4
$\begingroup$

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!!

  • 0
    If $T=0$, then $W(T)=\{0\}$ so $M$ doesn't need to be positive.2012-04-26
  • 0
    @DavideGiraudo: $T = -I$ would show $M$ can be strictly negative.2012-04-26

2 Answers 2

4

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 $m=\inf\limits_{x\in B}w(x)$, $M=\sup\limits_{x\in B}w(x)$.

  • 0
    How do you show that $B$ is connected?2012-04-26
  • 2
    @DavideGiraudo For each $x_1,x_2\in H$, $\operatorname{span}\{x_1,x_2\}\cap B$ is a circle. Which is, of course, connected.2012-04-26
  • 0
    In fact, if I'm not mistaking, it can be shown that the unit sphere is arcwise connected, using, for normed $x,y$, the map $f\colon t\mapsto \frac{tx+(1-t)y}{\lVert tx+(1-t)y\rVert}$.2012-04-26
  • 0
    @DavideGiraudo, I think you are not mistaken :) I know your style - you always want to put as much formulas as possible ;)2012-04-26
  • 1
    @Davide: So if $y=-x$, what happens at $t=1/2$?2012-04-26
  • 0
    @NateEldredge $f$ is not well defined, and my argument is wrong. Thanks!2012-04-26
  • 0
    Thank you @Norbert ! But I'm not very clear about how you arguement that B is connected. Why is every circle connected? And why does that imply the connectedness of B? Sorry I'm quite rusty about these concepts... If it is complicated to state, a name of a book would also be very helpful!2012-04-27
  • 0
    @Vokram To prove that every circle is conected it is enough to show that every interval is connected, see theorem 2.47 in Rudin's Principles of Mathematical Analysis. As for the fact that continuous image of connected space is continuous, you can find its proof in theorem 4.22 in the book mentioned above.2012-04-27
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.