Riesz's lemma

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 resultEdit

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(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.

The proof can be found in functional analysis texts such as Kreyszig. An online proof from Prof. Paul Garrett is available.

Some consequencesEdit

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   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 dimensionEdit

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

  for a constant 0 < α < 1, where Yn−1 is the linear span of {x1 ... xn−1} and  .

Clearly {xn} 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, ..., cnC such that


We claim that the finite dimensional subspace Y spanned by {ci} 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 CY. Now, by induction,


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 alsoEdit


  1. ^ Rynne, Bryan P.; Youngson, Martin A. (2008). Linear Functional Analysis (2nd ed.). London: Springer. p. 47. ISBN 978-1848000049.
  2. ^ Kreyszig (1978, Theorem 2.5-3, 2.5-5)
  3. ^ https://terrytao.wordpress.com/2011/05/24/locally-compact-topological-vector-spaces/
  4. ^ https://www.emis.de/journals/PM/51f2/pm51f205.pdf/
  • Kreyszig, Erwin (1978), Introductory functional analysis with applications, John Wiley & Sons, ISBN 0-471-50731-8