We know that if $A$ is a self-adjoint unbounded operator on a Hilbert space $(H;\left<.,.\right>)$ then $\sigma(A) \subset \mathbb R$. Now, how it can be shown that if $A$ is more positive i.e. $\left<Au,u\right> \geq 0$, then $\sigma(A) \subset \mathbb [0,+\infty[$ ?
Thank you in advance
First notice that $\overline{\mathrm{ran}(A-\lambda)}=\ker(A-\lambda)^\perp$ for $\lambda\in\mathbb{R}$. Thus, it suffices to show that $\ker(A-\lambda)=\{0\}$ and that $\mathrm{ran}(A-\lambda)$ is closed.
Let $\lambda<0$. We have $$ \|(A-\lambda)u\|\|u\|\geq\langle (A-\lambda) u,u\rangle\geq -\lambda\|u\|^2 $$ for all $u\in D(A)$.
Thus, $\ker(A-\lambda)=0$ and $A-\lambda$ is invertible as operator from $D(A)$ to $\mathrm{ran}(A-\lambda)$. It follows that $\mathrm{ran}(A-\lambda)$ is complete, hence closed.