Schmidt decomposition - Revision history

139.165.120.70: /* Proof */

2015-01-07T08:31:13Z

Proof

← Older revision		Revision as of 10:31, 7 January 2015
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	~~'''Riesz' lemma''' (after [[Frigyes Riesz]])~~ is ~~a [[lemma (mathematics)\|lemma]] in [[functional analysis]]. It specifies (often easy to check) conditions which guarantee that a [[Linear subspace\|subspace]] in a [[normed linear space]]~~ is ~~[[dense set\|dense]].~~

	~~== The result ==~~
	~~Before stating~~ the ~~result, we fix some notation~~. ~~Let ''X'' be a normed linear space with norm \|·\| and ''x'' be an element~~ of ~~''X''. Let ''Y'' be a closed subspace in ''X''. The distance between an element ''x'' and ''Y'' is defined by~~

	: ~~<math>d(x, Y) = \inf_{y \in Y} \|x - y\|.<~~/~~math>~~

	~~Now we can state the Lemma:~~

	<blockquote>'''Riesz's Lemma.''' Let ''X'' be a normed linear space, ''Y'' be a closed proper subspace of ''X'' and α be a real number with {{nowrap\|0 < α < 1.}} Then there exists an ''x'' in ''X'' with \|''x''\| = 1 such that \|''x'' − ''y''\| > α for all ''y'' in ''Y''.<ref>{{cite book\|last=Rynne\|first=Bryan P.\|title=Linear Functional Analysis\|year=2008\|publisher=Springer\|location=London\|isbn=978-1848000049\|edition=2nd\|coauthors=Youngson, Martin A.~~\|page=47}}<~~/~~ref><~~/~~blockquote>~~

	''Remark 1.'' For the finite-dimensional case, equality can be achieved. In other words, there exists ''x'' of unit norm such that ''d''(''x'', ''Y'') = 1. When dimension of ''X'' is finite, the unit ball ''B'' ⊂ ''X'' is compact. Also, the distance function ''d''(· , ''Y'') is continuous. Therefore its image on the unit ball ''B'' must be a compact subset of the real line, proving the claim.

	~~''Remark 2.'' The space ℓ<sub>∞<~~/~~sub> of all bounded sequences shows that~~ the ~~lemma does not hold~~ for~~ α = 1.~~

	~~== Converse ==~~
	~~Riesz's lemma can be applied directly to show that the [[unit ball]] of an infinite-dimensional normed space ''X''~~ is ~~never~~ [~~[compact set\|compact]]~~: ~~Take an element ''x''<sub>1<~~/~~sub> from the unit sphere. Pick ''x<sub>n<~~/~~sub>'' from the unit sphere such that~~

	~~:<math>d(x_n, Y_{n~~-~~1}) > k </math> for a constant 0 < ''k'' < 1, where ''Y''<sub>''n''−1</sub> is the linear span of {''x''<sub>1</sub>~~ ... ~~''x''<sub>''n''−1<~~/~~sub>}~~.

	~~Clearly {''x''<sub>''n''</sub>} contains no convergent subsequence and the noncompactness of the unit ball follows~~.

	~~The converse of this, in a more general setting, is also true~~. ~~If a [[topological vector space]~~] '~~'X'' is [[locally compact]], then~~ it ~~is finite dimensional~~. ~~Therefore local compactness characterizes finite-dimensionality~~. ~~This classical result is also attributed to Riesz~~. ~~A short proof can be sketched as follows: let ''C'' be a compact neighborhood of 0 ∈ ''X''. By compactness, there are ''c''~~<~~sub~~>1</~~sub>,~~ .~~.., ''c<sub>n<~~/~~sub>'' ∈ ''C'' such that~~

	~~:<math>C = \bigcup_{i=1}^n \; \left( c_i + \frac{1}{2} C \right).<~~/~~math>~~

	~~We claim that the finite dimensional subspace ''Y'' spanned by {''c<sub>i<~~/~~sub>''}, or equivalently, its closure, is ''X''. Since scalar multiplication is continuous, its enough to show ''C'' ⊂ ''Y''. Now, by induction,~~

	~~:<math>C \sub Y + \frac{1}{2^m} C<~~/~~math>~~

	~~for every ''m''. But compact sets are [[Bounded set (topological vector space)\|bounded]], so ''C'' lies in the closure of ''Y''. This proves the result.~~

	~~== Some consequences ==~~
	~~The [[spectral theory of compact operators\|spectral properties of compact operators]] acting on a Banach space are similar to those of matrices. Riesz's lemma is essential in establishing this fact.~~

	~~Riesz's lemma guarantees~~ that ~~any infinite~~-~~dimensional normed space contains~~ a ~~sequence of unit vectors {''x<sub>n</sub>''} with <math>\|x_n~~ - ~~x_m\| > k<~~/~~math> for 0 < ''k'' < 1. This is useful in showing the non-existence of certain [[measure (mathematics)\|measures]] on infinite-dimensional [[Banach space]]s.~~

	One can also use this lemma to demonstrate whether or not the normed vector space X is finite dimensional or otherwise: if the closed unit ball is compact, the X is finite dimensional ( proof by contradiction).

