---
title: Reflexive Spaces
parent: Functional Analysis Basics
nav_order: 4
description: >
  A normed space is said to be reflexive if the canonical embedding into its bidual is surjective.
---

# {{ page.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 $X''$.
{% enddefinition %}

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

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 $X'$.
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

$$
\abs{g_x(f)} = \abs{f(x)} \le \norm{f} \norm{x}
$$

holds for all $f \in X'$.
Moreover, this implies that $\norm{g_x} \le \norm{x}$.
Thanks to
[Hahn–Banach](/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.html#hahn-banach-theorem-existence-of-functionals),
we know that there exists a bounded linear functional
$f \in X'$ with $\norm{f} = 1$ such that $f(x) = \norm{x}$;
hence, $\norm{g_x} = \norm{x}$.
This means that the mapping $x \mapsto g_x$ is isometric.
Clearly, this mapping is also linear, and thus an embedding
of normed spaces.
{% endproof %}

{% definition Reflexivity %}
A normed space is said to be *reflexive*
if the canonical embedding into its bidual
is surjective.
{% enddefinition %}

If a normed space $X$ is reflexive,
then $X$ is isometrically 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.
{% endtheorem %}

{% proof %}
{% endproof %}

{% theorem %}
If a normed space $X$ is reflexive,
then the weak and weak$^*$ topologies on $X'$ agree.
{% endtheorem %}

{% 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.
{% endtheorem %}

{% 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 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.
{% endtheorem %}

{% theorem %}
Every Hilbert space is reflexive.
{% endtheorem %}