Compact, bounded, closed-range operators on Hilbert spaces have finite rank
I am trying to prove that if is a compact, bounded operator with closed range, where is an -dimensional Hilbert space, then has finite rank. I want to prove this without using the open-mapping theorem. is not necessarily linear.
Let denote the space of all bounded operators mapping , denote the space of all compact operators mapping , denote the space of all finite rank operators mapping .
The definitions I have in my lecture notes are:
* is compact if the closure of is a compact set.
* has finite rank if is finite-dimensional.
I'm not sure how to do the proof, but I think that the following propositions in my lecture notes could be useful:
* iff is the norm limit of a sequence of finite rank operators, i.e. is the closure of .
*Let . Then there is an orthonormal set s.t.
where are complex numbers.
Any help with the proof would be greatly appreciated.
Thank you in advance.
Re: Compact, bounded, closed-range operators on Hilbert spaces have finite rank
Suppose Range(T) is infinite dimensional. Since Range(T) is closed, the map
is a bijection between Hilbert spaces. Take a sequence of pairwise orthogonal elements such that . By the bounded inverse theorem, , so since T' is compact should have a convergent subsequence. This is a contradiction since has no subsequence that is Cauchy.