Atkinson's theorem


In operator theory, Atkinson's theorem gives a characterization of Fredholm operators.

The theorem

Let H be a Hilbert space and L the set of bounded operators on H. The following is the classical definition of a Fredholm operator: an operator T ∈ L is said to be a Fredholm operator if the kernel Ker is finite-dimensional, Ker is finite-dimensional, and the range Ran is closed.
Atkinson's theorem states:
In other words, an operator T ∈ L is Fredholm, in the classical sense, if and only if its projection in the Calkin algebra is invertible.

Sketch of proof

The outline of a proof is as follows. For the ⇒ implication, express H as the orthogonal direct sum
The restriction T : Ker⊥ → Ran is a bijection, and therefore invertible by the open mapping theorem. Extend this inverse by 0 on Ran⊥ = Ker to an operator S defined on all of H. Then I − TS is the finite-rank projection onto Ker, and I − ST is the projection onto Ker. This proves the only if part of the theorem.
For the converse, suppose now that ST = I + C2 for some compact operator C2. If x ∈ Ker, then STx = x + C2x = 0. So Ker is contained in an eigenspace of C2, which is finite-dimensional. Therefore Ker is also finite-dimensional. The same argument shows that Ker is also finite-dimensional.
To prove that Ran is closed, we make use of the approximation property: let F be a finite-rank operator such that ||F − C2|| < r. Then for every x in Ker,
Thus T is bounded below on Ker, which implies that T is closed. On the other hand, T is finite-dimensional, since Ker⊥ = Ran is finite-dimensional. Therefore Ran = T + T is closed, and this proves the theorem.
A more complete treatment of Atkinson's Theorem is in the reference by Arveson: it shows that if B is a Banach space, an operator is Fredholm iff it is invertible modulo a finite rank operator. For Banach spaces, a Fredholm operator is one with finite dimensional kernel and range of finite codimension. Note that the hypothesis that Ran is closed is redundant since a space of finite codimension that is also the range of a bounded operator is always closed ; this is a consequence of the open-mapping theorem.