Integral of secant cubed - Revision history

2601:2:4D00:27B:6CB3:835:833:4E05: /* Reduction to an integral of a rational function */

2014-06-23T23:28:40Z

Reduction to an integral of a rational function

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75.118.162.24: Undid revision 586078092 by 67.230.144.79 (talk)

2014-02-09T16:53:27Z

Undid revision 586078092 by 67.230.144.79 (talk)

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	~~In [[operator theory]], '''Atkinson's theorem''' (named for [[Frederick Valentine Atkinson]]) gives a characterization of [[Fredholm operator]]s.~~		Hello and welcome. My name is Ling. Her family members life in Delaware but she needs to transfer because of her family members. Managing individuals is how I make cash and it's some thing I really enjoy. Playing crochet is some thing that I've done for many years.<br><br>Here is my site - [http://www.sceltic.at/index.php?mod=users&action=view&id=51301 extended car warranty]

	~~== 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'' ∈ ''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''\|\|·\|\|''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''\|\|~~.

	~~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]~~]

67.230.144.79: /* Reduction to an integral of a rational function */

2013-12-14T18:34:38Z

Reduction to an integral of a rational function

← Older revision		Revision as of 20:34, 14 December 2013
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	~~Hello buddy~~. Let ~~me introduce myself~~. ~~I am Ron but I don~~'t [~~http~~:~~//www~~.~~allstate~~.~~com~~/~~tools-~~and~~-resources~~/~~car-insurance~~/~~whats-in-a-new-car-warranty.aspx extended] car warranty like when people use my full name. I am a production~~ and ~~distribution officer~~. ~~Her husband~~ and ~~her selected to reside~~ in ~~Delaware but she requirements to move simply because~~ of ~~her family~~. ~~One of [http~~://~~eric~~.~~jandjsitebuilders.com~~/~~UserProfile/tabid/109/userId/106643/Default~~.~~aspx auto warranty~~] ~~the things~~ I ~~adore most~~ is ~~climbing and~~ now I ~~have extended car [http:~~//~~Blog~~.~~caranddriver~~.~~com~~/~~buying-peace-~~of~~-mind-how-to-buy-~~a-~~used-car-warranty~~/ ~~warranty time] to consider~~ on ~~new things~~.<br><br>~~Also visit my site: [http:~~//~~www~~.~~alexsantosdesign~~.~~com/index~~.~~php?mod~~=~~users&action~~=~~view&id~~=~~8657 auto warranty~~] [[~~http~~:~~//Viumallorca.com/UserProfile/tabid/1900/userId/20193/language/en-US/Default.aspx my sources~~]]		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'' ∈ ''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''\|\|·\|\|''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''\|\|.

			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]]

en>Michael Hardy: /* Reduction to an integral of a rational function */

2012-08-09T18:29:41Z

Reduction to an integral of a rational function

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