Reflexive space

A Banach space ${\displaystyle X}$ is reflexive if it is linearly isometric to its bidual under this canonical embedding ${\displaystyle J.}$ 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 ${\displaystyle J}$ has codimension one in its bidual. [2] A Banach space ${\displaystyle X}$ is called quasi-reflexive (of order ${\displaystyle d}$) if the quotient ${\displaystyle X^{\prime \prime }/J(X)}$ has finite dimension ${\displaystyle d.}$

1. 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 ${\displaystyle J}$ from the definition is bijective, by the rank–nullity theorem.
2. The Banach space ${\displaystyle c_{0}}$ of scalar sequences tending to 0 at infinity, equipped with the supremum norm, is not reflexive. It follows from the general properties below that ${\displaystyle \ell ^{1}}$ and ${\displaystyle \ell ^{\infty }}$ are not reflexive, because ${\displaystyle \ell ^{1}}$ is isomorphic to the dual of ${\displaystyle c_{0}}$ and ${\displaystyle \ell ^{\infty }}$ is isomorphic to the dual of ${\displaystyle \ell ^{1}.}$
3. All Hilbert spaces are reflexive, as are the Lp spaces ${\displaystyle L^{p}}$ for ${\displaystyle 1<p<\infty .}$ More generally: all uniformly convex Banach spaces are reflexive according to the Milman–Pettis theorem. The ${\displaystyle L^{1}(\mu )}$ and ${\displaystyle L^{\infty }(\mu )}$ spaces are not reflexive (unless they are finite dimensional, which happens for example when ${\displaystyle \mu }$ is a measure on a finite set). Likewise, the Banach space ${\displaystyle C([0,1])}$ of continuous functions on ${\displaystyle [0,1]}$ is not reflexive.
4. The spaces ${\displaystyle S_{p}(H)}$ of operators in the Schatten class on a Hilbert space ${\displaystyle H}$ are uniformly convex, hence reflexive, when ${\displaystyle 1<p<\infty .}$ When the dimension of ${\displaystyle H}$ is infinite, then ${\displaystyle S_{1}(H)}$ (the trace class) is not reflexive, because it contains a subspace isomorphic to ${\displaystyle \ell ^{1},}$ and ${\displaystyle S_{\infty }(H)=L(H)}$ (the bounded linear operators on ${\displaystyle H}$) is not reflexive, because it contains a subspace isomorphic to ${\displaystyle \ell ^{\infty }.}$ In both cases, the subspace can be chosen to be the operators diagonal with respect to a given orthonormal basis of ${\displaystyle H.}$

Since every finite-dimensional normed space is a reflexive Banach space, only infinite-dimensional spaces can be non-reflexive.

If a Banach space ${\displaystyle Y}$ is isomorphic to a reflexive Banach space ${\displaystyle X}$ then ${\displaystyle Y}$ is reflexive.[3]

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.[4]

Let ${\displaystyle X}$ be a Banach space. The following are equivalent.

1. The space ${\displaystyle X}$ is reflexive.
2. The continuous dual of ${\displaystyle X}$ is reflexive.[5]
3. The closed unit ball of ${\displaystyle X}$ is compact in the weak topology. (This is known as Kakutani's Theorem.)[6]
4. Every bounded sequence in ${\displaystyle X}$ has a weakly convergent subsequence.[7]
5. The statement of Riesz's lemma holds when the real number[note 1] is exactly ${\displaystyle 1.}$[8] Explicitly, for every closed proper vector subspace ${\displaystyle Y}$ of ${\displaystyle X,}$ there exists some vector ${\displaystyle u\in X}$ of unit norm ${\displaystyle \|u\|=1}$ such that ${\displaystyle \|u-y\|\geq 1}$ for all ${\displaystyle y\in Y.}$
   • Using ${\displaystyle d(u,Y):=\inf _{y\in Y}\|u-y\|}$ to denote the distance between the vector ${\displaystyle u}$ and the set ${\displaystyle Y,}$ this can be restated in simpler language as: ${\displaystyle X}$ is reflexive if and only if for every closed proper vector subspace ${\displaystyle Y,}$ there is some vector ${\displaystyle u}$ on the unit sphere of ${\displaystyle X}$ that is always at least a distance of ${\displaystyle 1=d(u,Y)}$ away from the subspace.
   • For example, if the reflexive Banach space ${\displaystyle X=\mathbb {R} ^{3}}$ is endowed with the usual Euclidean norm and ${\displaystyle Y=\mathbb {R} \times \mathbb {R} \times \{0\}}$ is the ${\displaystyle x-y}$ plane then the points ${\displaystyle u=(0,0,\pm 1)}$ satisfy the conclusion ${\displaystyle d(u,Y)=1.}$ If ${\displaystyle Y}$ is instead the ${\displaystyle z}$-axis then every point belonging to the unit circle in the ${\displaystyle x-y}$ plane satisfies the conclusion.
6. Every continuous linear functional on ${\displaystyle X}$ attains its supremum on the closed unit ball in ${\displaystyle X.}$[9] (James' theorem)

