# Riesz's lemma

{{ safesubst:#invoke:Unsubst||$N=Refimprove |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} 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

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

Now we can state the Lemma:

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]

Remark 1. For the finite-dimensional case, equality can be achieved. In other words, there exists x of unit norm such that d(xY) = 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.

## 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

${\displaystyle d(x_{n},Y_{n-1})>k}$ 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 are c1, ..., cnC such that

${\displaystyle C=\bigcup _{i=1}^{n}\;\left(c_{i}+{\frac {1}{2}}C\right).}$

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,

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

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

## 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 {xn} with ${\displaystyle |x_{n}-x_{m}|>k}$ 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).

## References

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