Timeline for Uniqueness of differential adjoint operator
Current License: CC BY-SA 2.5
5 events
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| Feb 22, 2011 at 10:31 | comment | added | Theo Buehler | I apologize for the two typos in my previous comment. It should have been $L^{\dagger}$ and $f \mapsto (Lf,g)$. You might also want to have a look at en.wikipedia.org/wiki/Unbounded_operator | |
| Feb 22, 2011 at 8:51 | comment | added | Theo Buehler | No you haven't missed the point. The work-around is to restrict the domain of definition of $A^{\dagger}$ to the subspace consisting of $g$'s such that $g \mapsto (Lf,g)$ is continuous. However, this is not a research-level question. You should ask this on math.stackexchange.com instead where you'll get a detailed answer. | |
| Feb 22, 2011 at 8:49 | comment | added | Zen Harper | I think you're getting mixed up between "bounded" and "unbounded" operators. The standard theory works well for bounded operators, but most differential operators and others used in quantum mechanics, etc. are unbounded - they cannot be defined on the whole space in any nice way. The theory of unbounded operators is a lot more complicated and subtle than bounded ones. Yosida, Functional Analysis is a good reference, I believe. | |
| Feb 22, 2011 at 8:47 | comment | added | Matthew Daws | By "differential operator" do you mean: the operator given by differentiation? If so, then this is very far from been defined for all $L^2$ functions-- you need to restrict to those which can be differentiated! So you're really studying an unbounded, densely defined operator. There is a lot of literature about such things-- in particular, taking adjoints does make sense, but you need some technology, and you have to be careful. | |
| Feb 22, 2011 at 8:39 | history | asked | Phil | CC BY-SA 2.5 |