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Let $A$ be an unbounded selfadjouint operator in the Hilbert space $H$, having domain $D(A)$.

Denoting by $\sigma_A$ the spectrum of $A$, we have

$\inf \sigma_A \ = \ \inf_{u\in D(A),\|u\|=1} \ \langle u, A u\rangle$

My question is: is this still true if I take the infimum only over a dense subset (EDIT: here I mean a subspace, not just a subset) $C$ of $D(A)$ instead of the whole $D(A)$?

If $A$ was bounded this should be ok by continuity, but in the unbounded case?

EDIT: After some more reflection it seems to me that it is ok also in the unbounded case if $C$ is a core for $A$ (i.e. A is the closure of its restriction to $C$). If $C$ is not a core I start to doubt seriously that it is true in general (I'm thinking of Laplacian on bounded domain, the bottom of the spectrum is different in the Neumann and Dirichlet case...). Still I am not completely sure how to make this precise, and help is still appreciated.

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Presumably you mean $C$ to be a subspace, not just a subset. Otherwise the problem is trivial: e.g. you could take $C = \{v \in D(A): \langle v, A v \rangle > N \|v\|^2 \}$ which is dense in $D(A)$ if $\sigma_A$ is not bounded above. –  Robert Israel Oct 16 '12 at 21:13
@RobertIsrael : yes, I mean a subspace, thanks for pointing to this and for your example. –  Hans Oct 16 '12 at 21:32
The answer is quite simple: A core contains all the information while a merely dense domain may loose alot of information. –  Freeze_S Jul 9 at 11:44
More specific: There are closed symmetric operators having many different selfadjoint extensions. –  Freeze_S Jul 9 at 11:46

1 Answer 1

up vote 3 down vote accepted

Consider the unbounded operators $A_b = -\dfrac{d^2}{dx^2}$ for $b \in \mathbb R$, where the domain $D(A_b)$ is the space of functions $f \in L^2[0,1]$ such that $f$ is differentiable, $f'$ is absolutely continuous, $f'' \in L^2[0,1]$, $f(0) = 0$, $b f(1) + f'(1) = 0$. These are all self-adjoint, with different spectra: $\inf \sigma(A_b)$ is the least $\lambda > 0$ such that $b \sin(\sqrt{\lambda}) + \sqrt{\lambda} \cos(\sqrt{\lambda}) = 0$ (except for $b=-1$ where $\inf \sigma(A_0) = 0$), and this is a non-constant function of $b$. Take $C$ to be the space of functions $f \in L^2[0,1]$ such that $f$ is differentiable, $f'$ is absolutely continuous, $f'' \in L^2[0,1]$, $f(0) = 0$, $f(1) = 0$, $f'(1) = 0$. This is dense in all $D(A_b)$, and all $A_b$ agree on it, so $\inf_{u \in C, \|u\|=1} \langle u, A_b u \rangle$ is a constant which can't be $\inf \sigma(A_b)$ for more than one $b$.

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The sentence after "with different spectra:" is not clear to me, maybe it is a typo. What is the condition on $\lambda$ so that it equals the inf of the spectrum? –  Hans Oct 16 '12 at 23:59
Sorry, left out "$=0$". I corrected it. –  Robert Israel Oct 17 '12 at 1:18
Many, thanks. I always learn a lot from you! –  Hans Oct 19 '12 at 10:55

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