showing $\inf \sigma (T) \leq \mu \leq \sup \sigma (T)$, where $\mu \in V(T)$ I am trying to prove the following:
Let $H$ be a Hilbert space, and $T\in B(H)$ be a self-adjoint operator. Then for all $\mu \in V(T)$,
$\inf \left\{\lambda: \lambda \in \sigma (T)  \right\}\leq \mu \leq \sup \left\{\lambda : \lambda \in \sigma (T)\right\}$.
Recall that $V(T)=\left\{(Tx,x):\|x \|=1\right\}$ (this is called the $\textbf{numerical range}$ of $T$), and that $\sigma (T)=\left\{ \lambda \in \mathbb{R}: (T-\lambda I) \, \text{is not invertible} \right\}$ (the $\textbf{spectrum}$ of $T$).
I want to try to prove the statement by contradiction because I see no other way to do this. Let $\alpha=\inf \left\{\lambda: \lambda \in \sigma (T)  \right\}$ and let $\beta = \sup \left\{\lambda : \lambda \in \sigma (T)\right\}$. Suppose $\alpha> \mu$, where $\mu = (Tx,x)$ and $\|x \|=1$. * Here is where I really have no idea what to do...I was playing around with the following:
If $\lambda \in \sigma (T)$, then $\mu < \lambda$. Now notice
$((T-\lambda I)x,x)=(Tx,x)-\lambda (x,x)$. Does this do anything for me?? I would greatly appreciate some help. Thanks!!

For selfadjoint operators we know that 
$$
\inf\sigma(T)=\inf\{\langle Tx,x\rangle:x\in S_H\}\\
\sup\sigma(T)=\sup\{\langle Tx,x\rangle:x\in S_H\}
$$
It is remains to apply result of this answer.
