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 |
8
votes
Accepted
why is this a sufficient condition for a domain to be a core of an unbounded operator?
This is stated and proved as Proposition 2.20 here:
https://isem-mathematik.uibk.ac.at/isemwiki/images/9/94/ISEM15_Lecture2.pdf
2
votes
Example of a linear operator whose graph is not closed
On $L^2(R)$, consider the densely defined operator $u\mapsto \int u\,dx$, defined on $L^2\cap L^1$. This operator is neither closed nor closable. If you want the operator defined on all of X, with X i …
1
vote
Accepted
Smooth dependence of the spectrum on the operator
Look at Kato's book on Perturbation Theory for Linear Operators.