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

• May 26th 2012, 01:54 AM
UniversityMathsStudent
Compact, bounded, closed-range operators on Hilbert spaces have finite rank
Hi,

I am trying to prove that if $T:H\to{H}$ is a compact, bounded operator with closed range, where $H$ is an $\infty$-dimensional Hilbert space, then $T$ has finite rank. I want to prove this without using the open-mapping theorem. $T$ is not necessarily linear.

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 $H\to{H}$.

The definitions I have in my lecture notes are:

* $T\in{B(H)}$ 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.

I'm not sure how to do the proof, but I think that 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.

$T': (\ker T)^\perp \rightarrow Range(T)$
$T'(x) = T(x)$
is a bijection between Hilbert spaces. Take a sequence $\{ T' x_i \} \in Range(T)$ of pairwise orthogonal elements such that $||T' x_i || =1$ . By the bounded inverse theorem, $||x_i|| \leq || T'^{-1}|| \ ||T'x||= || T'^{-1}||$, so since T' is compact $\{ T' x_i \}$ should have a convergent subsequence. This is a contradiction since $\{ T' x_i \}$ has no subsequence that is Cauchy.