Since norm-closed convex subsets in a Banach space are weakly closed,[10] it follows from the third property that closed bounded convex subsets of a reflexive space ${\displaystyle X}$ are weakly compact. Thus, for every decreasing sequence of non-empty closed bounded convex subsets of ${\displaystyle X,}$ the intersection is non-empty. As a consequence, every continuous convex function ${\displaystyle f}$ on a closed convex subset ${\displaystyle C}$ of ${\displaystyle X,}$ such that the set

$${\displaystyle C_{t}=\{x\in C\,:\,f(x)\leq t\}}$$

is non-empty and bounded for some real number ${\displaystyle t,}$ attains its minimum value on ${\displaystyle C.}$

The promised geometric property of reflexive Banach spaces is the following: if ${\displaystyle C}$ is a closed non-empty convex subset of the reflexive space ${\displaystyle X,}$ then for every ${\displaystyle x\in X}$ there exists a ${\displaystyle c\in C}$ such that ${\displaystyle \|x-c\|}$ minimizes the distance between ${\displaystyle x}$ and points of ${\displaystyle C.}$ This follows from the preceding result for convex functions, applied to${\displaystyle f(y)+\|y-x\|.}$ Note that while the minimal distance between ${\displaystyle x}$ and ${\displaystyle C}$ is uniquely defined by ${\displaystyle x,}$ the point ${\displaystyle c}$ is not. The closest point ${\displaystyle c}$ is unique when ${\displaystyle X}$ 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 ${\displaystyle Y,}$ separability of the continuous dual ${\displaystyle Y^{\prime }}$ implies separability of ${\displaystyle Y.}$[11]

Informally, a super-reflexive Banach space ${\displaystyle X}$ has the following property: given an arbitrary Banach space ${\displaystyle Y,}$ if all finite-dimensional subspaces of ${\displaystyle Y}$ have a very similar copy sitting somewhere in ${\displaystyle X,}$ then ${\displaystyle Y}$ must be reflexive. By this definition, the space ${\displaystyle X}$ itself must be reflexive. As an elementary example, every Banach space ${\displaystyle Y}$ whose two dimensional subspaces are isometric to subspaces of ${\displaystyle X=\ell ^{2}}$ satisfies the parallelogram law, hence[12] ${\displaystyle Y}$ is a Hilbert space, therefore ${\displaystyle Y}$ is reflexive. So ${\displaystyle \ell ^{2}}$ is super-reflexive.

The formal definition does not use isometries, but almost isometries. A Banach space ${\displaystyle Y}$ is finitely representable[13] in a Banach space ${\displaystyle X}$ if for every finite-dimensional subspace ${\displaystyle Y_{0}}$ of ${\displaystyle Y}$ and every ${\displaystyle \epsilon >0,}$ there is a subspace ${\displaystyle X_{0}}$ of ${\displaystyle X}$ such that the multiplicative Banach–Mazur distance between ${\displaystyle X_{0}}$ and ${\displaystyle Y_{0}}$ satisfies

$${\displaystyle d\left(X_{0},Y_{0}\right)<1+\varepsilon .}$$

A Banach space finitely representable in ${\displaystyle \ell ^{2}}$ is a Hilbert space. Every Banach space is finitely representable in ${\displaystyle c_{0}.}$ The Lp space ${\displaystyle L^{p}([0,1])}$ is finitely representable in ${\displaystyle \ell ^{p}.}$

