Reflexive Spaces

Definition (Canonical Embedding)

Let $X$ be a normed space. The mapping

$$C:X⟶X^{\prime \prime },x\mapsto g_x,$$

where the functional $g_x$ on $X^{\prime }$ is defined by

$$g_x(f)=f(x)\text{for f∈X′,}$$

is called the canonical embedding of $X$ into its bidual $X^{\prime \prime }$.

The canonical embedding $C:X\to X^{\prime \prime }$ of a normed space into its bidual is well-defined and an embedding of normed spaces.

In particular, $C$ is isometric, hence injective.

Proof   We have to show that, for any given $x\in X$, $g_x$ is a bounded linear functional on $X^{\prime }$. Linearity follows from the fact that the vector space structure on $X^{\prime }$ is given by pointwise operations. To see that $g_x$ is bounded, observe that

$$|g_x(f)|=|f(x)|\le \Vert f\Vert \Vert x\Vert$$

holds for all $f\in X^{\prime }$. Moreover, this implies that $\Vert g_x\Vert \le \Vert x\Vert$. Thanks to Hahn–Banach, we know that there exists a bounded linear functional $f\in X^{\prime }$ with $\Vert f\Vert =1$ such that $f(x)=\Vert x\Vert$; hence, $\Vert g_x\Vert =\Vert x\Vert$. This means that the mapping $x\mapsto g_x$ is isometric. Clearly, this mapping is also linear, and thus an embedding of normed spaces. $\square$

Definition (Reflexivity)

A normed space is said to be reflexive if the canonical embedding into its bidual is surjective.

If a normed space $X$ is reflexive, then $X$ is isometrically isomorphic with $X^{\prime \prime }$, its bidual. James gives a counterexample for the converse statement.

If a normed space is reflexive, then it is complete; hence a Banach space.

If a normed space $X$ is reflexive, then the weak and weak${}^{\ast }$ topologies on $X^{\prime }$ agree.

Proof   By definition, the weak and weak${}^{\ast }$ topologies on $X^{\prime }$ are the initial topologies induced by the sets of functionals $X^{\prime \prime }$ and $C(X)$, respectively. Since $X$ is reflexive, those sets are equal. $\square$

The converse is true as well. Proof: TODO

If a normed space $X$ is reflexive, then its dual $X^{\prime }$ is reflexive.

Proof   Since $X$ is reflexive, the canonical embedding

$$C:X⟶X^{\prime \prime },C(x)(f)=f(x),x\in X,f\in X^{\prime },$$

is an isomorphism. Therefore, the dual map

$$C^{\prime }:X^{\prime \prime \prime }⟶X^{\prime },C^{\prime }(h)(x)=h(C(x)),x\in X,h\in X^{\prime \prime \prime },$$

is an isomorphism as well. A priori, it is not clear how $C^{\prime }$ is related to the canonical embedding

$$D:X^{\prime }⟶X^{\prime \prime \prime },D(f)(g)=g(f),f\in X^{\prime },g\in X^{\prime \prime }.$$

To show that $D$ is surjective, consider any element $h$ in $X^{\prime \prime \prime }$. We claim that $h=D(f)$ with $f=C^{\prime }(h)$. Let $g$ be any element of $X^{\prime \prime }$. It is of the form $g=C(x)$ with $x\in X$ unique, because $X$ is reflexive. We have

$$h(g)=h(C(x))=C^{\prime }(h)(x)=f(x)$$

by the definition of $C^{\prime }$. On the other hand,

$$D(f)(g)=g(f)=C(x)(f)=f(x)$$

by the definitions of $D$ and $C$. This shows that $D$ is surjective, hence $X^{\prime }$ is reflexive. In fact, we have shown more: $D=(C^{\prime }{)}^{-1}$. $\square$

Every finite-dimensional normed space is reflexive.

Every Hilbert space is reflexive.