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"("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" + "C"₁ and "ST" = "I" + "C"₂ for some bounded operator "S" and compact operators "C"₁ and "C"₂.

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.

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("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" + "C"₂ for some compact operator "C"₂. If "x" ∈ Ker("T"), then "STx" = "x" + "C"₂"x" = 0. So Ker("T") is contained in an eigenspace of "C"₂, 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" - "C"₂|| < "r". Then for every "x" in Ker("F"),

:||"S"||&middot;||"Tx"|| &ge; ||"STx"|| = ||"x" + "C"₂"x"|| = ||"x" + "Fx" +"C"₂"x" - "Fx"|| &ge; ||x|| - ||"C"₂ - "F"||&middot;||x|| &ge; (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.

References

*F.V. Atkinson, The normal solvability of linear equations in normed spaces, "Mat. Sb." 28 (70), 1951, 3-14.