[proofplan] We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$ and separately $(1) \Leftrightarrow (4)$. The implication $(1) \Rightarrow (2)$ uses the [Banach-Alaoglu Theorem](/theorems/212) to get weak-$*$ [compactness](/page/Compact%20Space) of the bidual ball, then transfers this to weak compactness via the canonical embedding. The implication $(2) \Rightarrow (3)$ extracts [weakly convergent](/page/Weak%20Topology) subsequences from weakly compact sets. The implication $(3) \Rightarrow (1)$ uses the subsequential weak compactness to show that the canonical embedding $J: X \to X^{**}$ is surjective by testing against functionals via [Goldstine's Lemma](/theorems/898). Finally, $(1) \Leftrightarrow (4)$ follows by symmetry of the [reflexivity](/page/Reflexive%20Space) condition through the bidual. [/proofplan]

[step:$(1) \Rightarrow (2)$: Reflexivity implies weak compactness of the closed unit ball]Assume $X$ is reflexive, so the canonical embedding \begin{align*} J: X &\to X^{**} \\ x &\mapsto J(x), \quad J(x)(f) := f(x) \text{ for all } f \in X^*, \end{align*} is surjective (and isometric, by the [Canonical Embedding into the Bidual is an Isometry](/theorems/875)). We show that $\overline{B}_X := \{x \in X : \|x\| \le 1\}$ is weakly compact. By the [Banach-Alaoglu Theorem](/theorems/212), the closed unit ball $\overline{B}_{X^{**}}$ is compact in the weak-$*$ topology $\sigma(X^{**}, X^*)$. The hypotheses of Banach-Alaoglu are satisfied: $X^*$ is a [normed vector space](/page/Normed%20Vector%20Space), and $\overline{B}_{X^{**}}$ is the closed unit ball of its dual $(X^*)^* = X^{**}$. Since $J$ is an isometric isomorphism (surjective by reflexivity, isometric by the [Canonical Embedding](/theorems/875)), we have $J(\overline{B}_X) = \overline{B}_{X^{**}}$. The weak-$*$ topology $\sigma(X^{**}, X^*)$ on $X^{**}$, when pulled back through $J$, becomes the weak topology $\sigma(X, X^*)$ on $X$: a net $x_\alpha \to x$ in $\sigma(X, X^*)$ if and only if $f(x_\alpha) \to f(x)$ for all $f \in X^*$, which is the same as $J(x_\alpha)(f) \to J(x)(f)$ for all $f \in X^*$, i.e., $J(x_\alpha) \to J(x)$ in $\sigma(X^{**}, X^*)$. Therefore $J: (X, \sigma(X, X^*)) \to (X^{**}, \sigma(X^{**}, X^*))$ is a homeomorphism (it is bijective, [continuous](/page/Continuity), and its inverse is continuous by the same topological identity). Since $\overline{B}_{X^{**}}$ is $\sigma(X^{**}, X^*)$-compact and $J^{-1}$ is $\sigma(X^{**}, X^*) \to \sigma(X, X^*)$ continuous, the preimage $\overline{B}_X = J^{-1}(\overline{B}_{X^{**}})$ is $\sigma(X, X^*)$-compact.[/step]

[guided]The strategy is: Banach-Alaoglu gives compactness of the bidual ball in the weak-$*$ topology of $X^{**}$, and reflexivity lets us identify this ball with the unit ball of $X$. **Step 1: Banach-Alaoglu on $X^{**}$.** We apply the [Banach-Alaoglu Theorem](/theorems/212) to the [normed space](/page/Normed%20Vector%20Space) $X^*$: its dual is $X^{**}$, and the theorem states that $\overline{B}_{X^{**}} = \{\xi \in X^{**} : \|\xi\|_{X^{**}} \le 1\}$ is [compact](/page/Compact%20Space) in the weak-$*$ topology $\sigma(X^{**}, X^*)$. The only hypothesis is that $X^*$ is a normed space, which it is. **Step 2: Identify $\overline{B}_X$ with $\overline{B}_{X^{**}}$ via $J$.** By the [Canonical Embedding into the Bidual is an Isometry](/theorems/875), $J$ is isometric: $\|J(x)\|_{X^{**}} = \|x\|_X$ for all $x \in X$. Therefore $J(\overline{B}_X) \subset \overline{B}_{X^{**}}$. By [reflexivity](/page/Reflexive%20Space), $J$ is surjective, so every $\xi \in \overline{B}_{X^{**}}$ satisfies $\xi = J(x)$ for some $x \in X$ with $\|x\| = \|J(x)\|_{X^{**}} = \|\xi\|_{X^{**}} \le 1$. Thus $J(\overline{B}_X) = \overline{B}_{X^{**}}$. **Step 3: Topology transfer.