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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!!

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    @DavideGiraudo: $T = -I$ would show $M$ can be strictly negative.2012-04-26

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

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    @Vokram To prove that every c$i$rcle 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
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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.