If $A$ is a bounded nonnegative (i.e. $\langle Ax,x\rangle_H\ge0$ for all $x\in H$) linear operator on a $\mathbb R$-Hilbert space $H$, are we able to show that the spectrum $\sigma(A)$ of $A$ is contained in $[0,\infty)$?
I know that the claim is true for bounded nonnegative linear operators on complex Hilbert spaces, but I wonder whether this might fail to hold in the real case. The situation should be difference, since in the complex setting nonnegativity already implies self-adjointness. But even with the additional assumption of self-adjointness I'm ensure whether there are any issues occurring.
Best Answer
Yes. By scaling $A$, it suffices to show that $-1\not\in\sigma(A)$. So, suppose $-1\in\sigma(A)$, i.e. $A+I$ is not invertible. If $A+I$ does not have dense image, let $x$ be a nonzero vector orthogonal to its image and observe that $$\langle Ax,x\rangle=\langle (A+I)x,x\rangle-\langle x,x\rangle=-\langle x,x\rangle<0,$$ a contradiction. Thus $A+I$ must have dense image. Since it is not invertible, there must exist unit vectors $x$ such that $(A+I)x$ is arbitrarily small. In particular, choosing a unit vector $x$ such that $\|(A+I)x\|<1$, we again have $$\langle Ax,x\rangle=\langle (A+I)x,x\rangle-\langle x,x\rangle<0$$ giving a contradiction.