5
$\begingroup$

An operator ideal $\mathcal{I}$ possesses the $\sum_{p}$-property, say $1<p<\infty$, if for arbitrary collections of Banach spaces $E_{m}, F_{n} (m,n=1,2,\ldots)$ the following holds:

if $T\in \mathcal{L}((\sum_{m} E_{m})_{p},(\sum_{n} F_{n})_{p})$ and $Q_{n}TJ_{m} \in \mathcal{I}(E_{m},F_{n}) \mbox { for every } m,n$ then $T\in \mathcal{I}((\sum_{m} E_{m})_{p},(\sum_{n} F_{n})_{p})$

($J_{m}, Q_{n}$ are the natural injections and projections), see [1, page 404].

Many things are known, see [1]. For example:

i) if the ideal has the $\sum_{p}$-property then it is closed.

ii) the operator ideals of weakly compact operators, Rosenthal, separable, Banach-Saks, Alternating sign Banach-Saks, Decomposing and others have the $\sum_{p}$-property.

iii) also are known some conditions that imply the $\sum_{p}$-property.

$\sum_{p}$-property is called p-stability by Piestch in his book, see [2, Epilogue].

Question 1: Does $\mathcal{I}$ closed and $\ell_{p}\in$ space($\mathcal{I}$) (that is, identity $1_{\ell_{p}}\in\mathcal{I}$) imply $\mathcal{I} \mbox{ has }\sum_{p}$-property?.

Question 2: If the answer to the above question is negative, does $\mathcal{I}$ closed + $\ell_{p}\in$ space($\mathcal{I}$) + something imply $\mathcal{I} \mbox{ has }\sum_{p}$-property?.

Question 3: Does someone know of a book or paper where this concept of $\sum_{p}$-property for ideals is studied?.

[1] S. Heinrich, Closed operator ideals and interpolation, J. Funct. Anal. $\mathbf{35}$ (1980), 397-411.

[2] A. Pietsch, Operator ideals, North Holland, Amsterdam, 1980.

$\endgroup$

1 Answer 1

3
$\begingroup$

The answer to your first question is no. To see this, one can consider for every ordinal $\xi$ the class $\mathcal{I}=\textbf{Sz}_\xi$ of operators whose Szlenk index does not exceed $\omega^\xi$. Then $\text{Space}(\mathcal{I})$ contains all $\ell_p$ spaces for any $\xi>0$, $1<p<\infty$ (these spaces are asymptotically uniformly smooth, and have Szlenk index $\omega$). However, it is known that there are ordinals for which $\textbf{Sz}_\xi$ does not have the $\Sigma_p$ property. The ordinal $\xi=1$ is such an ordinal, and so is every $\xi=\omega^\gamma$, $\gamma$ a limit ordinal of countable cofinality. This is because if $\textbf{Sz}_\xi$ has the $\Sigma_p$ property, then every operator with Szlenk index not exceeding $\omega^\xi$ factors through a space with Szlenk index not exceeding $\omega^\xi$, and it is known that when $\xi=1$ or when $\xi=\omega^\gamma$, $\gamma$ a limit ordinal with countable cofinality, there exists an operator with Szlenk index $\omega^\xi$ which cannot be factored through a space with Szlenk index $\omega^\xi$. All of these facts can be found in the work of Philip Brooker on Asplund operators.

One can do similar things with other indices besides the Szlenk index. Two such examples are the Bourgain $\ell_1$ index of an operator and the James weakly compact index of an operator. A recent work of Causey and Beanland gives counterexamples similar to the ones above for both of these other indices.

In particular, the class of super weakly compact operators is a closed operator ideal which contains each $I_{\ell_p}$, $1<p<\infty$, but which lacks the $\Sigma_p$ property. This is because there are super weakly compact operators which do not factor through superreflexive spaces. The compact diagonal operator $\sum a_n e_n\mapsto \sum \frac{a_n}{\log(n+1)} e_n$ on $c_0$ has no non-trivial Rademacher cotype, and so it cannot factor through space with non-trivial Rademacher cotype, and therefore does not factor through any superreflexive space.

As for the second question, it is a little vague for me to have a good answer. My above answers do have a common theme, but not anything which gives the $\Sigma_p$ property of $\mathcal{I}$. Basically, for Szlenk, Bourgain, and James indices, the fact that $\ell_p\in \text{Space}(\mathcal{I})$ means that $\ell_p$ is "very'' Asplund or "very'' not containing $\ell_1$ or "very'' reflexive. In all of those examples, what has been shown is that in this case, if $T:(\oplus X_n)_{\ell_p}\to (\oplus Y_n)_{\ell_p}$ has $Q_nTJ_m\in \mathcal{I}$ for all $n,m$, then $T$ cannot be "too much less'' Asplund or "too much more'' containing $\ell_1$ or "too much less'' reflexive. That is, each $Q_n TJ_m$ has some Szlenk or Bourgain or James index uniformly bounded by some ordinal $\xi$, and the index of $T$ cannot be too much more than $\xi$. But in each of the above cases, one cannot deduce exactly the same bound.

All I know of the $\Sigma_p$ properties I learned from the works of Heinrich and Brooker.

$\endgroup$
6
  • $\begingroup$ Why are you unregistered and anonymous? $\endgroup$ May 4, 2017 at 21:12
  • $\begingroup$ I came here to ask a question (which I ended up forgetting to do) and saw a topic I knew a little about. It seemed easiest to just answer as a guest. $\endgroup$
    – user496750
    May 5, 2017 at 12:02
  • $\begingroup$ @AlexiQuevedoS: Nice question. Check out those papers Ryan (user496750) mentioned in his answer. $\endgroup$ May 5, 2017 at 17:33
  • $\begingroup$ Thanks very much for your answer user496750. Nice and convincing counterexample, you have been very kind. I will check that paper of Philip Brooker on Asplund operators. $\endgroup$ May 6, 2017 at 14:05
  • $\begingroup$ Thanks Kevin Beanland. I will check the papers that user496750 mentioned in his answer. $\endgroup$ May 6, 2017 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.