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I need to prove the following statement: Let $\mathcal{A}$ be a unital $C^*-$Algebra, $A$ a self-adjoint element and $P$ a projection, so $P^2=P=P^*$.

Let $\delta :=\|P-A\|$. I want to prove that then: $\sigma(A)\subseteq [-\delta,\delta]\cup [1-\delta,1+\delta]$.

I want to show, that $(\lambda e_\mathcal{A}-A)$ is invertible for $\lambda$ outside the given set. Probably I need to use functional calculus but I'm not quite sure how to start. I know that since $P$ is a projection, $\sigma(P)\subseteq\{0,1\}$ and so $\|P\|=0$ or $\|P\|=1$.

Now to get $P$ involved,I thought to consider $$\|(\lambda e_\mathcal{A}-A-P+P)\|$$ and maybe use reverse triangle inequality but I don't see, how this could lead to the existence of $(\lambda e_\mathcal{A}-A)^{-1}$ defined via functional calculus.

Can someone help me?

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  • $\begingroup$ by definition of spectrum $A-\lambda e_{\mathcal A}$ is not invertible for any $\lambda\in\sigma(A)$. Hence it is surely invertible for any $\lambda\not\in[-\delta,\delta]\cup[1-\delta,1+\delta]$. $\endgroup$ – Phoenix87 Jan 7 '15 at 12:38
  • $\begingroup$ But I have to prove, that the given set is the spectrum. $\endgroup$ – TheoPhysicae Jan 7 '15 at 12:44
  • $\begingroup$ I edited the question, I hope, now it is clear, what shall be proven. $\endgroup$ – TheoPhysicae Jan 7 '15 at 12:58
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$$(\lambda e-A-P+P)=(\lambda e-P)(e-(\lambda e-P)^{-1}(A-P)).$$ Since $e-a$ is invertible for $\|a\|<1$ with $(e-a)^{-1}=\sum a^k$, it's enough to prove that $$\|(\lambda e-P)^{-1}(A-P)\|\le \delta\|(\lambda e-P)^{-1}\|<1.$$ By Gelfand-Neumark theorem, it's sufficient to consider $\mathcal{A}$ as a subalgebra of the space $L(H)$ of operators on some Hilbert space $H$. Then $H=KerP\oplus ImP$ and $(\lambda e-P)^{-1}=\lambda^{-1} e\oplus (\lambda-1)^{-1} e$. Thus $$\|(\lambda e-P)^{-1}\|\le max\{|\lambda^{-1}|,|(\lambda-1)^{-1}|\}<\delta^{-1},$$ when $\lambda$ is outside of the given set; q.e.d.

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    $\begingroup$ Observe that Gelfand-Naimark theorem can be avoided using functional calculus alone: $\Vert P-\lambda I\Vert=\sup_{\{0,1\}}|x - \lambda| = \max\{|\lambda|,|1-\lambda|\}$. $\endgroup$ – Phoenix87 Jan 7 '15 at 16:48

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