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Given a Hilbert space $\mathcal{H}$.

Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$

And its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int\lambda\mathrm{d}E(\lambda)$$

Then one has: $$\|T\|=\|\sigma(N)\|=r(E)$$

How can I prove this?

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The equality holds, including the infinity case. For normal bounded operators, since $\|T\|=\|T^*\|$ we verify that $$\|T\|^2=\sup_{\|\xi\|=1}\|T\xi\|^2=\sup_{\|\xi\|=1}\langle T\xi,T\xi\rangle =\sup_{\|\xi\|=1}\langle T^*T\xi,\xi\rangle =\|T^*T\|\leq\|T\|\|T^*\|=\|T\|^2,$$ and, $\|T^*T\|=\|T\|^2$. But we also have $$\|T^*T\xi\|^2=\langle T^*T\xi,T^*T\xi\rangle=\langle T^2\xi,T^2\xi\rangle=\|T^2\|^2,$$ hence $\|T\|^2=\|T^2\|$.

The spectral radius formula due to Gelfand states that $$r_\sigma(T):=\sup_{\lambda\in\sigma(T)}|\lambda|=\lim_{n\to\infty}\| T^n\|^{1/n}$$ But, $\|T^{2^n}\|=\|T\|^{2^n}$ and $$r_\sigma(T)=\lim_{n\to\infty}\|T^{2^n}\|^{1/2^n}=\|T\|,$$ which is the desired result. For the infinite case, consider restrictions of $T$ to bounded regions of the spectrum, and take the limit, to see that if $r_\sigma(T)=\infty$, $T$ must be unbounded.

Edit: If $r_\sigma(T)=\infty$, then $\forall M>0$, $\exists \lambda\in \sigma(T)$, such that $|\lambda|>M$. Take a sequence $(\lambda_n)\in \sigma(T)$, such that $|\lambda_n|>n$. Consider the sequence $T_{R_n}$ (as defined in the comment), with $R_n=|\lambda_n|+1.$ Then $T_{R_n}\to T$ strongly, but $\|T_{R_n}\|>n$, so $T$ must be unbounded.

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  • $\begingroup$ I'm having trouble verifying the argumentation for the infinite case. Can you specify in what sense the limit is taken precisely, please? $\endgroup$ – C-Star-W-Star Oct 16 '14 at 3:07
  • $\begingroup$ Consider the restriction $$T_R:=T\chi_{B(0,R)}(T)=\int_{B(0,R)}zdE(z)$$ It's not hard to see that each restriction $T_R$ is bounded with $\|T_R\|\leq R$. But for $x\in \mathcal{D}(T)$, $T_Rx\to Tx$, as $R\to \infty$, so use the strong limit. $\endgroup$ – Mateus Sampaio Oct 16 '14 at 12:13
  • $\begingroup$ But you will need that $\|T_R\|=R$ in order to conclude that $\|T_R\|\to\infty$... Can you - please - be more precise and put it also as answer? I'm sorry I haven't mentioned that the unbounded case is actually the part I'm struggling with... $\endgroup$ – C-Star-W-Star Oct 16 '14 at 13:38
  • $\begingroup$ Made the edit with a better explanation. See if it's clearer now. $\endgroup$ – Mateus Sampaio Oct 16 '14 at 14:08
  • $\begingroup$ I gave an answer myself - I hope you don't mind. $\endgroup$ – C-Star-W-Star Oct 21 '14 at 18:54
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For the spectrum: $$\sigma(N)=\operatorname{supp}E$$

But one has also: $$\|N\|<\infty\iff\|\mathrm{id}\|_\infty<\infty\iff\|\operatorname{supp}E\|<\infty$$

Concluding the assertion.

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