# Positive operator has a positive spectrum?

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?

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 $\sigma(T)\subset[0,\infty)$.
• @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. – DisintegratingByParts Mar 20 '16 at 5:09