Riesz's lemma

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Riesz's lemma (after Frigyes Riesz) is a lemma in functional analysis. It specifies (often easy to check) conditions that guarantee that a subspace in a normed vector space is dense. 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 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(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.^{[further explanation needed]}

The proof can be found in functional analysis texts such as Kreyszig. An 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 {x_n} with ${\displaystyle |x_{n}-x_{m}|>\alpha }$ for 0 < α < 1. This is useful in showing the non-existence of certain measures on infinite-dimensional Banach spaces. Riesz's lemma also shows that the identity operator on a Banach space X is compact if and only if X is finite-dimensional.^[2]

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 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

${\displaystyle d(x_{n},Y_{n-1})>\alpha }$ for a constant 0 < α < 1, where Y_n−1 is the linear span of {x₁ ... x_n−1} and ${\displaystyle d(x_{n},Y)=\inf _{y\in Y}|x_{n}-y|}$.

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

More generally, if a topological vector space X is locally compact, 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.^[3] 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, ..., c_n ∈ C such that

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

We claim that the finite dimensional subspace Y spanned by {c_i} 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,

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

for every m. But compact sets are bounded, so C lies in the closure of Y. This proves the result. For a different proof based on Hahn-Banach Theorem see.^[4]

See also[]

• F. Riesz's theorem

References[]

• Kreyszig, Erwin (1978), Introductory functional analysis with applications, John Wiley & Sons, ISBN 0-471-50731-8