A Banach space ${\displaystyle X}$ is super-reflexive if all Banach spaces ${\displaystyle Y}$ finitely representable in ${\displaystyle X}$ are reflexive, or, in other words, if no non-reflexive space ${\displaystyle Y}$ is finitely representable in ${\displaystyle X.}$ The notion of ultraproduct of a family of Banach spaces[14] allows for a concise definition: the Banach space ${\displaystyle X}$ 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.[13]

One of James' characterizations of super-reflexivity uses the growth of separated trees.[15] The description of a vectorial binary tree begins with a rooted binary tree labeled by vectors: a tree of height ${\displaystyle n}$ in a Banach space ${\displaystyle X}$ is a family of ${\displaystyle 2^{n+1}-1}$ vectors of ${\displaystyle X,}$ that can be organized in successive levels, starting with level 0 that consists of a single vector ${\displaystyle x_{\varnothing },}$ the root of the tree, followed, for ${\displaystyle k=1,\ldots ,n,}$ by a family of ${\displaystyle s^{k}}$2 vectors forming level ${\displaystyle k:}$

$${\displaystyle \left\{x_{\varepsilon _{1},\ldots ,\varepsilon _{k}}\right\},\quad \varepsilon _{j}=\pm 1,\quad j=1,\ldots ,k,}$$

that are the children of vertices of level ${\displaystyle k-1.}$ 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:

$${\displaystyle x_{\emptyset }={\frac {x_{1}+x_{-1}}{2}},\quad x_{\varepsilon _{1},\ldots ,\varepsilon _{k}}={\frac {x_{\varepsilon _{1},\ldots ,\varepsilon _{k},1}+x_{\varepsilon _{1},\ldots ,\varepsilon _{k},-1}}{2}},\quad 1\leq k<n.}$$

Given a positive real number ${\displaystyle t,}$ the tree is said to be ${\displaystyle t}$-separated if for every internal vertex, the two children are ${\displaystyle t}$-separated in the given space norm:

$${\displaystyle \left\|x_{1}-x_{-1}\right\|\geq t,\quad \left\|x_{\varepsilon _{1},\ldots ,\varepsilon _{k},1}-x_{\varepsilon _{1},\ldots ,\varepsilon _{k},-1}\right\|\geq t,\quad 1\leq k<n.}$$

Theorem.[15] The Banach space ${\displaystyle X}$ is super-reflexive if and only if for every ${\displaystyle t\in (0,2\pi ],}$ there is a number ${\displaystyle n(t)}$ such that every ${\displaystyle t}$-separated tree contained in the unit ball of ${\displaystyle X}$ has height less than ${\displaystyle n(t).}$

Uniformly convex spaces are super-reflexive.[15] Let ${\displaystyle X}$ be uniformly convex, with modulus of convexity ${\displaystyle \delta _{X}}$ and let ${\displaystyle t}$ be a real number in ${\displaystyle (0,2].}$ By the properties of the modulus of convexity, a ${\displaystyle t}$-separated tree of height ${\displaystyle n,}$ contained in the unit ball, must have all points of level ${\displaystyle n-1}$ contained in the ball of radius ${\displaystyle 1-\delta _{X}(t)<1.}$ By induction, it follows that all points of level ${\displaystyle n-k}$ are contained in the ball of radius

$${\displaystyle \left(1-\delta _{X}(t)\right)^{j},\ j=1,\ldots ,n.}$$

If the height ${\displaystyle n}$ was so large that

$${\displaystyle \left(1-\delta _{X}(t)\right)^{n-1}<t/2,}$$

then the two points ${\displaystyle x_{1},x_{-1}}$ of the first level could not be ${\displaystyle t}$-separated, contrary to the assumption. This gives the required bound ${\displaystyle n(t),}$ function of ${\displaystyle \delta _{X}(t)}$ only.

