Suppose X is a
normed vector space over the number field \mathbb{F} = \R or \mathbb{F} = \Complex (the
real numbers or the
complex numbers), with a norm \|\,\cdot\,\|. Consider its
dual normed space X^{\prime}, that consists of all
continuous linear functionals f : X \to \mathbb{F} and is equipped with the
dual norm \|\,\cdot\,\|^{\prime} defined by \|f\|^{\prime} = \sup \{ |f(x)| \,:\, x \in X, \ \|x\| = 1 \}. The dual X^{\prime} is a normed space (a
Banach space to be precise), and its dual normed space X^{\prime\prime} = \left(X^{\prime}\right)^{\prime} is called
bidual space for X. The bidual consists of all continuous linear functionals h : X^{\prime}\to \mathbb{F} and is equipped with the norm \|\,\cdot\,\|^{\prime\prime} dual to \|\,\cdot\,\|^{\prime}. Each vector x \in X generates a scalar function J(x) : X^{\prime} \to \mathbb{F} by the formula: J(x)(f) = f(x) \qquad \text{ for all } f \in X^{\prime}, and J(x) is a continuous linear functional on X^{\prime}, that is, J(x)\in X^{\prime\prime}. One obtains in this way a map J : X \to X^{\prime\prime} called
evaluation map, that is linear. It follows from the
Hahn–Banach theorem that J is injective and preserves norms: \text{ for all } x \in X \qquad \|J(x)\|^{\prime\prime} = \|x\|, that is, J maps X isometrically onto its image J(X) in X^{\prime\prime}. Furthermore, the image J(X) is closed in X^{\prime\prime}, but it need not be equal to X^{\prime\prime}. A normed space X is called
reflexive if it satisfies the following equivalent conditions: the evaluation map J : X \to X^{\prime\prime} is
surjective, the evaluation map J : X \to X^{\prime\prime} is an
isometric isomorphism of normed spaces, the evaluation map J : X \to X^{\prime\prime} is an
isomorphism of normed spaces. A reflexive space X is a Banach space, since X is then isometric to the Banach space X^{\prime\prime}.
Remark A Banach space X is reflexive if it is linearly isometric to its bidual under this canonical embedding J.
James' space is an example of a non-reflexive space which is linearly isometric to its
bidual. Furthermore, the image of James' space under the canonical embedding J has
codimension one in its bidual. A Banach space X is called
quasi-reflexive (of order d) if the quotient X^{\prime\prime} / J(X) has finite dimension d.
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
c_0 of scalar sequences tending to 0 at infinity, equipped with the supremum norm, is not reflexive. It follows from the general properties below that
\ell^1 and \ell^{\infty} are not reflexive, because \ell^1 is isomorphic to the dual of c_0 and \ell^{\infty} is isomorphic to the dual of \ell^1. • All
Hilbert spaces are reflexive, as are the
Lp spaces L^p for 1 More generally: all
uniformly convex Banach spaces are reflexive according to the
Milman–Pettis theorem. The L^1(\mu) and L^{\infty}(\mu) spaces are not reflexive (unless they are finite dimensional, which happens for example when \mu is a measure on a finite set). Likewise, the Banach space C([0, 1]) of continuous functions on [0, 1] 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 When the dimension of H is infinite, then S_1(H) (the
trace class) is not reflexive, because it contains a subspace isomorphic to \ell^1, and S_{\infty}(H) = L(H) (the bounded linear operators on H) is not reflexive, because it contains a subspace isomorphic to \ell^{\infty}. In both cases, the subspace can be chosen to be the operators diagonal with respect to a given orthonormal basis of H.
Properties Since every finite-dimensional normed space is a reflexive
Banach space, only infinite-dimensional spaces can be non-reflexive. If a Banach space Y is isomorphic to a reflexive Banach space X then Y is reflexive. Every
closed linear subspace of a reflexive space is reflexive. The continuous dual of a reflexive space is reflexive. Every
quotient of a reflexive space by a closed subspace is reflexive. Let X be a Banach space. The following are equivalent. The space X is reflexive. The continuous dual of X is reflexive. The closed unit ball of X is
compact in the
weak topology. (This is known as Kakutani's Theorem.) Every bounded sequence in X has a weakly convergent subsequence. The statement of
Riesz's lemma holds when the real number is exactly 1. Explicitly, for every closed proper vector subspace Y of X, there exists some vector u \in X of unit norm \|u\| = 1 such that \|u - y\| \geq 1 for all y \in Y. • Using d(u, Y) := \inf_{y \in Y} \|u - y\| to denote the distance between the vector u and the set Y, this can be restated in simpler language as: X is reflexive if and only if for every closed proper vector subspace Y, there is some vector u on the
unit sphere of X that is always at least a distance of 1 = d(u, Y) away from the subspace. • For example, if the reflexive Banach space X = \Reals^3 is endowed with the usual
Euclidean norm and Y = \Reals \times \Reals \times \{0\} is the x-y plane then the points u = (0, 0, \pm 1) satisfy the conclusion d(u, Y) = 1. If Y is instead the z-axis then every point belonging to the unit circle in the x-y plane satisfies the conclusion. Every continuous linear functional on X attains its supremum on the closed unit ball in X. (
James' theorem) Since norm-closed
convex subsets in a Banach space are weakly closed, it follows from the third property that closed bounded convex subsets of a reflexive space X are weakly compact. Thus, for every decreasing sequence of non-empty closed bounded convex subsets of X, the intersection is non-empty. As a consequence, every continuous
convex function f on a closed convex subset C of X, such that the set C_t = \{ x \in C \,:\, f(x) \leq t \} is non-empty and bounded for some real number t, attains its minimum value on C. The promised geometric property of reflexive Banach spaces is the following: if C is a closed non-empty
convex subset of the reflexive space X, then for every x \in X there exists a c \in C such that \|x - c\| minimizes the distance between x and points of C. This follows from the preceding result for convex functions, applied tof(y) + \|y - x\|. Note that while the minimal distance between x and C is uniquely defined by x, the point c is not. The closest point c is unique when X is uniformly convex. A reflexive Banach space is
separable if and only if its continuous dual is separable. This follows from the fact that for every normed space Y, separability of the continuous dual Y^{\prime} implies separability of Y.
