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

Hi,

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

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

The definitions I have in my lecture notes are:

*$\displaystyle T\in{B(H)}$ is compact if the closure of $\displaystyle T(B(0,1))$ is a compact set.

*$\displaystyle T\in{B(H)}$ has finite rank if $\displaystyle Range(T)=T(H)$ 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:

*$\displaystyle T\in{R(H)}$ iff $\displaystyle T\in{B(H)}$ is the norm limit of a sequence of finite rank operators, i.e. $\displaystyle K(H)$ is the closure of $\displaystyle R(H)$.

*Let $\displaystyle T\in{R(H)}$. Then there is an orthonormal set $\displaystyle \{e_1,...,e_L\}\in{H}$ s.t.

$\displaystyle Tu=\sum\limits_{i,j=1}^{L}{c_{ij}(u,e_j)e_i}$

where $\displaystyle c_{ij}$ 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

$\displaystyle T': (\ker T)^\perp \rightarrow Range(T)$

$\displaystyle T'(x) = T(x)$

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