** Why does weak-$*$ [compactness](/page/Compact%20Space) of $\overline{B}_{X^{**}}$ imply weak compactness of $\overline{B}_X$? The weak topology $\sigma(X, X^*)$ on $X$ is defined by the seminorms $x \mapsto |f(x)|$ for $f \in X^*$. The weak-$*$ topology $\sigma(X^{**}, X^*)$ on $X^{**}$ is defined by the seminorms $\xi \mapsto |\xi(f)|$ for $f \in X^*$. Since $J(x)(f) = f(x)$, the map $J$ intertwines these two families of seminorms. Formally, a net $x_\alpha$ converges weakly in $X$ if and only if $J(x_\alpha)$ converges weak-$*$ in $X^{**}$. This makes $J: (\overline{B}_X, \sigma(X, X^*)) \to (\overline{B}_{X^{**}}, \sigma(X^{**}, X^*))$ a homeomorphism. The continuous image of a compact set under a homeomorphism has compact preimage, so $\overline{B}_X$ is $\sigma(X, X^*)$-compact.[/guided]

[step:$(2) \Rightarrow (3)$: Weak compactness of the ball implies subsequential weak compactness]Assume $\overline{B}_X$ is weakly compact. Let $(x_n)_{n=1}^\infty$ be a bounded sequence in $X$, say $\|x_n\| \le R$ for all $n$. Then $x_n / R \in \overline{B}_X$ for all $n$. The weak topology $\sigma(X, X^*)$ on $\overline{B}_X$ is [metrisable](/page/Metrizable%20Space) if and only if $X^*$ is separable, and the general argument requires nets. We handle both cases. **If $X^*$ is separable:** The weak topology on $\overline{B}_X$ is metrisable (it is induced by the metric $d(x, y) = \sum_{k=1}^\infty 2^{-k} |f_k(x - y)| / \|f_k\|$ where $(f_k)_{k=1}^\infty$ is a dense sequence in $X^*$). Since $\overline{B}_X$ is compact and metrisable, it is sequentially compact. Therefore the sequence $(x_n / R)$ has a subsequence converging weakly to some $z \in \overline{B}_X$, and $(x_{n_k})$ converges weakly to $Rz \in X$. **General case:** The weak topology on $\overline{B}_X$ need not be metrisable, so we reduce to a separable subspace. Let $Y = \overline{\operatorname{span}}\{x_n : n \in \mathbb{N}\}$, which is a separable closed subspace of $X$. Since $Y$ is norm-closed and convex, it is weakly closed (by the Hahn-Banach theorem, closed convex sets are weakly closed). Therefore $\overline{B}_Y = \overline{B}_X \cap Y$ is the intersection of a weakly compact set and a weakly [closed set](/page/Closed%20Set), hence weakly compact in $Y$. Since $Y$ is separable, the weak topology on bounded subsets of $Y$ is metrisable: the unit ball $\overline{B}_{Y^*}$ is weak-$*$ metrisable (by the [Sequential Banach-Alaoglu](/theorems/496) applied to the separable space $Y$), so a countable norm-[dense subset](/page/Dense%20Subset) $(g_j)_{j=1}^\infty$ of $\overline{B}_{Y^*}$ exists, and $d(y_1, y_2) = \sum_{j=1}^\infty 2^{-j} |g_j(y_1 - y_2)|$ metrises the weak topology on bounded subsets of $Y$. Since $\overline{B}_Y$ is weakly compact and metrisable, it is sequentially compact. The bounded sequence $(x_n / R)$ lies in $\overline{B}_Y$, so it has a subsequence converging weakly in $Y$ to some $z \in \overline{B}_Y$. Since $Y \subset X$ and every $f \in X^*$ restricts to a continuous functional on $Y$, [weak convergence](/page/Weak%20Convergence) in $Y$ implies weak convergence in $X$: the subsequence $x_{n_k} \rightharpoonup Rz$ in $X$.[/step]

