Atkinson's theorem

From Wikipedia, the free encyclopedia

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

[edit] The theorem

Let H be a Hilbert space and L(H) the bounded operators on H. The following is the classical definition of a Fredholm operator: a T ∈ L(H) is said to be a Fredholm operator if the kernel of T Ker(T) is finite dimensional, Ker(T*) is finite dimensional, and the range of T Ran(T) is closed.

Atkinson's theorem states:

A T ∈ L(H) is a Fredholm operator if and only if T is invertible modulo compact perturbation, i.e. TS = I + C₁ and ST = I + C₂ for some bounded operator S and compact operators C₁ and C₂.

In other words, an operator T ∈ L(H) is Fredholm, in the classical sense, if and only if its projection in the Calkin algebra is invertible.

[edit] Sketch of proof

The outline of a proof is as follows. For the ⇒ implication, express H as the orthogonal direct sum

$H = \begin{matrix} \mbox{Ker}(T) ^{\perp} \\ \oplus \\ \mbox{Ker} (T) \end{matrix}.$

The restriction T : Ker(T)^⊥ → Ran(T) is a bijection, and therefore invertible by the open mapping theorem. Extend this inverse by 0 on Ran(T)^⊥ = Ker(T*) to an operator S defined on all of H. Then I - TS is the finite rank projection onto Ker(T*), and I - ST is the projection onto Ker(T). 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(T), then STx = x + C₂x = 0. So Ker(T) is contained in an eigenspace of C₂, which is finite dimensional (see spectral theory of compact operators). Therefore Ker(T) is also finite dimensional. The same argument shows that Ker(T*) is also finite dimensional.

To prove that Ran(T) 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(F),

||S||·||Tx|| ≥ ||STx|| = ||x + C₂x|| = ||x + Fx +C₂x - Fx|| ≥ ||x|| - ||C₂ - F||·||x|| ≥ (1 - r)||x||.

Thus T is bounded below on Ker(F), which implies that T(Ker(F)) is closed. On the other hand, T(Ker(F)^⊥) is finite dimensional, since Ker(F)^⊥ = Ran(F*) is finite dimensional. Therefore Ran(T) = T(Ker(F)) + T(Ker(F)^⊥) is closed, and this proves the theorem.