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.
