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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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    More specific: There are closed symmetric operators having many different selfadjoint extensions.2015-07-09

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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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    Many, thanks. I always learn a lot from you!2012-10-19