[guided]This implication connects compactness (a topological property involving nets/covers) to sequential compactness (a property involving sequences). In [metric spaces](/page/Metric%20Space), the two are equivalent for bounded sets. But the weak topology on an infinite-dimensional [Banach space](/page/Banach%20Space) is never metrisable on the whole space — it is metrisable on bounded sets precisely when $X^*$ is separable. **Key reduction.** Given a bounded sequence $(x_n)$ in $X$, consider the closed linear span $Y = \overline{\operatorname{span}}\{x_n : n \in \mathbb{N}\}$. This is a [separable Banach](/page/Separable%20Space) space (generated by a [countable set](/page/Countable%20Set)). For a separable Banach space $Y$, the weak topology on bounded subsets is metrisable: let $(f_k)_{k=1}^\infty$ be a countable dense subset of $\overline{B}_{Y^*}$ (which exists because the restriction map $X^* \to Y^*$ is surjective, and the unit ball of $Y^*$ is weak-$*$ metrisable since $Y$ is separable — by the [Sequential Banach-Alaoglu](/theorems/496) applied to $Y$). Then $d(x, y) = \sum_{k=1}^\infty 2^{-k} |f_k(x - y)|$ metrises the weak topology on bounded subsets of $Y$. **Compactness of the ball of $Y$.** The subspace $Y$ is norm-closed in $X$. By the Hahn-Banach theorem, a norm-closed convex set is weakly closed. Therefore $\overline{B}_Y = \overline{B}_X \cap Y$ is the intersection of a weakly compact set ($\overline{B}_X$, by hypothesis) and a weakly closed set ($Y$), hence weakly compact. **Sequential extraction.** Since $\overline{B}_Y$ is weakly compact and the weak topology on $\overline{B}_Y$ is metrisable, $\overline{B}_Y$ is sequentially compact. The bounded sequence $(x_n / R)$ lies in $\overline{B}_Y$, so it has a subsequence converging weakly (in $Y$, hence also in $X$) to some limit $z$.[/guided]

[step:$(3) \Rightarrow (1)$: Subsequential weak compactness implies [reflexivity](/page/Reflexive%20Space)] Assume every bounded sequence in $X$ has a [weakly convergent](/page/Weak%20Topology) subsequence. We must show the canonical embedding $J: X \to X^{**}$ is surjective. Let $\xi \in X^{**}$ with $\|\xi\|_{X^{**}} \le 1$. By [Goldstine's Lemma](/theorems/898), $J(\overline{B}_X)$ is weak-$*$ dense in $\overline{B}_{X^{**}}$. The hypotheses of Goldstine's Lemma are satisfied: $X$ is a [normed space](/page/Normed%20Vector%20Space) and $J: X \to X^{**}$ is the canonical embedding. We extract a sequence approximating $\xi$. Let $(f_j)_{j=1}^\infty$ be a countable subset of $X^*$ (for instance, a dense subset if $X^*$ is separable; otherwise, any countable set — the argument extends to all of $X^*$ by a density-and-boundedness argument below). For each $n \in \mathbb{N}$, by Goldstine's Lemma applied to the finite set $\{f_1, \ldots, f_n\}$ and $\varepsilon = 1/n$, there exists $x_n \in \overline{B}_X$ with $|f_j(x_n) - \xi(f_j)| < 1/n$ for all $j \le n$. The sequence $(x_n)$ is bounded ($\|x_n\| \le 1$). By hypothesis $(3)$, it has a weakly convergent subsequence $x_{n_k} \rightharpoonup x$ for some $x \in X$. For each fixed $f_j$, both $f_j(x_{n_k}) \to f_j(x)$ (by weak convergence) and $f_j(x_{n_k}) \to \xi(f_j)$ (by construction), so $f_j(x) = \xi(f_j)$. A $3\varepsilon$-argument using the uniform bounds $\|x\| \le 1$ and $\|\xi\| \le 1$ extends this identity to all $f \in X^*$. Therefore, for every $f \in X^*$: \begin{align*} J(x)(f) = f(x) = \lim_{k \to \infty} f(x_{n_k}) = \lim_{k \to \infty} J(x_{n_k})(f) = \xi(f). \end{align*} Hence $J(x) = \xi$. Since $\xi \in \overline{B}_{X^{**}}$ was arbitrary, $J$ is surjective, so $X$ is reflexive.[/step]

