Riesz's lemma

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

1 The result
- 1.1 Note
2 Some consequences

[edit] 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 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 reads as follows:

Let X be a normed linear space and Y be a subspace in X. If there exists 0 < r < 1 such that for every x ∈ X with |x| =1, one has d(x, Y) < r, then Y is dense in X.

In other words, for every proper closed subspace Y, one can always find a vector x on the unit sphere of X such that d(x, Y) is less than and arbitrarily close to 1.

Proof: A simple proof can be sketched as follows. Suppose Y is not dense in X, therefore the closure of Y, denoted by Y' , is a proper subspace of X. Take an element x not in Y' , then we have

$\; d(x, Y') = d > 0$

So, for any k > 1, there exists y₀ in Y' such that

$d \leq | x - y_0 | < kd .$

Consider the vector z = x - y₀. We can calculate directly

$\left|\frac{z}{|z|} - \frac{y}{|z|} \right| = \frac{1}{|x - y_0|} |(x - y_0) - y| > \frac{1}{|x - y_0|} d > \frac{1}{k}$

for any y in Y. Choosing k arbitrarily close to 1 finishes the proof.

[edit] 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 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, and this proves the claim.

On the other hand, the example of the space l^∞ of all bounded sequences shows that the lemma does not hold for k = 1.

[edit] Some consequences

[edit] Noncompactness of unit ball

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 x_n from the unit sphere such that

d (x n, Y n - 1) > k

for a constant 0 < k < 1, where Y_n-1 is the linear span of {x₁ ... x_n-1}.

Clearly {x_n} contains no convergent subsequence and the noncompactness of the unit ball follows.

[edit] Spectral theory of compact operators

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.

[edit] Other

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.