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+---
+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.