Reflexive Space - Examples

Examples

Every finite-dimensional normed space is reflexive, simply because in this case, the space, its dual and bidual all have the same linear dimension, hence the linear injection J from the definition is bijective, by the rank-nullity theorem.

The Banach space c₀ of scalar sequences tending to 0 at infinity, equipped with the supremum norm, is not reflexive. It follows from the general properties below that ℓ1 and ℓ∞ are not reflexive, because ℓ1 is isomorphic to the dual of c₀, and ℓ∞ is isomorphic to the dual of ℓ1.

All Hilbert spaces are reflexive, as are the Lp spaces for 1 < p < ∞. More generally: all uniformly convex Banach spaces are reflexive according to the Milman–Pettis theorem. The L1(μ) and L∞(μ) spaces are not reflexive (unless they are finite dimensional, which happens for example when μ is a measure on a finite set). Likewise, the Banach space C of continuous functions on is not reflexive.

The spaces S_p(H) of operators in the Schatten class on a Hilbert space H are uniformly convex, hence reflexive, when 1 < p < ∞. When the dimension of H is infinite, then S₁(H) (the trace class) is not reflexive, because it contains a subspace isomorphic to ℓ1, and S_∞(H) = L(H) (the bounded operators) is not reflexive, because it contains a subspace isomorphic to ℓ∞ (in both cases, the subspace can be chosen to be the operators diagonal with respect to a given orthonormal basis of H).

Every finite-dimensional Hausdorff topological vector space is reflexive, because J is bijective by linear algebra, and because there is a unique Hausdorff vector space topology on a finite dimensional vector space.

Montel spaces are reflexive locally convex topological vector spaces.

Every semi-reflexive normed space is reflexive. A (somewhat artificial) example of a semi-reflexive space, not reflexive, is obtained as follows: let Y be an infinite dimensional reflexive Banach space, and let X be the topological vector space (Y, σ(Y, Y ′)), that is, the vector space Y equipped with the weak topology. Then the continuous dual of X is the set Y ′ and bounded subsets of X are norm-bounded, hence the Banach space Y ′ is the strong dual of X. Since Y is reflexive, the continuous dual of X ′ = Y ′ is equal to the image J(X) of X under the canonical embedding J, but the topology on X is not the strong topology β(X, X ′), that is equal to the norm topology of Y.