Re: Are closed operators maximal?

On Jul 16, 1:02 am, rusty <mr.ru...@xxxxxxx> wrote:
fjbl...@xxxxxxxxx wrote:
Hi everyone,

I'm studying unbounded operators, and I'm trying to check my intuition
on something. I have this idea that the closure of an operator is in
some sense a maximal extension. That is, suppose X is a Banach space,
A is a closed densely defined operator, and B a closed extension of
A. Must A = B? I can't prove it, but I don't know of too many things
to try as counterexamples either.

Thanks in advance for any tips.

In a Hilbert space, you can take the Friedrichs extension B and
the von Neumann-Krein extensions C of a positive symmetric densely
defined operator A. Both are (densely defined) self-adjoint
operators and any other self-adjoint extension D of A satisfies
B <= D <= C. For a discussion with examples and references
please see:

http://en.wikipedia.org/wiki/Extensions_of_symmetric_operatorshttp://en.wikipedia.org/wiki/Friedrichs_extension

Thanks rusty, I'll have to digest those examples, but it's good to
know what the answer is.

.

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