Let 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 and not just the normal range?
or should I look to the resolvant to find my answer?
Let 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 and not just the normal range?
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 is in the closure of the range of f, but not in the range itself, then will have an inverse function defined on the disk, but the inverse will not be bounded. So will not have an inverse in .