[proofplan] We prove (i) $\Leftrightarrow$ (iii) directly using Goldstein's theorem and Banach-Alaoglu, then deduce (i) $\Leftrightarrow$ (ii) by applying the equivalence to $X^*$ and using that the canonical embedding into the bidual is closed-image. The forward direction (i) $\Rightarrow$ (iii) uses that under reflexivity the canonical map $J : X \to X^{**}$ is a surjective isometry, and the homeomorphism $J : (B_X, w) \to (B_{X^{**}}, w^*)$ transports Banach-Alaoglu compactness. The reverse direction (iii) $\Rightarrow$ (i) uses weak-compactness of $\hat{B}_X$ to show it is $w^*$-closed in $B_{X^{**}}$, hence equals $B_{X^{**}}$ by Goldstein, hence $J$ is surjective on unit balls and (by linearity and homogeneity) globally surjective. [/proofplan]

[step:Prove (i) $\Rightarrow$ (iii): reflexivity implies weak compactness of the unit ball] Assume $X$ is reflexive: the canonical embedding \begin{align*} J : X &\to X^{**}, \\ x &\mapsto \hat{x}, \quad \hat{x}(f) = f(x), \end{align*} is surjective. By the [Hahn-Banach (Normed Space Version)](/theorems/2629), $J$ is an isometry, so $J(B_X) = \hat{B}_X = B_{X^{**}}$. We claim the map $J : (B_X, w) \to (B_{X^{**}}, w^*)$ is a homeomorphism. Let $w = \sigma(X, X^*)$ and $w^* = \sigma(X^{**}, X^*)$. By the [Universal Property of Weak Topologies](/theorems/2645), continuity of a map $g : Z \to (Y, \sigma(Y, F))$ is equivalent to continuity of $\hat{f} \circ g$ for each $f \in F$. For continuity of $J$: take $f \in X^*$. Then $(\hat{f} \circ J)(x) = \hat{f}(\hat{x}) = \hat{x}(f) = f(x)$, which is $\sigma(X, X^*)$-continuous (by definition of $w$). So $J$ is $w$-to-$w^*$ continuous. For continuity of $J^{-1} : (B_{X^{**}}, w^*) \to (B_X, w)$: take $f \in X^*$. Then $f \circ J^{-1}(\hat{x}) = f(x) = \hat{x}(f) = \hat{f}(\hat{x})$, which is $w^*$-continuous (by definition of $w^*$). So $J^{-1}$ is $w^*$-to-$w$ continuous. Hence $J : (B_X, w) \to (B_{X^{**}}, w^*)$ is a homeomorphism. By the **Banach-Alaoglu Theorem**, $(B_{X^{**}}, w^*)$ is compact, so $(B_X, w)$ is compact. [/step]

[step:Prove (iii) $\Rightarrow$ (i): weak compactness of the unit ball implies reflexivity]Assume $(B_X, w)$ is compact. We show $J(B_X) = B_{X^{**}}$, which by linearity of $J$ and $J(\lambda B_X) = \lambda J(B_X)$ for $\lambda > 0$ yields $J(X) = X^{**}$, i.e.\ reflexivity. The argument from Step 1 (read with hypotheses reversed: now we know $J$ is an isometric injection, but not yet surjective) shows that $J : (B_X, w) \to (\hat{B}_X, w^*)$ is a homeomorphism, where $w^*$ here denotes the relative topology induced by $(X^{**}, w^*)$ on $\hat{B}_X = J(B_X)$. (The argument is identical: $J$ is $w$-to-$w^*$ continuous and its inverse, defined on its image $\hat{B}_X$, is $w^*$-to-$w$ continuous, by the same identifications $f(x) = \hat{x}(f) = \hat{f}(\hat{x})$.) Since $(B_X, w)$ is compact (hypothesis), $\hat{B}_X = J(B_X)$ is $w^*$-compact in $X^{**}$. As $(X^{**}, w^*)$ is Hausdorff (since the family $\{\hat{f} : f \in X^*\}$ separates points of $X^{**}$: if $\varphi \neq 0$ then there is $f$ with $\varphi(f) \neq 0$, i.e.\ $\hat{f}(\varphi) \neq 0$), every compact subset is closed. Hence $\hat{B}_X$ is $w^*$-closed in $X^{**}$. Therefore $\overline{\hat{B}_X}^{w^*} = \hat{B}_X$. By **Goldstein's Theorem**, $\overline{\hat{B}_X}^{w^*} = B_{X^{**}}$. So $\hat{B}_X = B_{X^{**}}$. Now extend to all of $X^{**}$. Take $\varphi \in X^{**}$ with $\varphi \neq 0$. Set $\psi := \varphi/\|\varphi\|_{X^{**}} \in B_{X^{**}}$. By the previous paragraph, $\psi = \hat{x}_0$ for some $x_0 \in B_X$. Then $\varphi = \|\varphi\| \cdot \hat{x}_0 = \widehat{\|\varphi\| x_0}$ (using linearity of $J$), so $\varphi \in J(X)$. Together with $0 = \hat{0} \in J(X)$, $J(X) = X^{**}$, so $X$ is reflexive.[/step]

