In the following statements, $\mathcal{L}$ is a Banach algebra. $\sigma(M)$ is the spectrum of the operator, $M$. $\rho(M)$ is the resolvent set of $M$.

Here we have a definition and a theorem for functions of operators.

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How to solve the following exercise?

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My attempt is to use part (iv) of theorem 5. I define $f(z)=\ln(z), z\in \mathbb{C}$, and $g(z)=e^z$. But I don't know whether $f(z)$ is analytic on an open set containing $\sigma(M)$. How to use the condition that 0 can be connected to $\infty$ by curve that lies in $\rho(M)$? Or am I in the wrong direction?

  • $\begingroup$ Suppose the curve is simple, and that the complement $\Omega$ of the curve is a simply-connected region. Define a function $g(z)=\int_{z_0}^{z}\frac{1}{w}dw$, where the integral is along any path from a fixed $z_0\in \Omega$ to the given $z\in\Omega$, while remaining in the region which is the complement of the given curve. Then $e^{g(z)}=Cz$ where $C$ is a constant. Then the theorem applies. $\endgroup$ – DisintegratingByParts Oct 2 '17 at 21:09
  • $\begingroup$ @DisintegratingByParts Which part of the theorem applies? $\endgroup$ – Xianjin Yang Oct 3 '17 at 7:33
  • $\begingroup$ Part (iv) applies. $\endgroup$ – DisintegratingByParts Oct 3 '17 at 10:45
  • $\begingroup$ @DisintegratingByParts So, $g(z)$ is analytic on $\sigma(M)$? How can we see this from the conditions? $\endgroup$ – Xianjin Yang Oct 3 '17 at 12:17
  • $\begingroup$ The integral $g$ (mentioned above) gives a holomorphic function on the plane minus the curve; this is because the region is simply connected. $\endgroup$ – DisintegratingByParts Oct 3 '17 at 16:36

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