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| In [[operator theory]], '''Atkinson's theorem''' (named for [[Frederick Valentine Atkinson]]) gives a characterization of [[Fredholm operator]]s.
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| == The theorem ==
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| 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_(linear_operator)|kernel]] Ker(''T'') is finite dimensional, Ker(''T*'') is finite dimensional (where ''T*'' denotes the [[Hermitian_adjoint|adjoint]] of ''T''), and the [[Range_(mathematics)|range]] Ran(''T'') is closed.
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| '''Atkinson's theorem''' states:
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| :A ''T'' ∈ ''L''(''H'') is a Fredholm operator if and only if ''T'' is invertible modulo compact perturbation, i.e. ''TS'' = ''I'' + ''C''<sub>1</sub> and ''ST'' = ''I'' + ''C''<sub>2</sub> for some bounded operator ''S'' and [[compact operator]]s ''C''<sub>1</sub> and ''C''<sub>2</sub>.
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| 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.
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| === Sketch of proof ===
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| The outline of a proof is as follows. For the ⇒ implication, express ''H'' as the orthogonal direct sum
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| :<math> H =
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| \begin{matrix}
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| \operatorname{Ker}(T)^\perp \\
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| \oplus \\
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| \operatorname{Ker} (T).
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| \end{matrix}
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| </math>
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| The restriction ''T'' : Ker(''T'')<sup>⊥</sup> → Ran(''T'') is a bijection, and therefore invertible by the [[open mapping theorem (functional analysis)|open mapping theorem]]. Extend this inverse by 0 on Ran(''T'')<sup>⊥</sup> = Ker(''T*'') to an operator ''S'' defined on all of ''H''. Then ''I'' − ''TS'' is the [[finite-rank operator|finite-rank]] projection onto Ker(''T*''), and ''I'' − ''ST'' is the projection onto Ker(''T''). This proves the only if part of the theorem.
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| For the converse, suppose now that ''ST'' = ''I'' + ''C''<sub>2</sub> for some compact operator ''C''<sub>2</sub>. If ''x'' ∈ Ker(''T''), then ''STx'' = ''x'' + ''C''<sub>2</sub>''x'' = 0. So Ker(''T'') is contained in an eigenspace of ''C''<sub>2</sub>, 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.
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| 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''<sub>2</sub>|| < ''r''. Then for every ''x'' in Ker(''F''),
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| :||''S''||·||''Tx''|| ≥ ||''STx''|| = ||''x'' + ''C''<sub>2</sub>''x''|| = ||''x'' + ''Fx'' +''C''<sub>2</sub>''x'' − ''Fx''|| ≥ ||x|| − ||''C''<sub>2</sub> − ''F''||·||x|| ≥ (1 − ''r'')||''x''||.
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| Thus ''T'' is bounded below on Ker(''F''), which implies that ''T''(Ker(''F'')) is closed. On the other hand, ''T''(Ker(''F'')<sup>⊥</sup>) is finite dimensional, since Ker(''F'')<sup>⊥</sup> = Ran(''F*'') is finite dimensional. Therefore Ran(''T'') = ''T''(Ker(''F'')) + ''T''(Ker(''F'')<sup>⊥</sup>) is closed, and this proves the theorem.
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| ==References==
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| * {{cite journal |first=F. V. |last=Atkinson |title=The normal solvability of linear equations in normed spaces |journal=Mat. Sb. |volume=28 |issue=70 |year=1951 |pages=3–14 |zbl=0042.12001 }}
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| [[Category:Fredholm theory]]
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| [[Category:Theorems in functional analysis]]
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