Evaluation map and reflexive space

For a normed space $X$, an isometric isomorphism can be defined from $X$ to its second dual space $X''$, i.e. $J: X \rightarrow X''$, such that for all $x \in X$, $J(x) = J_x$ with $J_x$ being defined as $J_x(x') = x'(x) \; (\forall x' \in X')$. This map $J$ is called the evaluation map. When the range of $J$ is equal to $X''$, we say $X$ is reflexive. In this post, we'll prove that

the evaluation map $J$ really maps an element in $X$ to an element in $X''$;
$J$ is an isometric isomorphism from $X$ to $J(X)$.

Part 1

To prove $J(x) = J_x \in X'' (\forall x \in X)$, we should show that $J_x$ is both linear and continuous.

For the linearity of $J_x$, let $x', y' \in X'$ and $a, b \in \mathbb{K}$. Due to the fact that $X'$ is itself a linear space with respect to operator addition and scalar product in the sense of point-wise evaluation at $x$, we have

\[
\begin{aligned}
J_x(ax' + by') &= (ax' + by')(x) = a x'(x) + b y'(x) \\
&= a J_x(x') + b J_x(y')
\end{aligned}.
\]

This proves $J_x$ is linear and this linearity actually inherits from the linear structure of $X'$.

For the continuity of $J_x$, we need to show it is a bounded functional.

Because $x' \in X'$ is bounded, for all $x' \in X'$,

\[
\abs{J_x(x')} = \abs{x'(x)} \leq \norm{x'}_{X'} \cdot \norm{x}_X.
\]

We can see the norm of $J_x$, i.e. $\norm{J_x}_{X''}$ is bounded by $\norm{x}_X$. Therefore, $J_x$ is continuous. To sum up, we have $J_x \in X''$.

Part 2

Next, we shall prove $J$ is isometric, viz. norm-preserving.

In the above, we've already shown that $\norm{J_x}_{X''} \leq \norm{x}_X$. If we can further prove $\norm{J_x}_{X''} \geq \norm{x}_X$ so that $\norm{J_x}_{X''} = \norm{x}_X$, $J$ must be norm-preserving. The proof of this depends on whether we can find an $x'$ in $X'$, such that

\[
\frac{\abs{J_x(x')}}{\norm{x'}_{X'}} = \norm{x}_X,
\]

which naturally leads to

\[
\norm{x}_X \leq \norm{J_x}_{X''}.
\]

Let $x_0$ be arbitrarily selected from $X$. We can define a functional $x'$ which at the moment can only be evaluated at $x_0$ as $x'(x_0) = \norm{x_0}_X$. Then we extend the domain of $x'$ to the subspace $M$ of $X$ spanned by $x_0$

\[
M = \span\{x_0\} = \{x = c x_0 \vert c \in \mathbb{K}\}
\]

and for all $x = c x_0 \in M$, define

\[
x'(x) = x'(c x_0) = c \norm{x_0}_X.
\]

It is obvious that the extended $x'$ on $M$ is linear. In addition, we have

\[
\abs{x'(x)} = \abs{x'(c x_0)} = \abs{c x'(x_0))} = \norm{c x_0}_X = \norm{x}_X,
\]

which indicates that $x'$ is bounded and $\norm{x'}_{X'} = 1$. Hence, $x'$ belongs to the dual space $M'$ of $M$.

Next, by applying the Hahn-Banach theorem, we can extend the domain of $x'$ from the subspace $M$ of $X$ to the whole space $X$, while preserving the norm $\norm{x'}_{X'} = 1$. Therefore, for this specific $x' \in X'$,

\[
\frac{\abs{J_{x_0}(x')}}{\norm{x'}_{X'}} = \frac{\abs{x'(x_0)}}{1} = \norm{x_0}_X,
\]

so that

\[
\norm{x_0}_X \leq \norm{J_{x_0}}_{X''} \leq \norm{x_0}_X.
\]

Because $x_0$ is arbitrarily selected from $X$, we've proved that $J: X \rightarrow X''$ is really an isometric map.

To prove $J$ is an isomorphism between $X$ and $J(X) \subset X''$, we should prove $J$ preserves the linear structure from $X$ to $X''$ and is also an injective map. For the preservation of linear structure, it has already been verified during the proof of the linearity of $J_x$ as above. To show $J$ is injective, let $x_1, x_2 \in X$ and $x_1 \neq x_2$. For sure we can find an $x'$ in $X'$ such that $x'(x_1) \neq x'(x_2)$. Then for this $x'$, we have $J_{x_1}(x') = x'(x_1)$ is different from $J_{x_2}(x') = x'(x_2)$, which indicates $J_{x_1} \neq J_{x_2}$. Hence $J$ is injective.

Conclusions

Summarizing the above proof, we arrive at the conclusion that $J$ is an isometric isomorphism between $X$ and $J(X) \subset X''$.

Remark The key step in the above is during the proof of isometry, where a specific functional $x'$ is firstly defined at a single point $x_0 \in X$ with its value equal to $\norm{x_0}_X$. Then its domain is extended to the span of $x_0$ and further to the whole space $X$ by using the Hahn-Banach theorem, which ensures the extension is both continuous and norm-preserving.