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Hello, folks,
A Sturm-Liouville operator is typically defined not on the whole space of [itex]C^2[/itex] functions,
but rather on some subspace described by boundary conditions. My question is: are those subspaces closed (hence complete, hence Hilbert) in [itex]L^2[/itex]? In case of an affirmative answer, how
can one prove that?
I've been searching for this answer in books, but apparently they don't even bother to consider
that operators are defined in subspaces of [itex]L^2[/itex] and go on applying spectral theory to
them as if they were defined on a Hilbert space.
Please, let me know if the question was not clear.
Thanks in advance.
A Sturm-Liouville operator is typically defined not on the whole space of [itex]C^2[/itex] functions,
but rather on some subspace described by boundary conditions. My question is: are those subspaces closed (hence complete, hence Hilbert) in [itex]L^2[/itex]? In case of an affirmative answer, how
can one prove that?
I've been searching for this answer in books, but apparently they don't even bother to consider
that operators are defined in subspaces of [itex]L^2[/itex] and go on applying spectral theory to
them as if they were defined on a Hilbert space.
Please, let me know if the question was not clear.
Thanks in advance.