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Let A be a bounded operator in complex Hilbert space. Prove that \sigma(Exp(A))=Exp(\sigma(A)).
It is known, that \sigma(P(A))=P(\sigma(A)), where P is a polynomial.
In addition, if an operator A has a bounded inverse, then for any operator B such that ||B||<1/||A^{-1}|| their sum A+B has a bounded inverse.
I managed to prove that \sigma(Exp(A))\supseteq Exp(\sigma(A)).
As
Exp(A)=I+A+\frac{A^2}{2}+\ldots,
let
P_n(x)=1+x+\cdots+\frac{x^n}{n!}
and
Q_n(x)=\frac{x^{n+1}}{(n+1)!}+\ldots.
Then ||Q_n(A)||\to 0.
Let \lambda\notin\sigma(Exp(A)). Then Exp(A)-\lambda I is invertible, and P_n(A)-\lambda I=Exp(A)-\lambda I-Q_n(A) is invertible for large n.
Moreover, P_n(A)-(\lambda+\epsilon)I is invertible for sufficiently small epsilon. So \lambda\notin Exp(\sigma(A)).
It is known, that \sigma(P(A))=P(\sigma(A)), where P is a polynomial.
In addition, if an operator A has a bounded inverse, then for any operator B such that ||B||<1/||A^{-1}|| their sum A+B has a bounded inverse.
I managed to prove that \sigma(Exp(A))\supseteq Exp(\sigma(A)).
As
Exp(A)=I+A+\frac{A^2}{2}+\ldots,
let
P_n(x)=1+x+\cdots+\frac{x^n}{n!}
and
Q_n(x)=\frac{x^{n+1}}{(n+1)!}+\ldots.
Then ||Q_n(A)||\to 0.
Let \lambda\notin\sigma(Exp(A)). Then Exp(A)-\lambda I is invertible, and P_n(A)-\lambda I=Exp(A)-\lambda I-Q_n(A) is invertible for large n.
Moreover, P_n(A)-(\lambda+\epsilon)I is invertible for sufficiently small epsilon. So \lambda\notin Exp(\sigma(A)).
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