[proofplan]
Parts (i) and (ii) are direct applications of the [Dual of Weak Topology](/theorems/2649) result to two specific dual pairs: $(E, F) = (X, X^*)$ for the weak topology $w = \sigma(X, X^*)$, and $(E, F) = (X^*, \hat{X})$ for the weak-$*$ topology $w^* = \sigma(X^*, \hat{X})$, where $\hat{X} \subseteq X^{**}$ is the canonical image of $X$ in its bidual under the evaluation map $x \mapsto \hat{x}$ defined by $\hat{x}(f) = f(x)$. For (iii), the equivalence between reflexivity and the coincidence of $w$ and $w^*$ on $X^*$ follows because $\sigma(X^*, A) = \sigma(X^*, B)$ for two separating subspaces $A, B \subseteq (X^*)^\#$ forces $A = B$ when both are taken to be the dual of $(X^*, \sigma(X^*, A))$, which by (i) and (ii) gives $\hat{X} = X^{**}$.
[/proofplan]
[step:Set up the canonical embedding $\hat{X} \hookrightarrow X^{**}$ and the topologies $w, w^*$]
Let $X$ be a normed space and $X^*$ its topological dual (the space of bounded linear functionals on $X$). Define the canonical evaluation map
\begin{align*}
J : X &\to X^{**} \\
x &\mapsto \hat{x}, \qquad \hat{x}(f) = f(x) \text{ for } f \in X^*.
\end{align*}
Each $\hat{x}$ is linear in $f$ (immediate) and bounded with $|\hat{x}(f)| \leq \|f\|\|x\|$, so $\hat{x} \in X^{**}$ with $\|\hat{x}\| \leq \|x\|$. The Hahn-Banach theorem (norming functional version) gives equality $\|\hat{x}\| = \|x\|$, so $J$ is an isometry. Write $\hat{X} := J(X) \subseteq X^{**}$.
By definition:
- **Weak topology** on $X$: $w = \sigma(X, X^*)$, the coarsest topology on $X$ making each $f \in X^*$ continuous.
- **Weak-$*$ topology** on $X^*$: $w^* = \sigma(X^*, \hat{X})$, the coarsest topology on $X^*$ making each $\hat{x} : X^* \to \mathbb{R}$ (for $x \in X$) continuous.
$X$ is **reflexive** iff $J : X \to X^{**}$ is surjective, i.e. $\hat{X} = X^{**}$.
[/step]
[step:Prove (i) by applying Dual of Weak Topology to $(E, F) = (X, X^*)$]
Set $E := X$ and $F := X^*$. Then $E$ is a vector space, and $F$ is a vector space of linear functionals on $E$ that separates points of $E$ (this last point follows from the Hahn-Banach theorem: for any $x \neq 0$, the norming functional gives $f \in X^*$ with $f(x) = \|x\| \neq 0$).
The hypotheses of the [Dual of Weak Topology](/theorems/2649) are met: $E, F$ form a separating dual pair, and $\sigma(X, X^*) = w$ by definition.
The result gives: a linear $f : X \to \mathbb{R}$ is $\sigma(X, X^*)$-continuous if and only if $f \in X^*$. Equivalently,
\begin{align*}
(X, w)^* = X^*.
\end{align*}
[/step]
[step:Prove (ii) by applying Dual of Weak Topology to $(E, F) = (X^*, \hat{X})$]
Set $E := X^*$ and $F := \hat{X}$. Then $E$ is a vector space, and $\hat{X} \subseteq (X^*)^\#$ is a vector space of linear functionals on $X^*$.
We check $\hat{X}$ separates points of $X^*$: if $f \in X^*$ satisfies $\hat{x}(f) = f(x) = 0$ for all $x \in X$, then $f = 0$ (a functional vanishing at every point is the zero functional).
By definition, $\sigma(X^*, \hat{X}) = w^*$. Applying the [Dual of Weak Topology](/theorems/2649) to the pair $(E, F) = (X^*, \hat{X})$ gives: a linear $\varphi : X^* \to \mathbb{R}$ is $w^*$-continuous if and only if $\varphi \in \hat{X}$, i.e. $\varphi = \hat{x}$ for some $x \in X$. Equivalently,
\begin{align*}
(X^*, w^*)^* = \hat{X}.
\end{align*}
[/step]
[step:Reduce reflexivity to coincidence of dual topologies on $X^*$]
We prove (iii). Recall that on $X^*$ we have two natural weak-type topologies:
- $w^* = \sigma(X^*, \hat{X})$, generated by evaluations against $\hat{X} \subseteq X^{**}$,
- $w = \sigma(X^*, X^{**})$, the weak topology on the Banach space $X^*$, generated by evaluations against the full bidual $X^{**}$.
Note: $X^{**}$ separates points of $X^*$ (Hahn-Banach applied to the Banach space $X^*$), and $\hat{X} \subseteq X^{**}$ also separates points of $X^*$ (Step 3). Both topologies are well-defined.
Since $\hat{X} \subseteq X^{**}$, the topology $w^*$ is coarser than $w$ on $X^*$: every $w^*$-open set is $w$-open. The two coincide iff every $w$-open set is $w^*$-open, iff the identity map $(X^*, w^*) \to (X^*, w)$ is continuous.
[/step]
[step:Forward direction of (iii): $X$ reflexive implies $w = w^*$ on $X^*$]
Assume $X$ is reflexive, so $\hat{X} = X^{**}$. Then by definition,
\begin{align*}
w^* = \sigma(X^*, \hat{X}) = \sigma(X^*, X^{**}) = w
\end{align*}
on $X^*$. (Two topologies generated by sub-bases drawn from the same family of functionals are identical.)
[/step]
[step:Converse direction of (iii): $w = w^*$ on $X^*$ implies $X$ is reflexive]
Assume $w^* = w$ as topologies on $X^*$. Then $(X^*, w^*) = (X^*, w)$ as topological vector spaces, so they have the same continuous dual:
\begin{align*}
(X^*, w^*)^* = (X^*, w)^*.
\end{align*}
By Step 3 (part (ii) applied to $X$), $(X^*, w^*)^* = \hat{X}$.
By Step 2 (part (i) applied to the Banach space $X^*$ in place of $X$), $(X^*, w)^* = X^{**}$.
Equating the two: $\hat{X} = X^{**}$. Hence $J : X \to X^{**}$ is surjective, i.e. $X$ is reflexive.
Combining Steps 5 and 6: $X$ is reflexive iff $w = w^*$ on $X^*$, which completes the proof of (iii) and of all three parts of the theorem.
[/step]
