Positive operator has a positive spectrum?

Solution 1:

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 $\sigma(T)\subset[0,\infty)$.