Is the spectrum of an unbounded self-adjoint operator always an unbounded set?

Question

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?

Answer 1

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.

Answer 2

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.