Timeline for What is the "correct" generalization of operator norms for nonlinear operators?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2023 at 20:22 | answer | added | Ben Francis | timeline score: 0 | |
| Apr 8, 2011 at 7:34 | comment | added | Suvrit | @Bill: That sounds very promising; thanks. | |
| Apr 7, 2011 at 13:33 | comment | added | Bill Johnson | Look at the book "Geometric nonlinear functional analysis" by Benyamini and Lindenstrauss. | |
| Apr 6, 2011 at 21:28 | comment | added | Helge | Depends on what you want to do. But for most things, e.g. convergence: An appropriate topology. | |
| Apr 6, 2011 at 21:20 | comment | added | Suvrit | @Mark: indeed the correct perspective, but I am currently just exploring these notions, and do not know what properties will actually matter. Essentially (also @Dirk), am just trying to understand how to nicely characterize the "magnitude" of the action of a nonlinear operator, so that within an iterative algorithm I can guarantee that subsequent iterates remain "feasible." However, even though asking for these norms might be a bad approach for such problems, I am still interested in knowing more about how one handles such operators at all! | |
| Apr 6, 2011 at 19:39 | comment | added | Dirk | Usually a norm is used to turn a vector space into a normed space. Do you have a special vector space of nonlinear operators in mind? Probably the space of all nonlinear operators between two (normed?) spaces is too large... | |
| Apr 6, 2011 at 15:38 | comment | added | Mark Meckes | A useful bit of perspective on this question is provided by recalling that for a linear operator $A$ between normed spaces, the following are all equivalent: $A$ is continuous, $A$ is uniformly continuous, $A$ is Lipschitz, $\| A \|$ is finite. Without linearity the equivalences all fail. As abatkai points out, how to generalize $\| A \|$ appropriately depends on the context, in particular which properties of $A$ are actually important for you. | |
| Apr 6, 2011 at 15:36 | comment | added | Suvrit | Thanks Mikael; I guess linearizing takes us to Lipschitz? But for functions that are only locally Lipschitz? or which flavors of Lipschitz? Perhaps you'd care to expand a bit on your comment? | |
| Apr 6, 2011 at 15:29 | answer | added | András Bátkai | timeline score: 10 | |
| Apr 6, 2011 at 14:41 | comment | added | Mikael de la Salle | Maybe the Lispschitz constant? | |
| Apr 6, 2011 at 14:37 | history | asked | Suvrit | CC BY-SA 2.5 |