Is there a "right" definition of the nuclear operator in the nonlinear framework ? Of course, such an operator must be compact, while a linear operator should be "nonlinearly" nuclear iff it is nuclear as usual. [These are naive conditions, of course.] Since here the Frechet derivative is useless. For example, the "cubing" self-map of $\ell^{2}$, i.e., $(x_{n})$ $\rightarrow\(x_{n}^{3})$ , has nuclear derivative at each point, yet it is clearly noncompact. OTOH, it seems that a [nonlinear] Lipschitz $p$-summing operator need not to be power compact. Some thoughts ?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Funny you should ask. My former student, Bentuo Zheng, and my visitor, Dongyang Chen, are in the process of developing this theory. The "right" definition involves the Pietsch factorization diagram, and is an off-shoot off what Farmer and I did for p-summing and p-integral operators (get the paper from my home page). Send me an email and I'll put you in contact with them.
On a related topic, my current student Alejandro ChavezDominguez is developing the operator ideal theory connected to non-linear p-summing operators and related mappings. This is something I have been interested in for a long time but did not see what to do (even the duality theory). In Javier's hands, the theory is developing very well.
answered Feb 17, 2010 at 2:23
-
1$\begingroup$ Are you interested in answers, Ady? If not, why do you ask a question? $\endgroup$ Mar 18, 2010 at 17:37
-
$\begingroup$ How far is this project along? Probably there are references to papers now? $\endgroup$ Aug 30, 2015 at 15:59
-
$\begingroup$ Yes. Some papers have appeared and I think all are on the ArXiv. Or you can check my former students' home pages (Bentuo is at the University of Memphis and Alejandro just moved to the University of Oklahoma). $\endgroup$ Aug 31, 2015 at 17:54