[guided]The strategy: weak compactness of $B_X$ transfers via $J$ to $w^*$-compactness, hence $w^*$-closedness, of $\hat{B}_X$ in $X^{**}$. Goldstein then identifies $\hat{B}_X$ with all of $B_{X^{**}}$. *Step (a) — Transfer compactness via $J$.* The argument that $J : (B_X, w) \to (\hat{B}_X, w^*)$ is a homeomorphism is the same as in Step 1, with the only difference that here $\hat{B}_X$ is *a priori* a subset of $B_{X^{**}}$ rather than equal to it. The continuity arguments only use the identification $f(x) = \hat{x}(f)$, not surjectivity of $J$. So $\hat{B}_X$ is the homeomorphic image of the (assumed) compact $(B_X, w)$, hence $w^*$-compact in $X^{**}$. *Step (b) — From compactness to closedness.* In a Hausdorff space, every compact set is closed (the standard topology fact: separate the compact set from each external point by disjoint open sets, then close up). The space $(X^{**}, w^*)$ is Hausdorff because $w^*$ is the initial topology generated by the family $\{\hat{f} : f \in X^*\}$, which separates points of $X^{**}$: given $\varphi \neq \psi$ in $X^{**}$, $\varphi - \psi \neq 0$ in $X^{**}$, so by the dual norm formula $\sup_f |( \varphi-\psi)(f)| > 0$, i.e.\ there is $f$ with $\varphi(f) \neq \psi(f)$, i.e.\ $\hat{f}(\varphi) \neq \hat{f}(\psi)$. So $\hat{B}_X$ is $w^*$-closed in $X^{**}$. *Step (c) — Apply Goldstein.* The hypotheses of **Goldstein's Theorem** are vacuously satisfied for any normed space $X$. The conclusion: $\overline{\hat{B}_X}^{w^*} = B_{X^{**}}$. Since $\hat{B}_X$ is already $w^*$-closed, $\overline{\hat{B}_X}^{w^*} = \hat{B}_X$. Combining: $\hat{B}_X = B_{X^{**}}$. *Step (d) — Extend to $X^{**}$.* The map $J$ is linear, so $J(\lambda B_X) = \lambda J(B_X) = \lambda \hat{B}_X = \lambda B_{X^{**}}$ for all $\lambda > 0$. Since $X^{**} = \bigcup_{\lambda > 0} \lambda B_{X^{**}} \cup \{0\}$ and $0 = J(0)$, we get $X^{**} = J(X)$.[/guided]

