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In operator theory, 'Atkinson's theorem' gives a characterization of Fredholm operators.
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.
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''||·||''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.
★ F.V. Atkinson, The normal solvability of linear equations in normed spaces, ''Mat. Sb.'' '28' (70), 1951, 3-14.
| Contents |
| The theorem |
| Sketch of proof |
| Reference |
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''||·||''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.
Reference
★ F.V. Atkinson, The normal solvability of linear equations in normed spaces, ''Mat. Sb.'' '28' (70), 1951, 3-14.
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