# Weak compactness in the James space and its dual

It is known that there are characterizations of weak compactness in most of classical non-reflexive spaces (e.g. $L_{1}$-spaces and $C(K)$-spaces). I wonder whether there are characterizations of weak compactness in James space $J$ or its dual $J^{*}$. Can we establish a criterion for it if there is no? Thank you!

• I am not an expert on Banach spaces, but I am not sure it is accurate to say weak compactness has been characterized in "most classical non-reflexive spaces". $L_1$ and $C(K)$ are pretty special cases, it seems to me May 5, 2016 at 16:21

James space is a commutative Banach algebra with pointwise operations. $$J^{\ast\ast} = J\oplus\mathbb{C}$$ is just $$J$$ with a unit attached [https://doi.org/10.4153/CJM-1980-083-7].

Second, since $$J$$ contains no copy of $$\ell^1$$, every bounded sequence $$(x_n)$$ in $$J$$ has a weakly Cauchy subsequence, say $$(u_n)$$, that converges weak* in $$J^{**}$$ to an element $$x+\lambda 1\in J^{**}$$. If $$\lambda\neq 0$$, then let $$e_n = \lambda^{-1}(u_n-x)$$ so that $$(e_n)$$ is a bounded approximate identity for $$J$$. If a subset $$C\subseteq J$$ is not weakly compact, then it contains a weakly Cauchy sequence $$(u_n)$$ with $$\lambda\neq 0$$.

Consequently, a subset $$C\subseteq J$$ is weakly compact if and only if $$y+\lambda C$$ contains no (sequential) bounded approximate identity for all $$y\in J$$ and all $$\lambda\in\mathbb{C}\backslash\{0\}$$.

About the weak$$^\ast$$ convergence in $$J^{\ast\ast}$$: The spectrum $$\sigma(J)=\{\delta_n:n\in\mathbb{N}\}$$ consists of coordinate functionals, defined by $$\delta_n(x_m)_{m\in\mathbb{N}} = x_n$$ [Proposition 2.7 in https://doi.org/10.4153/CJM-1980-083-7]. Furthermore, $$J^{\ast}$$ is the closed linear span of $$\sigma(J)$$ since $$J$$ is semisimple, $$J$$ is an ideal (in fact, maximal) in $$J^{\ast\ast}$$, and $$J$$ has a bounded approximate identity. Consequently, weak* convergence in $$J^{\ast\ast}$$ is just the pointwise convergence.

Thus, $$C\subseteq J$$ is weakly compact if and only if $$C$$ is closed under the pointwise convergence of sequences, i.e., whenever $$(x_n)\subseteq C$$ is a sequence such that $$x_n\to x$$ pointwise, then $$x\in C$$.

I'm not sure this deserves to be posted as an "answer," but it is far too long for a comment. But since there is already an answer posted, I do not believe this will detract from your question getting attention. Anyway...

First, note that $J$ has a unique (up to equivalence) spreading basis, which we shall call "the" spreading basis for $J$. As every seminormalized weakly null sequence in $J$ contains a subsequence equivalent to the $\ell_2$ basis, the spreading basis cannot be weakly null. This was all established here: http://www.mscand.dk/article/viewFile/11904/9920

From the same paper comes the following (as part (b) from Theorem 2.1).

Theorem. Let $(z_n)_{n=1}^\infty$ be a seminormalized sequence in $J$ having no weak cluster point. Then there is a subsequence $(z_{n_k})_{k=1}^\infty$ equivalent to the spreading basis of $J$, such that its closed linear span $[z_{n_k}]_{k=1}^\infty$ is complemented in $J$.

Corollary. Let $C\subset J$ be a subset which is weakly closed. Then the following are equivalent.

(i) $C$ is not weakly compact.

(ii) $C$ contains a sequence $(c_n)_{n=1}^\infty$ equivalent to the spreading basis of $J$ such that its closed linear span $[c_n]_{n=1}^\infty$ is complemented in $J$.

(iii) $C$ contains a basic sequence $(c_n)_{n=1}^\infty$ which is not weakly null.

Proof. (i) $\Rightarrow$ (ii): By Eberlein-Smulian we can find $(c_n)_{n=1}^\infty\subset C$ with no weak cluster point in $C$, and hence (by weak closure of $C$) no weak cluster point in $J$. Now apply the above theorem to obtain a subsequence equivalent to $J$ and complemented in $J$.

(ii) $\Rightarrow$ (iii): This follows from the fact that the spreading basis is not weakly null.

(iii) $\Rightarrow$ (i): By passing to a subsequence if necessary we may assume that the weak closure of $(c_n)_{n=1}^\infty$ (as a set) fails to contain zero. Hence, by a well-known criterion of Kadets-Pelczynski (e.g. Theorem 1.5.6 in the Albiac/Kalton book here), $(c_n)_{n=1}^\infty$ is not relatively weakly compact, and by Eberlein-Smulian, has no weakly convergent subsequence. Thus $C$ is not weakly compact. $\square$

Note that the equivalence (i) $\Leftrightarrow$ (iii) holds for arbitrary Banach space in place of $J$. So, the only special thing here is (ii).

• You are very welcome. I am also interested in the properties of the James space, so please feel free to post anything else you learn. the same goes for the generalized $p$-James spaces. May 7, 2016 at 15:11

One of the things which you can do is: to combine the following theorem of James [James, Robert C. Weakly compact sets. Trans. Amer. Math. Soc. 113 1964 129–140]: A weakly closed subset $C$ of a Banach space $B$ is weakly compact if and only if each member of $B^*$ attains a maximum on $C$ with the description of the dual space of $J$ and $J^*$.

Another thing: analyze when, say, the weak$^*$-limit of a sequence in $J$, considered as a sequence in $J^{**}$, is outside $J$. Do the same for the pair $J^*$ and $J^{***}$

I think that one can definitely get something on these lines. Possibly this has already been done somewhere, but it could be difficult to find.

• I search characterizations of weak compactness in $J$ by google, but fail. May 7, 2016 at 1:15
• Such results are difficult to find either because nobody ever needed such results or because they were published as auxiliary results in a paper devoted to something else. You can look for large papers dealing with the James space (e.g. James, Robert C. Banach spaces quasi-reflexive of order one. Studia Math. 60 (1977), no. 2, 157–177), you can also check the book Fetter, Helga; Gamboa de Buen, Berta, The James forest. Cambridge University Press, Cambridge, 1997 (if you have not yet done this). May 7, 2016 at 3:16
• Thank you for your suggestions, Mikhail. I'll look for it. May 7, 2016 at 4:20