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For a bounded self-adjoint operator $A$ on a Hilbert space $H$ we know that the spectrum is a compact subset of $\mathbb R$ with spectral radius given by the operator norm of $A$. Now assume that $A$ is a densely defined unbounded self-adjoint operator. Can we conclude that the spectrum of $A$ is an unbounded set of the reals?

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Yes, we can. It is a well-known result that a self-adjoint operator is bounded if and only if its spectrum is compact. Since the spectrum is always closed, you get what is desired.

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  • $\begingroup$ Could you give me a source for a proof of this statement? $\endgroup$
    – Muzi
    Commented Jul 12, 2018 at 18:31
  • $\begingroup$ There are probably many books where can you can find it. As a student I used "Weidmann, Joachim. Linear operators in Hilbert spaces. Vol. 68. Springer Science & Business Media, 2012". See Section 7.4 Theorem 7.22 and its corollaries. $\endgroup$
    – Paul Frost
    Commented Jul 12, 2018 at 18:41
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If $A : \mathcal{D}(A)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ is densely-defined and selfadjoint on a complex Hilbert space $\mathcal{H}$, and if the spectrum of $A$ is contained in $[-R,R]$ for some $R$, then $$ I = \frac{1}{2\pi i}\oint_{|z|=R+\epsilon}(\lambda I-A)^{-1} d\lambda\\ Ax = \frac{1}{2\pi i}\oint_{|z|=R+\epsilon}\lambda(\lambda I-A)^{-1}x\;d\lambda,\;\;\; x\in\mathcal{D}(A). $$ This last integral defines a bounded operator because $\lambda(\lambda I-A)^{-1}$ is uniformly bounded in $\lambda$ on $|\lambda|=R+\epsilon$. So, $A$ has a bounded extension. Because $A$ is closed, then $A$ is bounded and defined everywhere.

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