergent subsequence. By reflexivity and weak-to-weak continuity of $T$, $(Tx_n)$ has a weakly convergent subsequence. Subtracting the limit, assume that this weak limit is null. Extract a furter subsequence, so that $(Tx^\prime_n)$ spans a copy of $\ell_p$. Then $T|_{[x_n^\prime]}$ is a non-compact operator from a Hilbert space to $\ell_p$ ($p<2$); contradiction with Pitt's theorem.


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