	~~==References==~~

	~~{{Reflist}}~~

	~~[[Category:Functional analysis]]~~
	~~[[Category:Lemmas]~~]

129.7.128.30: /* Schmidt rank and entanglement */ Changed "can not" to "cannot"

2013-06-05T21:02:00Z

Schmidt rank and entanglement: Changed "can not" to "cannot"

← Older revision		Revision as of 23:02, 5 June 2013
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			'''Riesz' lemma''' (after [[Frigyes Riesz]]) is a [[lemma (mathematics)\|lemma]] in [[functional analysis]]. It specifies (often easy to check) conditions which guarantee that a [[Linear subspace\|subspace]] in a [[normed linear space]] is [[dense set\|dense]].

			== The result ==
			Before stating the result, we fix some notation. Let ''X'' be a normed linear space with norm \|·\| and ''x'' be an element of ''X''. Let ''Y'' be a closed subspace in ''X''. The distance between an element ''x'' and ''Y'' is defined by

			: <math>d(x, Y) = \inf_{y \in Y} \|x - y\|.</math>

			Now we can state the Lemma:

			<blockquote>'''Riesz's Lemma.''' Let ''X'' be a normed linear space, ''Y'' be a closed proper subspace of ''X'' and α be a real number with {{nowrap\|0 < α < 1.}} Then there exists an ''x'' in ''X'' with \|''x''\| = 1 such that \|''x'' − ''y''\| > α for all ''y'' in ''Y''.<ref>{{cite book\|last=Rynne\|first=Bryan P.\|title=Linear Functional Analysis\|year=2008\|publisher=Springer\|location=London\|isbn=978-1848000049\|edition=2nd\|coauthors=Youngson, Martin A.\|page=47}}</ref></blockquote>

			''Remark 1.'' For the finite-dimensional case, equality can be achieved. In other words, there exists ''x'' of unit norm such that ''d''(''x'', ''Y'') = 1. When dimension of ''X'' is finite, the unit ball ''B'' ⊂ ''X'' is compact. Also, the distance function ''d''(· , ''Y'') is continuous. Therefore its image on the unit ball ''B'' must be a compact subset of the real line, proving the claim.

			''Remark 2.'' The space ℓ<sub>∞</sub> of all bounded sequences shows that the lemma does not hold for α = 1.

			== Converse ==
			Riesz's lemma can be applied directly to show that the [[unit ball]] of an infinite-dimensional normed space ''X'' is never [[compact set\|compact]]: Take an element ''x''<sub>1</sub> from the unit sphere. Pick ''x<sub>n</sub>'' from the unit sphere such that

			:<math>d(x_n, Y_{n-1}) > k </math> for a constant 0 < ''k'' < 1, where ''Y''<sub>''n''−1</sub> is the linear span of {''x''<sub>1</sub> ... ''x''<sub>''n''−1</sub>}.

			Clearly {''x''<sub>''n''</sub>} contains no convergent subsequence and the noncompactness of the unit ball follows.

			The converse of this, in a more general setting, is also true. If a [[topological vector space]] ''X'' is [[locally compact]], then it is finite dimensional. Therefore local compactness characterizes finite-dimensionality. This classical result is also attributed to Riesz. A short proof can be sketched as follows: let ''C'' be a compact neighborhood of 0 ∈ ''X''. By compactness, there are ''c''<sub>1</sub>, ..., ''c<sub>n</sub>'' ∈ ''C'' such that

			:<math>C = \bigcup_{i=1}^n \; \left( c_i + \frac{1}{2} C \right).</math>

			We claim that the finite dimensional subspace ''Y'' spanned by {''c<sub>i</sub>''}, or equivalently, its closure, is ''X''. Since scalar multiplication is continuous, its enough to show ''C'' ⊂ ''Y''. Now, by induction,

			:<math>C \sub Y + \frac{1}{2^m} C</math>

			for every ''m''. But compact sets are [[Bounded set (topological vector space)\|bounded]], so ''C'' lies in the closure of ''Y''. This proves the result.

			== Some consequences ==
			The [[spectral theory of compact operators\|spectral properties of compact operators]] acting on a Banach space are similar to those of matrices. Riesz's lemma is essential in establishing this fact.

			Riesz's lemma guarantees that any infinite-dimensional normed space contains a sequence of unit vectors {''x<sub>n</sub>''} with <math>\|x_n - x_m\| > k</math> for 0 < ''k'' < 1. This is useful in showing the non-existence of certain [[measure (mathematics)\|measures]] on infinite-dimensional [[Banach space]]s.

			One can also use this lemma to demonstrate whether or not the normed vector space X is finite dimensional or otherwise: if the closed unit ball is compact, the X is finite dimensional ( proof by contradiction).

			==References==

			{{Reflist}}

			[[Category:Functional analysis]]
			[[Category:Lemmas]]

en>Li3939108: /* Proof */ add a comma

2012-07-21T10:08:52Z

Proof: add a comma

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