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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]].
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| == The result ==
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| 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
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| : <math>d(x, Y) = \inf_{y \in Y} |x - y|.</math> | |
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| Now we can state the Lemma:
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| <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>
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| ''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.
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| ''Remark 2.'' The space ℓ<sub>∞</sub> of all bounded sequences shows that the lemma does not hold for α = 1.
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| == Converse ==
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| 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
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| :<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>}.
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| Clearly {''x''<sub>''n''</sub>} contains no convergent subsequence and the noncompactness of the unit ball follows.
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| 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
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| :<math>C = \bigcup_{i=1}^n \; \left( c_i + \frac{1}{2} C \right).</math>
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| 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,
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| :<math>C \sub Y + \frac{1}{2^m} C</math>
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| 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.
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| == Some consequences ==
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| 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.
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| 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.
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| 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).
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| ==References==
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| {{Reflist}}
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| [[Category:Functional analysis]]
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| [[Category:Lemmas]]
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