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In operator theory, Atkinson's theorem (named for Frederick Valentine Atkinson) gives a characterization of Fredholm operators.

The theorem

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

Atkinson's theorem states:

A TL(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 TL(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

$$H = \operatorname{Ker}(T)^\perp \oplus \operatorname{Ker} (T).$$

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 ITS is the finite-rank projection onto Ker(T*), and IST 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 ||FC2|| < r. Then for every x in Ker(F),

||S||·||Tx|| ≥ ||STx|| = ||x + C2x|| = ||x + Fx +C2xFx|| ≥ ||x|| − ||C2F||·||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.

References

Atkinson, F. V. (1951). "The normal solvability of linear equations in normed spaces". Mat. Sb. 28 (70): 3–14. Zbl 0042.12001.