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 [itex]T\to{H}[/itex] 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.