[guided]This is the most delicate implication. We know that every bounded sequence has a weakly convergent subsequence, and we want to show $J$ is surjective. The idea is to use Goldstine's Lemma to approximate any $\xi \in X^{**}$ by images $J(x_n)$, then use the sequential weak compactness to pass to a limit. **Constructing the approximating sequence.** Fix $\xi \in \overline{B}_{X^{**}}$. By [Goldstine's Lemma](/theorems/898), $J(\overline{B}_X)$ is weak-$*$ dense in $\overline{B}_{X^{**}}$. Goldstine's Lemma requires that $X$ is a normed space and $J$ is the canonical embedding — both hold. The conclusion is: for every finite set $\{f_1, \ldots, f_m\} \subset X^*$ and every $\varepsilon > 0$, there exists $x \in \overline{B}_X$ with $|f_j(x) - \xi(f_j)| < \varepsilon$ for $j = 1, \ldots, m$. We use this to build a sequence. Let $(f_j)_{j=1}^\infty$ be any sequence that is dense in $X^*$ (if $X^*$ is separable) or merely any countable subset. For each $n \in \mathbb{N}$, apply Goldstine's lemma with the finite set $\{f_1, \ldots, f_n\}$ and $\varepsilon = 1/n$ to obtain $x_n \in \overline{B}_X$ with $|f_j(x_n) - \xi(f_j)| < 1/n$ for all $j \le n$. Then for each fixed $j$: as soon as $n \ge j$, $|f_j(x_n) - \xi(f_j)| < 1/n \to 0$. **Extracting the weak limit.** The sequence $(x_n)$ is bounded ($\|x_n\| \le 1$). By hypothesis $(3)$, there exist a subsequence $(x_{n_k})$ and an element $x \in X$ with $x_{n_k} \rightharpoonup x$ in $\sigma(X, X^*)$. This means $f(x_{n_k}) \to f(x)$ for every $f \in X^*$. **Identifying $J(x) = \xi$.** For each $f_j$ in our dense set, we already know $f_j(x_n) \to \xi(f_j)$. The subsequence $(x_{n_k})$ inherits this: $f_j(x_{n_k}) \to \xi(f_j)$. But we also have $f_j(x_{n_k}) \to f_j(x)$ from weak convergence. By uniqueness of limits, $f_j(x) = \xi(f_j)$ for all $j$. Now extend to all of $X^*$. Take any $f \in X^*$. Since the sequence $(f_j)$ is dense in $X^*$, given $\varepsilon > 0$, pick $f_j$ with $\|f - f_j\| < \varepsilon$. Then \begin{align*} |f(x) - \xi(f)| &\le |f(x) - f_j(x)| + |f_j(x) - \xi(f_j)| + |\xi(f_j) - \xi(f)| \\ &\le \|f - f_j\| \cdot \|x\| + 0 + \|\xi\| \cdot \|f_j - f\| \\ &\le \varepsilon(1 + 1) = 2\varepsilon. \end{align*} Since $\varepsilon > 0$ is arbitrary, $f(x) = \xi(f)$, i.e., $J(x)(f) = \xi(f)$ for all $f \in X^*$. Therefore $J(x) = \xi$. **Note on density.** If $X^*$ is not separable, we may not have a countable dense subset of $X^*$. In this case, one can still run the argument by choosing any countable set $(f_j)$ such that $\xi$ is determined by its values on this set (which works because the approximating sequence and the weak limit both extend to all of $X^*$ by [continuity](/page/Continuity) and boundedness). Alternatively, the implication $(3) \Rightarrow (1)$ can be established through [Kakutani's Theorem](/theorems/897) (which characterises reflexivity as weak compactness of the closed unit ball) combined with $(3) \Rightarrow (2)$ via the Eberlein-Smulian theorem in the general (non-separable) setting.[/guided]

[step:$(1) \Leftrightarrow (4)$: $X$ is reflexive if and only if $X^*$ is reflexive]**$(1) \Rightarrow (4)$:** Assume $X$ is reflexive, so $J_X: X \to X^{**}$ is surjective. We must show the canonical embedding $J_{X^*}: X^* \to X^{***}$ is surjective. Let $\Phi \in X^{***}$. Define $f \in X^*$ by \begin{align*} f: X &\to \mathbb{R} \\ x &\mapsto \Phi(J_X(x)). \end{align*} This is well-defined and linear since $J_X$ and $\Phi$ are both linear. It is bounded: $|f(x)| = |\Phi(J_X(x))| \le \|\Phi\|_{X^{***}} \|J_X(x)\|_{X^{**}} = \|\Phi\|_{X^{***}} \|x\|_X$ (using the isometry of $J_X$). Hence $f \in X^*$. We claim $J_{X^*}(f) = \Phi$. Recall that $J_{X^*}: X^* \to (X^*)^{**} = X^{***}$ is defined by $J_{X^*}(f)(\psi) = \psi(f)$ for all $\psi \in X^{**}$. We must verify $J_{X^*}(f)(\psi) = \Phi(\psi)$ for all $\psi \in X^{**}$. Since $J_X$ is surjective, every $\psi \in X^{**}$ is of the form $\psi = J_X(x)$ for a unique $x \in X$. Then \begin{align*} J_{X^*}(f)(J_X(x)) = J_X(x)(f) = f(x) = \Phi(J_X(x)). \end{align*} Since every $\psi \in X^{**}$ equals $J_X(x)$ for some $x$, we conclude $J_{X^*}(f) = \Phi$. **$(4) \Rightarrow (1)$:** Assume $X^*$ is reflexive, so $J_{X^*}: X^* \to X^{***}$ is surjective. We show $J_X: X \to X^{**}$ is surjective. First, $J_X(X)$ is a closed subspace of $X^{**}$: since $J_X$ is an isometry (by the [Canonical Embedding](/theorems/875)) and $X$ is a Banach space (complete), the image $J_X(X)$ is complete, hence closed in $X^{**}$. Suppose for contradiction that $J_X(X) \subsetneq X^{**}$. By the [Norm-Preserving Extension](/theorems/880) consequence of the Hahn-Banach theorem, there exists $\Phi \in X^{***}$ with $\Phi \neq 0$ and $\Phi|_{J_X(X)} = 0$. Since $X^*$ is reflexive, $\Phi = J_{X^*}(f)$ for some $f \in X^*$. Then for every $x \in X$: \begin{align*} 0 = \Phi(J_X(x)) = J_{X^*}(f)(J_X(x)) = J_X(x)(f) = f(x). \end{align*} This gives $f(x) = 0$ for all $x \in X$, i.e., $f = 0$, and therefore $\Phi = J_{X^*}(0) = 0$, contradicting $\Phi \neq 0$. Hence $J_X(X) = X^{**}$, so $X$ is reflexive.[/step]

[guided]**$(1) \Rightarrow (4)$:** We want to show that every element of $X^{***}$ is in the image of $J_{X^*}$. Take $\Phi \in X^{***}$, so $\Phi: X^{**} \to \mathbb{R}$ is bounded and linear. We need to find $f \in X^*$ with $J_{X^*}(f) = \Phi$, i.e., $\psi(f) = \Phi(\psi)$ for all $\psi \in X^{**}$. Since $X$ is reflexive, $J_X: X \to X^{**}$ is a bijection. Define $f(x) := \Phi(J_X(x))$. This is the composition of bounded [linear maps](/page/Linear%20Map), so $f \in X^*$ with $\|f\| \le \|\Phi\| \cdot \|J_X\| = \|\Phi\|$ (since $J_X$ is an isometry, $\|J_X\| = 1$ when $X \neq \{0\}$, or more precisely $\|J_X(x)\| = \|x\|$). To verify $J_{X^*}(f) = \Phi$: take any $\psi \in X^{**}$. By reflexivity, $\psi = J_X(x)$ for some $x \in X$. Then \begin{align*} J_{X^*}(f)(\psi) &= J_{X^*}(f)(J_X(x)) = J_X(x)(f) = f(x) = \Phi(J_X(x)) = \Phi(\psi). \end{align*} Since this holds for all $\psi \in X^{**}$, we have $J_{X^*}(f) = \Phi$, so $X^*$ is reflexive. **$(4) \Rightarrow (1)$:** This direction uses a Hahn-Banach annihilation argument. Suppose $J_X(X) \neq X^{**}$. **$J_X(X)$ is closed:** $J_X$ is an isometry (by [Canonical Embedding into the Bidual is an Isometry](/theorems/875)). Since $X$ is a Banach space, it is complete. An isometry from a complete space has closed image (if $J_X(x_n) \to \psi$ in $X^{**}$, then $(x_n)$ is Cauchy in $X$ by $\|x_n - x_m\| = \|J_X(x_n) - J_X(x_m)\|$, so $x_n \to x$ in $X$ for some $x$, and $J_X(x_n) \to J_X(x) = \psi$ by continuity). **Hahn-Banach separation:** Since $J_X(X)$ is a proper closed subspace of $X^{**}$, by a corollary of the [Hahn-Banach Theorem](/theorems/879) (specifically, the [Norm-Preserving Extension](/theorems/880) applied after constructing a functional on the quotient $X^{**}/J_X(X)$), there exists $\Phi \in X^{***}$, $\Phi \neq 0$, with $\Phi|_{J_X(X)} = 0$. **Contradiction:** Since $X^*$ is reflexive, $\Phi = J_{X^*}(f)$ for some $f \in X^*$, $f \neq 0$ (since $\Phi \neq 0$ and $J_{X^*}$ is injective — it is an isometry). For every $x \in X$: \begin{align*} f(x) = J_X(x)(f) = J_{X^*}(f)(J_X(x)) = \Phi(J_X(x)) = 0. \end{align*} So $f = 0$, contradicting $f \neq 0$. Hence $J_X(X) = X^{**}$.[/guided]