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# Atkinson's theorem

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 *T* ∈ *L*(*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
*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

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

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 *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*||·||*Tx*|| ≥ ||*STx*|| = ||*x*+*C*_{2}*x*|| = ||*x*+*Fx*+*C*_{2}*x*−*Fx*|| ≥ ||x|| − ||*C*_{2}−*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.

References

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

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