A compact, bounded, closed-range operator on a Hilbert space has finite rank

[SiennaB]

Homework Statement

Let H be an \infty-dimensional Hilbert space and T<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\to{H} be an operator.

Show that if T is compact, bounded and has closed range, then T has finite rank. Do not use the open-mapping theorem.

Let B(H) denote the space of all bounded operators mapping H\to{H}, K(H) denote the space of all compact operators mapping H\to{H}, R(H) denote the space of all finite rank operators mapping H\to{H}.

2. Relevant defintions

*T\in{B(H)} is compact if the closure of T(B(0,1)) is a compact set.
*T\in{B(H)} has finite rank if Range(T)=T(H) is finite-dimensional.

The Attempt at a Solution

I'm not sure how to do the proof, but I think the following propositions in my lecture notes could be useful:

*T\in{R(H)} iff T\in{B(H)} is the norm limit of a sequence of finite rank operators, i.e. K(H) is the closure of R(H).
*Let T\in{R(H)}. Then there is an orthonormal set {{e_1,...,e_L}}\in{H} s.t. Tu=\sum\limits_{i,j=1}^{L}{c_{ij}(u,e_j)e_i} where c_{ij} are complex numbers.

Any help with the proof would be greatly appreciated.

Thank you in advance.

 

[micromass]

Do you know that compact operators can never be invertible on Hilbert spaces??

Given your compact operator T with closed range, can you modify the domain and codomain a bit such that T becomes a bijection between Hilbert spaces??

 

[SiennaB]

Do you know that compact operators can never be invertible on Hilbert spaces??

Given your compact operator T with closed range, can you modify the domain and codomain a bit such that T becomes a bijection between Hilbert spaces??

Thank you for your reply.

The next part of the question is to show that T is not invertible, so I can't use that fact.

T would be surjective if its image equalled its range, but the question defines these to be equal (under heading 2 in the original post), so T is already surjective. I'm not sure how to make T injective.