Does formally self-adjoint imply self-adjoint and vice versa? Thanks.
Definition of (formally) self-adjoint
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3More context, please? – 2011-10-21
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0@HenningMakholm: I'd like to know if this is generally true. But for a specific example say for a differential operator of the form $L=\sum a_n(x)D^n$ if I want to find (iff) conditions s.t. $L$ is self-adjoint, is it enough to check that it is formally self-adjoint? I know that for "self-adjointness", I have to do lots of integration by parts. :( – 2011-10-21
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2craig: You should add that to your question. Remember: speaking of self adjointness of an operator (defined, say, on a dense subspace of $L^2$) does *not* make sense *per se*, you need to specify its domain (and for an operator $L = \sum a_n(x)D^n$ the domain of the adjoint depends drastically on the domain of $L$ you're considering). Formal self-adjointness makes sense for operators that are defined on the smooth functions and the formula you write down for the formal adjoint is actually the one you obtain by integrating by parts (as it should be...) – 2011-10-21
1 Answers
The comment by t.b. is essentially an answer:
Speaking of self adjointness of an operator (defined, say, on a dense subspace of $L^2$) does not make sense per se, you need to specify its domain (and for an operator $L=\sum a_n(x)D^n $ the domain of the adjoint depends drastically on the domain of $L$ you're considering). Formal self-adjointness makes sense for operators that are defined on the smooth functions and the formula you write down for the formal adjoint is actually the one you obtain by integrating by parts.
A more condensed version: Self-adjointness is a strictly stronger property, because it also requires the operator to be closed and its adjoint to have the same domain as the operator itself.
Recommended reading: Chapter 1 of the book An Initiation to Logarithmic Sobolev Inequalities by Gilles Royer is called Self-adjoint operators and is freely available.