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Riesz's lemma (after
Frigyes Riesz Frigyes Riesz ( hu, Riesz Frigyes, , sometimes spelled as Frederic; 22 January 1880 – 28 February 1956) was a HungarianEberhard Zeidler: Nonlinear Functional Analysis and Its Applications: Linear monotone operators. Springer, 199/ref> mathematic ...
) is a
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in
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. It specifies (often easy to check) conditions that guarantee that a subspace in a
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is
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. The lemma may also be called the Riesz lemma or Riesz inequality. It can be seen as a substitute for orthogonality when one is not in an inner product space.

# The result

Riesz's Lemma. Let ''X'' be a normed space, ''Y'' be a closed proper subspace of ''X'' and α be a real number with Then there exists an ''x'' in ''X'' with , ''x'', = 1 such that , ''x'' − ''y'',  ≥ α for all ''y'' in ''Y''.
''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 of all bounded sequences shows that the lemma does not hold for α = 1. The proof can be found in functional analysis texts such as Kreyszig. A
online proof from Prof. Paul Garrett
is available.

# Some consequences

The 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 with $, x_n - x_m, > \alpha$ for 0 < ''α'' < 1. This is useful in showing the non-existence of certain measures on infinite-dimensional
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s. Riesz's lemma also shows that the identity operator on a Banach space ''X'' is compact if and only if ''X'' is finite-dimensional. One can also use this lemma to characterize finite dimensional normed spaces: if X is a normed vector space, then X is finite dimensional if and only if the closed unit ball in X is compact.

## Characterization of finite dimension

Riesz's lemma can be applied directly to show that the
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of an infinite-dimensional normed space ''X'' is never
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: Take an element ''x''1 from the unit sphere. Pick ''xn'' from the unit sphere such that :$d\left(x_n, Y_\right) > \alpha$ for a constant 0 < ''α'' < 1, where ''Y''''n''−1 is the linear span of and $d\left(x_n, Y\right) = \inf_ , x_n - y,$. Clearly contains no convergent subsequence and the noncompactness of the unit ball follows. More generally, if a
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''X'' is
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, then it is finite dimensional. The converse of this is also true. Namely, if a topological vector space is finite dimensional, it is locally compact. 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''1, ..., ''cn'' ∈ ''C'' such that :$C \sub \bigcup_^n \; \left\left( c_i + \frac C \right\right).$ We claim that the finite dimensional subspace ''Y'' spanned by is dense in ''X'', or equivalently, its closure is ''X''. Since ''X'' is the union of scalar multiples of ''C'', it is sufficient to show that ''C'' ⊂ ''Y''. Now, by induction, :$C \sub Y + \frac C$ for every ''m''. But compact sets are bounded, so ''C'' lies in the closure of ''Y''. This proves the result. For a different proof based on Hahn-Banach Theorem see.https://www.emis.de/journals/PM/51f2/pm51f205.pdf