# Spectrum of an unbounded operator $A$

Are there any reference that we can find the following

Theorem: Let $A$ be a positive (resp. negative) self-adjoint unbounded operator on a Hilbert space $H$. Then, the spectrum of $A$ is contained in the positive real-axis (resp. negative real-axis).

• Why do you ask about "unbounded operator" in the title? – Artem Aug 20 '16 at 13:32
• Yes, it's an unbounded operator. I edited my question. Thanks – Z. Alfata Aug 20 '16 at 13:41
• Strictly saying, the definition of a self-adjoint operator includes being "bounded." If your operator is unbounded then it is not self-adjoint. Anyway, a very readable and rigorous discussion is given in Linear Operator Theory in Engineering and Science. – Artem Aug 20 '16 at 13:43
• @Artem : If $A : \mathcal{D}(A)\subseteq H \rightarrow H$ is a closed, densely-defined linear operator on a Hilbert space $H$, then the adjoint $A^*$ is also a closed, densely-defined linear operator. And it is quite possible to have $A=A^*$, which is what selfadjoint is defined to be in this context. – DisintegratingByParts Aug 20 '16 at 14:13
• @TrialAndError Then it looks like it depends on the definition. Say in the book I referenced above selfadjoint must be defined everywhere, and the operator that you are talking about is called symmetric. – Artem Aug 20 '16 at 17:33

## 1 Answer

Assume that $A : \mathcal{D}(A) \subset H\rightarrow H$ is an unbounded selfadjoint operator on a real or complex Hilbert space $H$ such that $(Ax,x) \ge 0$ for all $x\in\mathcal{D}(A)$. For $\lambda > 0$, $$\lambda(x,x) \le ((A+\lambda I)x,x)\le \|(A+\lambda I)x\|\|x\|,\;\;\; x\in\mathcal{D}(A).$$ Therefore, $$\lambda\|x\| \le \|(A+\lambda I)x\|,\;\; x\in\mathcal{D}(A).$$ Such an inequality forces the range of $A+\lambda I$ to be closed because because, if $(A+\lambda I)x_n$ converges to $y$, then $\{ (A+\lambda I)x_n \}$ is a Cauchy sequence and the above forces $\{ x_n \}$ to be a Cauchy sequence, which must converge to some $x$ because $H$ is complete. Then, because $A$ is closed, $y = (A+\lambda I)x$. So the range of $A+\lambda I$ is closed. The range of $A+\lambda I$ must be all of $X$ because it is closed and $y \in \mathcal{R}(A+\lambda I)^{\perp}$ iff $$((A+\lambda I)x,y)=0,\;\;\; x\in\mathcal{D}(A).$$ By the definition of adjoint, $y \in \mathcal{D}(A^*)=\mathcal{D}(A)$ and $(A+\lambda I)y=0$. However, $\mathcal{N}(A+\lambda I)=\{0\}$ which gives $y=0$. Hence, $-\lambda\in\rho(A)$, as was to be shown.