Let $H$ be a complex Hilbert space and $T$ be self-adjoint operator. Prove that $ \Vert T\Vert = \sup\{|\lambda|:\lambda \in W(T)\} $ We are supposed to use the following exercise and the fact that $r(T) = \Vert T\Vert$. How does this function?
A question on self adjoint operator and numerical range.
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functional-analysis
hilbert-spaces
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0I suggest you add these definitions to the original question, for the convenience of readers. – 2012-07-03
1 Answers
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1) For all $\lambda\in W(T)$ we have $\lambda=(Tx,x)$ for some $x\in H$ with $\Vert x\Vert=1$. Hence $ |\lambda|=|(Tx,x)|\leq\Vert Tx\Vert\Vert x\Vert=\Vert Tx\Vert\leq\Vert T\Vert\Vert x\Vert=\Vert T\Vert $ This implies $ \sup\{|\lambda|:\lambda\in W(T)\}\leq\Vert T\Vert\tag{1} $
2) From exercise you mentioned above it follows that $\sigma(T)\subset\overline{W(T)}$.
3) For all subsets $A\subset \mathbb{C}$ we have $\sup\{|\lambda|:\lambda\in A\}=\sup\{|\lambda|:\lambda\in \overline{A}\}$.
4) From paragraphs 2 and 3 we conclude $ \sup\{|\lambda|:\lambda\in W(T)\}=\sup\{|\lambda|:\lambda\in \overline{W(T)}\}\geq\sup\{|\lambda|:\lambda\in \sigma(T)\}=r(T)=\Vert T\Vert \tag{2} $
5) The result follows from inequalities $(1)$ and $(2)$