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Let $T : \operatorname{dom}(T) \rightarrow H $ be a positive self-adjoint operator, is it then true that $\sigma(T) \subset [0,\infty)$? This is something that sounds natural and I guess that it is true. For bounded operator, this can be easily shown, but the only proof I know cannot be adapted to unbounded operator, so I was wondering if anybody here knows how to show this?

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The proof should be the same. For $\lambda < 0$, $$ ((T-\lambda I)x,x) \ge -\lambda (x,x)=|\lambda|\|x\|^{2},\;\;\; x\in\mathcal{D}(T). $$ Therefore, $$ |\lambda|\|x\|^{2} \le \|(T-\lambda I)x\|\|x\|,\\ |\lambda|\|x\| \le \|(T-\lambda I)x\|,\;\;\; x\in\mathcal{D}(T). $$ This implies that $\mathcal{N}(T-\lambda I)=\{0\}$ and $(T-\lambda I)^{-1}$ is bounded on its range. So the range of $T-\lambda I$ is closed because if $\{ (T-\lambda I)x_{n} \}$ converges to some $y$, then $(T-\lambda I)^{-1}(T-\lambda I)x_{n}=x_{n}$ converges to some $x$ and, because $T$ is closed, $x\in\mathcal{D}(T)$ with $(T-\lambda I)x=y$. Finally, $$ \mathcal{R}(T-\lambda I)=\mathcal{N}(T-\lambda I)^{\perp} = \{0\}^{\perp} = H. $$ Therefore $\lambda \in\rho(T)$. This is for all such $\lambda < 0$, which implies $\sigma(T)\subset[0,\infty)$.

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  • $\begingroup$ I don't see why the null space of the adjoint of the operator has to be trivial. Would you mind elaborating? $\endgroup$ – Jacob Denson Mar 20 '16 at 3:40
  • $\begingroup$ @JacobDenson : The assumption in the post is that $T^{\star}=T$. And I showed that $\mathcal{N}(T-\lambda I)=\{0\}$ with $\lambda$ real. So $(T-\lambda I)^{\star}=T^{\star}-\lambda I=T-\lambda I$ has $\{0\}$ null space. $\endgroup$ – DisintegratingByParts Mar 20 '16 at 5:09

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