[step:Prove (i) $\Rightarrow$ (ii): reflexivity of $X$ implies reflexivity of $X^*$] Assume $X$ is reflexive. We show $(B_{X^*}, w_{X^*})$ is compact, where $w_{X^*} = \sigma(X^*, X^{**})$ is the weak topology on $X^*$. The equivalence (i) $\Leftrightarrow$ (iii) applied to the Banach space $X^*$ (a dual is always Banach) then yields reflexivity of $X^*$. By reflexivity of $X$, $J : X \to X^{**}$ is an isometric isomorphism. Therefore the families \begin{align*} \{\hat{x} : x \in X\} = \{J(x) : x \in X\} = J(X) = X^{**} \end{align*} agree as subsets of $X^{**}$ (the first family sits inside $X^{**}$ by definition). The two topologies on $X^*$ in question are: - $w^*_{X^*} = \sigma(X^*, X)$, generated by evaluations $\{\hat{x} : x \in X\}$ acting on $X^*$, where here $\hat{x}(f) := f(x)$. - $w_{X^*} = \sigma(X^*, X^{**})$, generated by evaluations $\{\varphi : \varphi \in X^{**}\}$ acting on $X^*$. Under $J$ surjective, the second family $\{\varphi : \varphi \in X^{**}\}$ equals the first family $\{\hat{x} : x \in X\}$ as sets of functions $X^* \to \mathbb{C}$ (or $\mathbb{R}$): every $\varphi \in X^{**}$ is of the form $\hat{x}$ for a unique $x \in X$, and $\varphi(f) = \hat{x}(f) = f(x)$. Hence the two topologies coincide: \begin{align*} w_{X^*} = \sigma(X^*, X^{**}) = \sigma(X^*, X) = w^*_{X^*}. \end{align*} By the **Banach-Alaoglu Theorem** applied to $X^*$, $(B_{X^*}, w^*_{X^*})$ is compact. So $(B_{X^*}, w_{X^*})$ is compact. By the (iii) $\Rightarrow$ (i) implication (Step 2) applied to $X^*$, $X^*$ is reflexive. [/step]

[step:Prove (ii) $\Rightarrow$ (i): reflexivity of $X^*$ implies reflexivity of $X$]Assume $X^*$ is reflexive. By the implication (i) $\Rightarrow$ (ii) (Step 3) applied to $X^*$, $X^{**}$ is reflexive. By the implication (i) $\Rightarrow$ (iii) (Step 1) applied to $X^{**}$, $(B_{X^{**}}, w_{X^{**}})$ is compact, where $w_{X^{**}} = \sigma(X^{**}, X^{***})$. Now we transfer this to weak compactness of $(B_X, w)$. Consider $\hat{B}_X = J(B_X) \subseteq B_{X^{**}}$. [claim:$\hat{B}_X$ is $w_{X^{**}}$-closed in $X^{**}$] The set $\hat{B}_X$ is closed in $(X^{**}, w_{X^{**}})$. [/claim] [proof] The set $\hat{B}_X$ is convex: $J$ is linear and $B_X$ is convex, so $J(B_X) = \hat{B}_X$ is convex. The set $\hat{B}_X$ is norm-closed in $X^{**}$: $J$ is an isometry from $X$ onto $\hat{X} := J(X)$, $X$ is complete, so $\hat{X}$ is complete, hence norm-closed in $X^{**}$. The set $B_X$ is norm-closed in $X$, hence $\hat{B}_X = J(B_X)$ is norm-closed in $\hat{X}$, hence in $X^{**}$. For convex sets in a Banach space, norm-closed iff weakly closed (this is **Mazur's theorem**, a standard consequence of Hahn-Banach geometric separation: if a convex set is not weakly closed, separate the convex closure from an external point by a continuous functional, getting a norm half-space containing the set, contradicting norm-closedness). Applied in the Banach space $X^{**}$ with weak topology $w_{X^{**}} = \sigma(X^{**}, X^{***})$: the convex norm-closed set $\hat{B}_X$ is $w_{X^{**}}$-closed. [/proof] By the claim, $\hat{B}_X$ is $w_{X^{**}}$-closed in $X^{**}$, in particular in the compact set $(B_{X^{**}}, w_{X^{**}})$. A closed subset of a compact set is compact: $(\hat{B}_X, w_{X^{**}})$ is compact. The relative topology on $\hat{B}_X$ from $w_{X^{**}}$ corresponds to the weak topology on $B_X$ via $J$: by the same universal-property argument as in Step 1 (with $w_X$ on $X$ and $w_{X^{**}}$ on $X^{**}$, both initial topologies), $J : (B_X, w) \to (\hat{B}_X, w_{X^{**}})$ is a homeomorphism. Concretely, for $\varphi \in X^{***}$, the composition $\varphi \circ J : X \to \mathbb{C}$ is a continuous linear functional (norm-continuity of $J$ plus norm-continuity of $\varphi$), i.e.\ $\varphi \circ J \in X^*$. Conversely, every $f \in X^*$ extends to $X^{**}$ via the bidual embedding $X^* \to X^{***}$, $f \mapsto \tilde{f}$ where $\tilde{f}(\psi) = \psi(f)$, and $\tilde{f} \circ J = f$. So the families of generating functionals match under $J$. Hence $(B_X, w)$ is compact, and by (iii) $\Rightarrow$ (i) (Step 2) applied to $X$, $X$ is reflexive.[/step]

