showing $\inf \sigma (T) \leq \mu \leq \sup \sigma (T)$, where $\mu \in V(T)$

Question

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!!

Answer 1

It is enough to show that $\alpha_{T}=\inf \{ (Tx,x) : \|x\|=1\}$ and $\beta_{T}=\sup\{(Tx,x) :\|x\|=1\}$ are in the spectrum of $T$.

Suppose $A \in \mathcal{B}(H)$ is selfadjoint with $(Ax,x) \ge 0$ for all $x \in H$. Equivalently $(Ax,x) \ge 0$ for all unit vectors $x$. Then $(x,y)_{A}$ defines a pseudo inner-product (i.e., is positive but maybe not positive definite.) So the Cauchy-Schwarz inequality holds for $(x,y)_{A}$, which gives $$ |(Ax,y)|^{2} \le (Ax,x)(Ay,y) \le (Ax,x)\|Ay\|\|y\|\le \|A\|(Ax,x)\|y\|^{2}. $$ Letting $y=Ax$ leads to the following (regardless of whether or not $x=0$): $$ \|Ax\|^{2} \le \|A\|(Ax,x). $$

By Assumption $((T-\alpha_{T}I)x,x) \ge 0$, and there exists a sequence of unique vectors $\{ x_{n} \}$ such that $((T-\alpha_{T}I)x_{n},x_{n})\rightarrow 0$. Hence, $$ \|(T-\alpha_{T}I)x_{n}\|^{2} \le \|T-\alpha_{T}I\|((T-\alpha_{T}I)x_{n},x_{n})\rightarrow 0. $$ The above implies that $\alpha_{T}\in\sigma(T)$ because $(T-\alpha_{T}I)$ cannot have a bounded inverse. Similarly, $\beta_{T}\in\sigma(T)$ is found to also hold by considering $(\beta_{T}I-T)$.

Answer 2

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.