Thread: Compact, bounded, closed-range operators on Hilbert spaces have finite rank

Hi,

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.

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.