Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Spectrum, resolvent, numerical range, functional calculus, operator semigroups. Special classes of operators: compact, Fredholm, dissipative, differential, integral, pseudodifferential, etc.
2
votes
2
answers
184
views
Show that $(S^1)^*=B(\ell^2)$ knowing $(\ell^1)^*=l^\infty$
Is there a way to show that dual of trace class operators, $S^1$, is $B(\ell^2)$, bounded operators on $\ell^2$, knowing that dual of $\ell^1$ is $\ell^\infty$?