# Riesz's lemma

Riesz' lemma (after Frigyes Riesz) is a lemma in functional analysis. It specifies (often easy to check) conditions which guarantee that a subspace in a normed linear space is 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

$d(x, Y) = \inf_{y \in Y} |x - y|.$

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 0 < α < 1. Then there exists an x in X with |x| = 1 such that |x − y| > α for all y in Y.[1]

### Note

For the finite dimensional case, equality can be achieved. In other words, there exists x of unit norm such that d(x, Y) is 1. When dimension of X is finite, the unit ball BX 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, and this proves the claim.

On the other hand, the example of the space $\ell^\infty$ of all bounded sequences shows that the lemma does not hold for $\alpha = 1$.

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## Converse

Riesz's lemma can be applied directly to show that the unit ball of an infinite-dimensional normed space X is never compact: Take an element x1 from the unit sphere. Pick xn from the unit sphere such that

$d(x_n, Y_{n-1}) > k$ for a constant 0 < k < 1, where Yn−1 is the linear span of {x1 ... xn−1}.

Clearly {xn} 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 is c1 ... cnC such that

$C = \bigcup_{i=1}^n \; \left( c_i + \frac{1}{2} C \right).$

We claim that the finite dimensional subspace Y spanned by {ci}, or equivalently, its closure, is X. Since scalar multiplication is continuous, its enough to show CY. Now, by induction,

$C \sub Y + \frac{1}{2^m} C$

for every m. But compact sets are bounded, so C lies in the closure of Y. This proves the result.

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## 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 $x_n$ with $|x_n - x_m| > k$ for 0 < k < 1. This is useful in showing the non-existence of certain measures on infinite-dimensional Banach spaces.

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

1. ^ Rynne, Bryan P.; Youngson, Martin A. (2008). Linear Functional Analysis (2nd ed.). London: Springer. p. 47. ISBN 978-1848000049.
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