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8 votes
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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
Michael Renardy's user avatar
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 …
Michael Renardy's user avatar
1 vote
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Smooth dependence of the spectrum on the operator

Look at Kato's book on Perturbation Theory for Linear Operators.
Michael Renardy's user avatar