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sfire

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**Let A be a bounded operator in complex Hilbert space. Prove that [tex]\sigma(Exp(A))=Exp(\sigma(A))[/tex].**

It is known, that [tex]\sigma(P(A))=P(\sigma(A))[/tex], where P is a polynomial.

In addition, if an operator A has a bounded inverse, then for any operator B such that [tex]||B||<1/||A^{-1}||[/tex] their sum A+B has a bounded inverse.

I managed to prove that [tex]\sigma(Exp(A))\supseteq Exp(\sigma(A))[/tex].

As

[tex]Exp(A)=I+A+\frac{A^2}{2}+\ldots,[/tex]

let

[tex]P_n(x)=1+x+\cdots+\frac{x^n}{n!}[/tex]

and

[tex]Q_n(x)=\frac{x^{n+1}}{(n+1)!}+\ldots.[/tex]

Then [tex]||Q_n(A)||\to 0[/tex].

Let [tex]\lambda\notin\sigma(Exp(A))[/tex]. Then [tex]Exp(A)-\lambda I[/tex] is invertible, and [tex]P_n(A)-\lambda I=Exp(A)-\lambda I-Q_n(A)[/tex] is invertible for large n.

Moreover, [tex]P_n(A)-(\lambda+\epsilon)I[/tex] is invertible for sufficiently small epsilon. So [tex]\lambda\notin Exp(\sigma(A))[/tex].

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