# Reflexive Spaces

Definition (Canonical Embedding)

Let $X$ be a normed space. The mapping

$C:X⟶X_{′′},x↦g_{x},$where the functional $g_{x}$ on $X_{′}$ is defined by

$g_{x}(f)=f(x)forf∈X_{′},$is called the

canonical embeddingof $X$ into its bidual $X_{′′}$.

The canonical embedding $C:X→X_{′′}$ 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 $x∈X$, $g_{x}$ is a bounded linear functional on $X_{′}$. Linearity follows from the fact that the vector space structure on $X_{′}$ is given by pointwise operations. To see that $g_{x}$ is bounded, observe that

$∣g_{x}(f)∣=∣f(x)∣≤∥f∥∥x∥$holds for all $f∈X_{′}$. Moreover, this implies that $∥g_{x}∥≤∥x∥$. Thanks to Hahn–Banach, we know that there exists a bounded linear functional $f∈X_{′}$ with $∥f∥=1$ such that $f(x)=∥x∥$; hence, $∥g_{x}∥=∥x∥$. This means that the mapping $x↦g_{x}$ is isometric. Clearly, this mapping is also linear, and thus an embedding of normed spaces. $□$

Definition (Reflexivity)

A normed space is said to be

reflexiveif the canonical embedding into its bidual is surjective.

If a normed space $X$ is reflexive, then $X$ is isometrically isomorphic with $X_{′′}$, 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$_{∗}$ topologies on $X_{′}$ agree.

Proof By definition, the weak and weak$_{∗}$ topologies on $X_{′}$ are the initial topologies induced by the sets of functionals $X_{′′}$ and $C(X)$, respectively. Since $X$ is reflexive, those sets are equal. $□$

The converse is true as well. Proof: TODO

If a normed space $X$ is reflexive, then its dual $X_{′}$ is reflexive.

Proof Since $X$ is reflexive, the canonical embedding

$C:X⟶X_{′′},C(x)(f)=f(x),x∈X,f∈X_{′},$is an isomorphism. Therefore, the dual map

$C_{′}:X_{′′′}⟶X_{′},C_{′}(h)(x)=h(C(x)),x∈X,h∈X_{′′′},$is an isomorphism as well. A priori, it is not clear how $C_{′}$ is related to the canonical embedding

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

$h(g)=h(C(x))=C_{′}(h)(x)=f(x)$by the definition of $C_{′}$. 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 $X_{′}$ is reflexive. In fact, we have shown more: $D=(C_{′})_{−1}$. $□$

Every finite-dimensional normed space is reflexive.

Every Hilbert space is reflexive.