Why is the numerical range of a self-adjoint operator an interval?

Question

I was reviewing for a test for functional analysis when I came across the following statement:

Let $T$ be a bounded self-adjoint operator on a Hilbert space $H$. Then the numerical range of it is an interval $[m, M]$ with $M>0$.

Is the above statement correct? How can I prove it?

Thank you!!

Answer 1

For self adjoint bounded operator $T\in\mathcal{B}(H)$ we have well defined continuous function $$ w:H\to\mathbb{R}:x\mapsto\langle Tx,x\rangle $$ Denote $B=\{x\in H:\Vert x\Vert=1\}$, which obviously connected and bounded. Since $T$ is bounded then $W(T)=w(B)$ is bounded. Since $w$ is continuous then $W(T)$ is connected as contiuous image of connected space $B$. Thus $W(T)$ is bounded and connected, then it is of the form $(m,M)$ or $[m,M)$ or $(m,M]$ where $m=\inf\limits_{x\in B}w(x)$, $M=\sup\limits_{x\in B}w(x)$.

Answer 2

Denote $W(T):=\{\langle Tx,x\rangle,x\in H,\lVert x\rVert=1\}$

• First, since $T$ is self-adjoint and bounded, $W(T)$ is a bounded subset of the real line.
• Let $t_1,t_2\in W(T)$, we have to show that for each $0<\lambda<1$, $\lambda t_1+(1-\lambda)t_2\in W(T)$. We have $t_1=\langle Tx_1,x_1\rangle$ and $t_2=\langle Tx_2,x_2\rangle$ for some normed $x_1,x_2\in H$. Let $F:=\operatorname{Span}\{f_1,f_2\}$, then $F$ is a closed subspace. Denote $P_F$ the projection over $F$. For $x\in F$, we have $$\langle Tx,x\rangle=\langle Tx,Px\rangle=\langle Tx,P^*x\rangle=\langle PTx,x\rangle,$$ so it's enough to show the result when $T$ is a $2\times 2$ complex matrix.