Integral of secant cubed: Difference between revisions

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In [[operator theory]], '''Atkinson's theorem''' (named for [[Frederick Valentine Atkinson]]) gives a characterization of [[Fredholm operator]]s.
 
== 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_(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.
 
'''Atkinson's theorem''' states:
 
:A ''T'' &isin; ''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>.
 
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
 
:<math> H =
\begin{matrix}
\operatorname{Ker}(T)^\perp \\
\oplus \\
\operatorname{Ker} (T).
\end{matrix}
</math>
 
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.
 
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.
 
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''),
 
:||''S''||&middot;||''Tx''|| &ge; ||''STx''|| = ||''x'' + ''C''<sub>2</sub>''x''|| = ||''x'' + ''Fx'' +''C''<sub>2</sub>''x'' − ''Fx''|| &ge; ||x|| − ||''C''<sub>2</sub> − ''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'')<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.
 
==References==
* {{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 }}
 
[[Category:Fredholm theory]]
[[Category:Theorems in functional analysis]]

Revision as of 19:34, 14 December 2013

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=Ker(T)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

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