Super-reflexive space Informally, a super-reflexive Banach space X has the following property: given an arbitrary Banach space Y, if all finite-dimensional subspaces of Y have a very similar copy sitting somewhere in X, then Y must be reflexive. By this definition, the space X itself must be reflexive. As an elementary example, every Banach space Y whose two dimensional subspaces are
isometric to subspaces of X = \ell^2 satisfies the
parallelogram law, hence Y is a Hilbert space, therefore Y is reflexive. So \ell^2 is super-reflexive. The formal definition does not use isometries, but almost isometries. A Banach space Y is
finitely representable in a Banach space X if for every finite-dimensional subspace Y_0 of Y and every \epsilon > 0, there is a subspace X_0 of X such that the multiplicative
Banach–Mazur distance between X_0 and Y_0 satisfies d\left(X_0, Y_0\right) A Banach space finitely representable in \ell^2 is a Hilbert space. Every Banach space is finitely representable in c_0. The
Lp space L^p([0, 1]) is finitely representable in \ell^p. A Banach space X is
super-reflexive if all Banach spaces Y finitely representable in X are reflexive, or, in other words, if no non-reflexive space Y is finitely representable in X. The notion of
ultraproduct of a family of Banach spaces allows for a concise definition: the Banach space X is super-reflexive when its ultrapowers are reflexive. James proved that a space is super-reflexive if and only if its dual is super-reflexive. The description of a vectorial binary tree begins with a
rooted binary tree labeled by vectors: a tree of
height n in a Banach space X is a family of 2^{n+1} - 1 vectors of X, that can be organized in successive levels, starting with level 0 that consists of a single vector x_{\varnothing}, the
root of the tree, followed, for k = 1, \ldots, n, by a family of s^k2 vectors forming level k: \left\{ x_{\varepsilon_1, \ldots, \varepsilon_k} \right\}, \quad \varepsilon_j = \pm 1, \quad j = 1, \ldots, k, that are the
children of vertices of level k - 1. In addition to the
tree structure, it is required here that each vector that is an
internal vertex of the tree be the midpoint between its two children: x_\emptyset = \frac{x_1 + x_{-1}}{2}, \quad x_{\varepsilon_1, \ldots, \varepsilon_k} = \frac{x_{\varepsilon_1, \ldots, \varepsilon_k, 1} + x_{\varepsilon_1, \ldots, \varepsilon_k, -1}} {2}, \quad 1 \leq k Given a positive real number t, the tree is said to be
t-separated if for every internal vertex, the two children are t-separated in the given space norm: \left\|x_1 - x_{-1}\right\| \geq t, \quad \left\|x_{\varepsilon_1, \ldots, \varepsilon_k, 1} - x_{\varepsilon_1, \ldots, \varepsilon_k, -1}\right\| \geq t, \quad 1 \leq k
Theorem. The Banach space X is super-reflexive if and only if for every t \in (0, 2 \pi], there is a number n(t) such that every t-separated tree contained in the unit ball of X has height less than n(t).
Uniformly convex spaces are super-reflexive. Let X be uniformly convex, with
modulus of convexity \delta_X and let t be a real number in (0, 2]. By the
properties of the modulus of convexity, a t-separated tree of height n, contained in the unit ball, must have all points of level n - 1 contained in the ball of radius 1 - \delta_X(t) By induction, it follows that all points of level n - k are contained in the ball of radius \left(1 - \delta_X(t)\right)^j, \ j = 1, \ldots, n. If the height n was so large that \left(1 - \delta_X(t)\right)^{n-1} then the two points x_1, x_{-1} of the first level could not be t-separated, contrary to the assumption. This gives the required bound n(t), function of \delta_X(t) only. Using the tree-characterization,
Enflo proved{{cite journal that super-reflexive Banach spaces admit an equivalent uniformly convex norm. Trees in a Banach space are a special instance of vector-valued
martingales. Adding techniques from scalar martingale theory,
Pisier improved Enflo's result by showing{{cite journal \delta_X(t) \geq c \, t^q, \quad \text{ whenever } t \in [0, 2]. == Reflexive locally convex spaces ==