Timeline for Norm inequality for stochastic maps
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2010 at 5:42 | vote | accept | Mateus Araújo | ||
| Sep 15, 2010 at 21:19 | answer | added | Martin Argerami | timeline score: 1 | |
| Sep 14, 2010 at 1:25 | history | edited | Mateus Araújo | CC BY-SA 2.5 |
Expanded, clarified question.; added 1 characters in body
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| Sep 14, 2010 at 1:02 | comment | added | Owen Sizemore | Could you also clarify what $\Lambda$ is acting on. Is it all of B(H)? Or is it some subalgebra?. The reason I ask this is to clarify what you mean by the 2-norm. Do you mean the Hilbert-Schmidt norm? If so this means that we must restrict to those B with finite Hilbert Schmidt norm. | |
| Sep 13, 2010 at 21:33 | comment | added | Mateus Araújo | Yes! Sorry, I've been working only with positive maps, so that I forgot to specify. | |
| Sep 13, 2010 at 21:31 | history | edited | Mateus Araújo | CC BY-SA 2.5 |
corrected question
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| Sep 13, 2010 at 21:05 | comment | added | Owen Sizemore | I'm confused about your first claim. If stochastic just means that it maps the identity to the identity then this does not imply that $\Lambda$ has norm 1. On the 2x2 matrices consider, $\Lambda$= $2(x)-Tr(x)I$. Where Tr is the normalized trace. This sends I to I but has norm larger than 2. You're statement requires that $\Lambda$ be a positive operator. Under this situation (that $\Lambda$ is positive) then this should still work for all p-norms. | |
| Sep 13, 2010 at 20:32 | history | asked | Mateus Araújo | CC BY-SA 2.5 |