Let $D$ be a closed unbounded operator densely defined on a Hilbert space $\cal{H}$. If $D$ is self-adjoint, then it is clear that its eigenvalues are real. Is it also clear that the elements of its spectrum are real?

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    $\begingroup$ Yes, because $\Im \lambda \|x\|^2 = \Im \langle (\lambda I-D)x,x\rangle$ forces $\|x\||\lambda| \le \|(\lambda I-D)x\|$. For $\lambda\notin\mathbb{R}$, this means $\lambda I-D$ is continuously invertible on the closure of its range, which is all of $\mathcal{H}$ because $\overline{\mathcal{R}(\lambda I-D)}=\mathcal{N}(\lambda I-D)^{\perp}=\{0\}^{\perp}$. $\endgroup$ Dec 15, 2017 at 0:09
  • $\begingroup$ Could you explain this in a bit more detail please. I'm not getting any of the steps. $\endgroup$ Dec 15, 2017 at 9:53

1 Answer 1


I'll carry out the argument for you based on a bounded selfadjoint operator $D$, though the argument does carry through for an unbounded selfadjoint operator as well. First note that $D-\lambda I$ is injective for all $\lambda\notin\mathbb{R}$ because $\langle Dx,x\rangle$ is real for all $x$, which leads to $$ \Im \langle (D-\lambda I)x,x\rangle = -\Im\lambda \langle x,x\rangle=-\Im\lambda \|x\|^2. $$ By the Cauchy-Schwarz inequality, $$ |\Im\lambda|\|x\|^2 \le |\Im\langle (D-\lambda I)x,x\rangle| \\ \le |\langle (D-\lambda I)x,x\rangle| \\ \le \|(D-\lambda I)x\|\|x\| \\ \implies |\Im\lambda|\|x\| \le \|(D-\lambda I)x\|. $$ The above shows that (a) $D-\lambda I$ has $\{0\}$ null space for $\lambda\notin\mathbb{R}$, and (b) the inverse operator of $D-\lambda I$ that is defined on the range of $D-\lambda I$ is bounded in norm by $1/|\Im\lambda|$. The range of $D-\lambda I$ is dense in $H$ because $$ \mathcal{R}(D-\lambda I)^{\perp} = \mathcal{N}((D-\lambda I)^*)=\mathcal{N}(D-\overline{\lambda}I))=\{0\},\;\;\lambda\notin\mathbb{R}. $$ And the range is closed because if $\{ y_n=(D-\lambda I)x_n \}$ converges to $y$, then $\{ y_n\}$ is Cauchy, which forces $\{x_n\}$ to be Cauchy because $$ \|x_n-x_m\| \le \frac{1}{|\Im\lambda|}\|(D-\lambda I)(x_n-x_m)\|=\frac{1}{|\Im\lambda|}\|y_n-y_m\|. $$ Hence $y=(D-\lambda I)x$. (This argument works if $D$ is just closed.) The conclusion is that $D-\lambda I$ is injective, surjective, and has a bounded inverse if $\lambda\notin\mathbb{R}$, which proves that $\lambda$ is in the resolvent set for $\lambda\notin\mathbb{R}$.


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