# Spectrum of operator's exponent

1. May 7, 2008

### sfire

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 $$\lambda\notin Exp(\sigma(A))$$.

Last edited: May 7, 2008
2. May 12, 2008

No ideas...