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Let $H$ be a complex Hilbert space. If $B$ is a bounded self-adjoint operator on $H$ then its spectrum is a closed and bounded subset of the real line and we can find its extremes in terms of the quadratic form $(B\psi, \psi)$:

$$\min (\sigma(B))=\inf_{\psi \in H} \frac{(B\psi, \psi)}{\lVert \psi\rVert^2}, \quad \max(\sigma(B))=\sup_{\psi \in H} \frac{(B\psi, \psi)}{\lVert \psi\rVert^2}$$

(cf. Brézis, Functional analysis, Sobolev spaces and PDE, §6.4, Proposition 6.9, p. 165 - link)

Question 1. Does this extend to the unbounded case? Specifically, I think it is true that, given a self-adjoint operator $A\colon D(A)\subset H \to H$, we have

$$\inf (\sigma(A))=\inf_{\psi \in D(A)} \frac{(A\psi, \psi)}{\lVert \psi\rVert^2}, \quad \sup(\sigma(A))=\sup_{\psi \in D(A)} \frac{(A\psi, \psi)}{\lVert \psi\rVert^2};$$

where of course we allow for infinite inf's and sup's.

Question 2. If the answer to 1. is affirmative, can we replace

$$\inf_{\psi \in D(A)} \frac{(A\psi, \psi)}{\lVert \psi\rVert^2} \quad \text{and}\quad \sup_{\psi \in D(A)} \frac{(A\psi, \psi)}{\lVert \psi\rVert^2}$$


$$\inf_{\psi \in D} \frac{(A\psi, \psi)}{\lVert \psi\rVert^2} \quad \text{and}\quad \sup_{\psi \in D} \frac{(A\psi, \psi)}{\lVert \psi\rVert^2}$$

where $D\subset D(A)$ is a dense subset? I guess that we cannot unless $D$ is a core for $A$, that is a domain of essential self-adjointness.

Thank you.

share|cite|improve this question
Thank you for the corrections, t.b.! – Giuseppe Negro Apr 8 '12 at 10:48
If you are still interested, the answer to the question 1 is yes, and to the question 2, yes, if $D$ is a core. To prove it you can use the spectral theorem in terms of multiplication operators. – Yurii Savchuk Nov 5 '13 at 14:10

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