At the moment I'm studying positive elements in $C^*$-algebras. Let $H$ be a complex Hilbert space, $L(H)$ the linear bounded operators $H\to H$ and $T^*=T$. Then:
$$\sigma(T)\subseteq [0,\|T\| ]\; \iff \; \langle Tx,x\rangle\ge 0\; \forall x\in H.$$
I have an idea how to prove $"\Leftarrow"$, but I have no idea how to prove $"\Rightarrow"$.
My try for $"\Leftarrow"$: T is self-adjoint, therefore it is $\sigma(T)\subseteq \overline{ \{\langle Tx,x\rangle\ge 0; \|x\|=1\}}$. $\Rightarrow \sigma(T)\subseteq \overline{ \{\langle Tx,x\rangle\ge 0; \|x\|=1\}}\subseteq [0,\infty)$. T self-adoint implies that $T$ is normal, therefore it is $\sup\limits_{\lambda\in\sigma(T)}|\lambda|=\|T\|$ and by compactness of $\sigma(T)$, it is $\sup\limits_{\lambda\in\sigma(T)}|\lambda|=\max\limits_{\lambda\in\sigma(T)}|\lambda|=\|T\|$. Then we have $\sigma(T)\subseteq [0,\|T\| ]$. Is it correct?
How to prove the other direction? I only have: $T$ is self-adjoint, i.e. $\langle Tx,x\rangle=\langle x,Tx\rangle =\overline{\langle Tx,x\rangle}\Rightarrow \langle x,Tx\rangle=\overline{\langle Tx,x\rangle}\Rightarrow \langle Tx,x\rangle\in \mathbb{R}$.
Edit: Now the other direction is clear too, see comments below. Regards