1. ## spectrum

Let $\mathcal{A}$ be the space of all bounded analytic functions in the open unit disk in the complex plane.

My question is why is die spectrum the closure of the range of $f\in\mathcal{A}$ and not just the normal range?

$\sigma(f):=\{\lambda\in \mathbb{C}: (\lambda 1 -f) \notin Inv(\mathcal{A}) \}$

or should I look to the resolvant to find my answer?

2. Originally Posted by Mauritzvdworm
Let $\mathcal{A}$ be the space of all bounded analytic functions in the open unit disk in the complex plane.

My question is why is die spectrum the closure of the range of $f\in\mathcal{A}$ and not just the normal range?

$\sigma(f):=\{\lambda\in \mathbb{C}: (\lambda 1 -f) \notin Inv(\mathcal{A}) \}$

or should I look to the resolvant to find my answer?
One way to answer this is to say that the spectrum is a closed (in fact, compact) set. The range of f on the open disk will not be closed.

Another way to look at it is to notice that if $\lambda$ is in the closure of the range of f, but not in the range itself, then $\lambda 1 -f$ will have an inverse function defined on the disk, but the inverse will not be bounded. So $\lambda 1 -f$ will not have an inverse in $\mathcal{A}$.