Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!!

share|cite|improve this question
up vote 1 down vote accepted

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 $(\beta_{T}I-T)$.

share|cite|improve this answer

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.

share|cite|improve this answer
Could you tell me what is $S_H$? Thanks – Sarah Apr 7 '14 at 22:59
$S_H=\{x\in H: \Vert x\Vert=1\}$ – Norbert Apr 7 '14 at 23:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.