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.

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[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] 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(xn,Yn − 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 ... cn ∈ C such that

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

We claim that the finite dimenional subspace Y spanned by {ci}, or equivalently, its closure, is X. Since scalar multiplication is continuous, its enough to show C ⊂ Y. 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.

[edit] 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 | xn − xm | > k for 0 < k < 1. This is useful in showing the non existence of certain measures on infinite-dimensional Banach spaces.

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