Spectrum of nonnegative operator Let $A$ be a bounded, nonnegative operator on a complex Hilbert space $H$. Prove that the spectrum $$\sigma(A)\subset[0,+\infty].$$
We say that an operator $A$ is nonnegative if it is self adjoint and 
$$ \langle Au,u\rangle \geq 0 \ \ \  \forall u \in H.$$
It is exercise 9.5 page 240 from https://www.math.ucdavis.edu/~hunter/book/ch9.pdf.
A more general question, there is a theorem that for a self-adjoint operator $$ \sigma(A)\subset\left[-\|A\|,\|A\|\,\right],$$
but is it true that 
$$ \sigma(A)\subset[\underset{\|u\|=1}{\inf}\langle Au,u\rangle,\underset{\|u\|=1}{\sup}\langle Au,u\rangle ]?$$
 A: So we want to show that if $A-\lambda I$ is not invertible, then $\lambda\geq0$. There are three ways in which $A-\lambda I$ may fail to be invertible:


*

*$\ker(A-\lambda I)\ne\{0\}$. In this case $\lambda $ is an eigenvalue. So there exists a unit vector $v\in H$ with $Av=\lambda v$. Then $$\lambda=\langle\lambda v,v\rangle=\langle Av,v\rangle\geq0.$$

*$\ker(A-\lambda I)=\{0\}$, but $A-\lambda I$ is not bounded below. In this case, there exists a sequence $v_n$ of unit vectors with $(A-\lambda I)v_n\to0$. Then $\lambda=\langle\lambda v_n,v_n\rangle$, so $$\lambda=\lim\langle Av_n,v_n\rangle\geq0.$$

*$A-\lambda I$ is bounded below, but not surjective. This case has nothing to do with the sign of $\lambda$; it is just the fact that the residual spectrum of a selfadjoint operator is empty. As the image of a bounded below operator is closed, we get $$\ker (A-\lambda I)=\ker(A^*-\bar\lambda I)=\text{ran}\,(A-\lambda I)^\perp\ne\{0\};$$ this is a contradiction, since a bounded-below operator is injective. So no such $\lambda  $ exists.  


As for your last question, yes. For a selfadjoint operator (normal, actually), the  convex hull of the spectrum is equal to the closure of the numerical range:
$$
{\text{conv}}\,\sigma(A)=\overline{\{\langle Av,v\rangle:\ \|v\|=1\}.}
$$ 
