Re: Are closed operators maximal?
- From: fjblurt@xxxxxxxxx
- Date: Mon, 16 Jul 2007 10:52:10 -0700
On Jul 15, 11:21 pm, William Elliot <ma...@xxxxxxxxxxxxxxxxxx> wrote:
On Sun, 15 Jul 2007 fjbl...@xxxxxxxxx wrote:
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.
What's the closure of a function between two topological spaces?
Hi William,
Well, it isn't just a function between any two topological spaces. An
unbounded operator on a Banach space X is actually a linear map A :
D(A) -> X where the domain D(A) is some (usually not closed) subspace
of X. It is said to be closed if its graph, { (x, Ax) : x in D(A) }
is closed in X x X. B is an extension of A if D(B) contains D(A) and
A=B on D(A). The closure of A is the extension of A whose graph is
the closure of the graph of A, provided this closure is again a
graph. (That is, if (x,y) and (x,z) are in the closure, we must have
y=z.) The closure is certainly the minimal closed extension of A, but
I couldn't think of a case where there could be more than one, if D(A)
was dense.
.
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