Using the tree-characterization, Enflo proved[16] 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[17] that a super-reflexive space ${\displaystyle X}$ admits an equivalent uniformly convex norm for which the modulus of convexity satisfies, for some constant ${\displaystyle c>0}$ and some real number ${\displaystyle q\geq 2,}$

$${\displaystyle \delta _{X}(t)\geq c\,t^{q},\quad {\text{ whenever }}t\in [0,2].}$$

If ${\displaystyle X}$ is a Hausdorff locally convex space then the following are equivalent:

1. ${\displaystyle X}$ is semireflexive;
2. The weak topology on ${\displaystyle X}$ had the Heine-Borel property (that is, for the weak topology ${\displaystyle \sigma \left(X,X^{\prime }\right),}$ every closed and bounded subset of ${\displaystyle X_{\sigma }}$ is weakly compact).[1]
3. If linear form on ${\displaystyle X^{\prime }}$ that continuous when ${\displaystyle X^{\prime }}$ has the strong dual topology, then it is continuous when ${\displaystyle X^{\prime }}$ has the weak topology;[24]
4. ${\displaystyle X_{\tau }^{\prime }}$ is barreled;[24]
5. ${\displaystyle X}$ with the weak topology ${\displaystyle \sigma \left(X,X^{\prime }\right)}$ is quasi-complete.[24]

If ${\displaystyle X}$ is a Hausdorff locally convex space then the following are equivalent:

1. ${\displaystyle X}$ is reflexive;
2. ${\displaystyle X}$ is semireflexive and infrabarreled;[23]
3. ${\displaystyle X}$ is semireflexive and barreled;
4. ${\displaystyle X}$ is barreled and the weak topology on ${\displaystyle X}$ had the Heine-Borel property (that is, for the weak topology ${\displaystyle \sigma \left(X,X^{\prime }\right),}$ every closed and bounded subset of ${\displaystyle X_{\sigma }}$ is weakly compact).[1]
5. ${\displaystyle X}$ is semireflexive and quasibarrelled.[25]

If ${\displaystyle X}$ is a normed space then the following are equivalent:

1. ${\displaystyle X}$ is reflexive;
2. The closed unit ball is compact when ${\displaystyle X}$ has the weak topology ${\displaystyle \sigma \left(X,X^{\prime }\right).}$[26]
3. ${\displaystyle X}$ is a Banach space and ${\displaystyle X_{b}^{\prime }}$ is reflexive.[27]
4. Every sequence ${\displaystyle \left(C_{n}\right)_{n=1}^{\infty },}$ with ${\displaystyle C_{n+1}\subseteq C_{n}}$ for all ${\displaystyle n}$ of nonempty closed bounded convex subsets of ${\displaystyle X}$ has nonempty intersection.[28]

Normed spaces

A normed space that is semireflexive is a reflexive Banach space.[30] A closed vector subspace of a reflexive Banach space is reflexive.[23]

Let ${\displaystyle X}$ be a Banach space and ${\displaystyle M}$ a closed vector subspace of ${\displaystyle X.}$ If two of ${\displaystyle X,M,}$ and ${\displaystyle X/M}$ are reflexive then they all are.[23] This is why reflexivity is referred to as a three-space property.[23]

Topological vector spaces

If a barreled locally convex Hausdorff space is semireflexive then it is reflexive.[1]

The strong dual of a reflexive space is reflexive.[31]Every Montel space is reflexive.[26] And the strong dual of a Montel space is a Montel space (and thus is reflexive).[26]

A locally convex Hausdorff reflexive space is barrelled. If ${\displaystyle X}$ is a normed space then ${\displaystyle I:X\to X^{\prime \prime }}$ is an isometry onto a closed subspace of ${\displaystyle X^{\prime \prime }.}$[30] This isometry can be expressed by:

$${\displaystyle \|x\|=\sup _{\stackrel {x^{\prime }\in X^{\prime },}{\|x^{\prime }\|\leq 1}}\left|\left\langle x^{\prime },x\right\rangle \right|.}$$

Suppose that ${\displaystyle X}$ is a normed space and ${\displaystyle X^{\prime \prime }}$ is its bidual equipped with the bidual norm. Then the unit ball of ${\displaystyle X,}$ ${\displaystyle I(\{x\in X:\|x\|\leq 1\})}$ is dense in the unit ball ${\displaystyle \left\{x^{\prime \prime }\in X^{\prime \prime }:\left\|x^{\prime \prime }\right\|\leq 1\right\}}$ of ${\displaystyle X^{\prime \prime }}$ for the weak topology ${\displaystyle \sigma \left(X^{\prime \prime },X^{\prime }\right).}$[30]

