I am currently reading a Paper and I don't quite get one part. The Paper is about Function of matrices and in one part he says:
"Since $g(z)$ is a holomorphic function on $\Omega$ and never vanishes, $g(A)$ must be invertible" (for some open $\Omega$ that contains the spectrum of $A$ and a linear bounded operator $A$)
I understand that since $g(z)$ never vanishes $h(z):=\frac{1}{g(z)}$ must also be holomorphic and therefor $h(A)$ exists. But how do I know that $h(A)$ is actually equal to $g(A)^{-1}$?
There is this theorem in spectral theory that says (if $g$ is continuous and defined on the spectrum of linear bounded $A$) $$ \sigma(g(A)) = g(\sigma(A)). $$ $\sigma$ of course denotes the spectrum. Since $g$ does not vanish, $0$ cannot be in the spectrum of $g(A)$. It follows that $g(A)$ is invertible.