---
title: Reflexive Spaces
parent: Functional Analysis Basics
nav_order: 2
# cspell:words
---

# {{ page.title }}

{: .definition-title }
> Definition (Canonical Embedding)
>
> Let $X$ be a normed space.
> The mapping
>
> $$
> C : X \longrightarrow X'', \quad x \mapsto g_x,
> $$
>
> where the functional $g_x$ on $X'$ is defined by
>
> $$
> g_x(f) = f(x) \quad \text{for $f \in X'$,} 
> $$
>
> is called the *canonical embedding* of $X$ into its bidual $X''$.

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

{% proof %}
{% endproof %}

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

{: .definition-title }
> 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 isomorphic with $X''$, its bidual.
James gives a counterexample for the converse statement.

{: .theorem }
> If a normed space is reflexive,
> then it is complete; hence a Banach space.

{% proof %}
{% endproof %}

{: .theorem }
> 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.
{% endproof %}

The converse is true as well. Proof: TODO

{: .theorem }
> If a normed space $X$ is reflexive,
> then its dual $X'$ is reflexive.

{% proof %}
Since $X$ is reflexive,
the canonical embedding

$$
C : X \longrightarrow X'', \quad C(x)(f) = f(x), \quad x \in X, f \in X',
$$

is an isomorphism.
Therefore, the the dual map

$$
C' : X''' \longrightarrow X', \quad C'(h)(x) = h(C(x)), \quad x \in X, h \in X''',
$$

is an isomorphism as well.
A priori, it is not clear how $C'$ is related to
the canonical embedding

$$
D : X' \longrightarrow X''', \quad D(f)(g) = g(f), \quad f \in X', g \in X''.
$$

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 $x \in X$ 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: $D = (C')^{-1}$.
{% endproof %}

{: .theorem }
> Every finite-dimensional normed space is reflexive.
>

{: .theorem }
> Every Hilbert space is reflexive.