1. Every finite-dimensional Hausdorff topological vector space is reflexive, because ${\displaystyle J}$ is bijective by linear algebra, and because there is a unique Hausdorff vector space topology on a finite dimensional vector space.
2. A normed space ${\displaystyle X}$ is reflexive as a normed space if and only if it is reflexive as a locally convex space. This follows from the fact that for a normed space ${\displaystyle X}$ its dual normed space ${\displaystyle X^{\prime }}$ coincides as a topological vector space with the strong dual space ${\displaystyle X_{b}^{\prime }.}$ As a corollary, the evaluation map ${\displaystyle J:X\to X^{\prime \prime }}$ coincides with the evaluation map ${\displaystyle J:X\to \left(X_{b}^{\prime }\right)_{b}^{\prime },}$ and the following conditions become equivalent:
   1. ${\displaystyle X}$ is a reflexive normed space (that is, ${\displaystyle J:X\to X^{\prime \prime }}$ is an isomorphism of normed spaces),
   2. ${\displaystyle X}$ is a reflexive locally convex space (that is, ${\displaystyle J:X\to \left(X_{b}^{\prime }\right)_{b}^{\prime }}$ is an isomorphism of topological vector spaces[18]),
   3. ${\displaystyle X}$ is a semi-reflexive locally convex space (that is, ${\displaystyle J:X\to \left(X_{b}^{\prime }\right)_{b}^{\prime }}$ is surjective).
3. A (somewhat artificial) example of a semi-reflexive space that is not reflexive is obtained as follows: let ${\displaystyle Y}$ be an infinite dimensional reflexive Banach space, and let ${\displaystyle X}$ be the topological vector space ${\displaystyle \left(Y,\sigma \left(Y,Y^{\prime }\right)\right),}$ that is, the vector space ${\displaystyle Y}$ equipped with the weak topology. Then the continuous dual of ${\displaystyle X}$ and ${\displaystyle Y^{\prime }}$ are the same set of functionals, and bounded subsets of ${\displaystyle X}$ (that is, weakly bounded subsets of ${\displaystyle Y}$) are norm-bounded, hence the Banach space ${\displaystyle Y^{\prime }}$ is the strong dual of ${\displaystyle X.}$ Since ${\displaystyle Y}$ is reflexive, the continuous dual of ${\displaystyle X^{\prime }=Y^{\prime }}$ is equal to the image ${\displaystyle J(X)}$ of ${\displaystyle X}$ under the canonical embedding ${\displaystyle J,}$ but the topology on ${\displaystyle X}$ (the weak topology of ${\displaystyle Y}$) is not the strong topology ${\displaystyle \beta \left(X,X^{\prime }\right),}$ that is equal to the norm topology of ${\displaystyle Y.}$
4. Montel spaces are reflexive locally convex topological vector spaces. In particular, the following functional spaces frequently used in functional analysis are reflexive locally convex spaces:[32]
   • the space ${\displaystyle C^{\infty }(M)}$ of smooth functions on arbitrary (real) smooth manifold ${\displaystyle M,}$ and its strong dual space ${\displaystyle \left(C^{\infty }\right)^{\prime }(M)}$ of distributions with compact support on ${\displaystyle M,}$
   • the space ${\displaystyle {\mathcal {D}}(M)}$ of smooth functions with compact support on arbitrary (real) smooth manifold ${\displaystyle M,}$ and its strong dual space ${\displaystyle {\mathcal {D}}^{\prime }(M)}$ of distributions on ${\displaystyle M,}$
   • the space ${\displaystyle {\mathcal {O}}(M)}$ of holomorphic functions on arbitrary complex manifold ${\displaystyle M,}$ and its strong dual space ${\displaystyle {\mathcal {O}}^{\prime }(M)}$ of analytic functionals on ${\displaystyle M,}$
   • the Schwartz space ${\displaystyle {\mathcal {S}}\left(\mathbb {R} ^{n}\right)}$ on ${\displaystyle \mathbb {R} ^{n},}$ and its strong dual space ${\displaystyle {\mathcal {S}}^{\prime }\left(\mathbb {R} ^{n}\right)}$ of tempered distributions on ${\displaystyle \mathbb {R} ^{n}.}$

• There exists a non-reflexive locally convex TVS whose strong dual is reflexive.[33]

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