[guided]We chain implications: $X^*$ reflexive $\Rightarrow$ $X^{**}$ reflexive $\Rightarrow$ $(B_{X^{**}}, w)$ compact $\Rightarrow$ $\hat{B}_X$ weakly compact in $X^{**}$ $\Rightarrow$ $(B_X, w)$ compact $\Rightarrow$ $X$ reflexive. The substantive step is showing $\hat{B}_X$ is closed (hence compact) in the weak topology of $X^{**}$, given that we know $X^{**}$ is reflexive (so its weak topology coincides with the weak topology induced by $X^*$ embedded in $X^{***}$). *Why $\hat{B}_X$ is norm-closed in $X^{**}$.* The image $\hat{X} = J(X)$ of a Banach space under an isometry is itself a Banach space (a complete metric subspace of a metric space is closed). So $\hat{X}$ is norm-closed in $X^{**}$, and $\hat{B}_X$ — the image of the closed unit ball $B_X$ under the isometry $J$ — is closed in $\hat{X}$, hence norm-closed in $X^{**}$. *Why norm-closed convex implies weakly closed (Mazur).* This is a hyperplane-separation argument. Given a norm-closed convex set $C$ and a point $\varphi \notin C$, the geometric Hahn-Banach theorem in the Banach space $X^{**}$ produces a norm-continuous linear functional $\Lambda \in X^{***}$ with $\Lambda(\varphi) > \alpha > \sup_C \Lambda$ for some $\alpha$. The half-space $\{\Lambda > \alpha\}$ is a weak neighbourhood of $\varphi$ (by definition of $w_{X^{**}} = \sigma(X^{**}, X^{***})$, every $\Lambda \in X^{***}$ is weakly continuous) disjoint from $C$. So $\varphi$ is not in the weak closure. Contrapositive: every point in the weak closure of $C$ lies in $C$, i.e.\ $C$ is weakly closed. *Why $J$ is a $w$-to-$w_{X^{**}}$ homeomorphism on unit balls.* Both topologies are initial topologies for compatible families of functionals. The weak topology on $X$ is $\sigma(X, X^*)$. The weak topology on $X^{**}$ is $\sigma(X^{**}, X^{***})$. Under $J$, the generating functionals on $X$ are $X^*$, and those on $X^{**}$ restricted to $\hat{X}$ are $X^{***}|_{\hat{X}}$. The pullback of $X^{***}$ to $X$ (via $J$) is exactly $X^*$ — this requires checking that every $\varphi \in X^{***}$ pulled back via $J$ gives an element $\varphi \circ J \in X^*$ (true: composition of norm-continuous maps is norm-continuous), and every $f \in X^*$ arises this way (true: extend $f$ to $X^{**}$ via $\tilde{f}(\psi) := \psi(f)$, an element of $X^{***}$, and $\tilde{f}(J(x)) = J(x)(f) = f(x)$). So the two initial topologies match under $J$, and $J : (B_X, w) \to (\hat{B}_X, w_{X^{**}})$ is a homeomorphism. Compactness of the codomain (closed in compact = compact) transfers to compactness of the domain.[/guided]

[step:Conclude] Steps 1 and 2 prove (i) $\Leftrightarrow$ (iii). Steps 3 and 4 prove (i) $\Leftrightarrow$ (ii). Together, all three statements are equivalent. [/step]