[proofplan] The inclusion $\overline{\hat{B}_X}^{w^*} \subseteq B_{X^{**}}$ follows because $\hat{B}_X \subseteq B_{X^{**}}$ and $B_{X^{**}}$ is $w^*$-closed (its complement is a union of $w^*$-open half-spaces). For the reverse inclusion we argue by contradiction: a hypothetical point $\varphi \in B_{X^{**}} \setminus \overline{\hat{B}_X}^{w^*}$ is separated from the $w^*$-closed convex set $\overline{\hat{B}_X}^{w^*}$ by a $w^*$-continuous functional via the geometric Hahn-Banach theorem in the locally convex space $(X^{**}, w^*)$. Since $w^*$-continuous functionals on $X^{**}$ are precisely evaluations at elements of $X^*$, the separating functional has the form $\hat\Lambda \mapsto \hat\Lambda(f) = \Lambda(f)$ for some $f \in X^*$. The supremum of this functional on $\hat{B}_X$ equals $\|f\|$, while it strictly exceeds $\|f\|$ at $\varphi$, forcing $\|\varphi\| > 1$, contradicting $\varphi \in B_{X^{**}}$. [/proofplan] [step:Show $\overline{\hat{B}_X}^{w^*} \subseteq B_{X^{**}}$] Recall the canonical embedding \begin{align*} J : X &\to X^{**}, \\ x &\mapsto \hat{x}, \quad \hat{x}(f) := f(x) \text{ for } f \in X^*. \end{align*} By the [Hahn-Banach (Normed Space Version)](/theorems/2629), $\|\hat{x}\|_{X^{**}} = \|x\|_X$, so $J$ is an isometry. In particular $\hat{B}_X = J(B_X) \subseteq B_{X^{**}}$. We show $B_{X^{**}}$ is $w^*$-closed. The complement is \begin{align*} X^{**} \setminus B_{X^{**}} = \{\varphi \in X^{**} : \|\varphi\|_{X^{**}} > 1\} = \bigcup_{f \in X^*,\, \|f\| \leq 1} \{\varphi : |\varphi(f)| > 1\}, \end{align*} where the second equality uses $\|\varphi\| = \sup_{\|f\| \leq 1} |\varphi(f)|$ (definition of operator norm on $X^{**} = (X^*)^*$). Each set $\{\varphi : |\varphi(f)| > 1\}$ is $w^*$-open: by definition of $w^* = \sigma(X^{**}, X^*)$, the evaluation $\hat{f} : \varphi \mapsto \varphi(f)$ is $w^*$-continuous for each $f \in X^*$, so the preimage of the open set $\{|z| > 1\} \subseteq \mathbb{C}$ (or $\mathbb{R}$) is $w^*$-open. A union of $w^*$-open sets is $w^*$-open, so $X^{**} \setminus B_{X^{**}}$ is $w^*$-open and $B_{X^{**}}$ is $w^*$-closed. Therefore $\overline{\hat{B}_X}^{w^*} \subseteq \overline{B_{X^{**}}}^{w^*} = B_{X^{**}}$. [/step] [step:Suppose for contradiction $\varphi \in B_{X^{**}} \setminus \overline{\hat{B}_X}^{w^*}$ and prepare separation] Assume there exists $\varphi \in B_{X^{**}}$ with $\varphi \notin \overline{\hat{B}_X}^{w^*}$. We will derive a contradiction by separating $\{\varphi\}$ from $\overline{\hat{B}_X}^{w^*}$ using a $w^*$-continuous functional. The set $\overline{\hat{B}_X}^{w^*}$ is convex: $\hat{B}_X$ is convex (image of the convex set $B_X$ under the linear map $J$), and the closure of a convex set in a topological vector space is convex (by continuity of addition and scalar multiplication, which holds in $(X^{**}, w^*)$ since $w^*$ is a vector topology). The set $\overline{\hat{B}_X}^{w^*}$ is $w^*$-closed (by definition). The singleton $\{\varphi\}$ is $w^*$-compact (a single point is compact). The two sets are disjoint by assumption. [guided] Before invoking separation, we list the structural properties needed. The space $(X^{**}, w^*)$ is a locally convex topological vector space: $w^* = \sigma(X^{**}, X^*)$ is the initial topology generated by the family of linear functionals $\{\hat{f} : f \in X^*\}$ where $\hat{f}(\varphi) = \varphi(f)$. Initial topologies generated by linear functionals on a vector space are vector topologies, and they are locally convex (a sub-base of $w^*$-neighbourhoods of $0$ is given by $\{|\hat{f}_1| < \varepsilon, \ldots, |\hat{f}_n| < \varepsilon\}$, which are convex). Now we check the four hypotheses needed for **separation by a $w^*$-continuous functional** (the geometric Hahn-Banach theorem in locally convex spaces, applied to a compact convex set and a closed convex set): 1. $\{\varphi\}$ is convex (singleton) and $w^*$-compact (singleton). 2. $\overline{\hat{B}_X}^{w^*}$ is convex. - $\hat{B}_X = J(B_X)$ is the image of the convex set $B_X$ under the linear map $J$, hence convex. - The $w^*$-closure of a convex set is convex: addition and scalar multiplication are $w^*$-continuous (any vector topology has this property), and these continuous operations send the closure of a set $A$ into the closure of $A + A$, $\lambda A$, etc. Concretely, given $\psi_1, \psi_2 \in \overline{\hat{B}_X}^{w^*}$ and $t \in [0,1]$, take nets $\psi_1^\alpha \to \psi_1$ and $\psi_2^\alpha \to \psi_2$ in $\hat{B}_X$ (we may pass to the same indexing set by product net); then $t\psi_1^\alpha + (1-t)\psi_2^\alpha \to t\psi_1 + (1-t)\psi_2$ by continuity, and each term is in $\hat{B}_X$ by convexity. So $t\psi_1 + (1-t)\psi_2 \in \overline{\hat{B}_X}^{w^*}$. 3. $\overline{\hat{B}_X}^{w^*}$ is $w^*$-closed by construction. 4. $\{\varphi\} \cap \overline{\hat{B}_X}^{w^*} = \varnothing$ by assumption. [/guided] [/step] [step:Apply Hahn-Banach separation to find a $w^*$-continuous separating functional] We apply the [Hahn-Banach for Locally Convex Spaces](/theorems/2638) (geometric form) to the compact convex set $\{\varphi\}$ and the disjoint $w^*$-closed convex set $\overline{\hat{B}_X}^{w^*}$ in the locally convex space $(X^{**}, w^*)$. There exist a $w^*$-continuous $\mathbb{R}$-linear functional $\Psi : X^{**} \to \mathbb{R}$ and $\alpha \in \mathbb{R}$ such that \begin{align*} \operatorname{Re} \Psi(\varphi) > \alpha > \sup_{\psi \in \overline{\hat{B}_X}^{w^*}} \operatorname{Re} \Psi(\psi). \end{align*} (In the real case, drop $\operatorname{Re}$; in the complex case the geometric separation is initially with respect to $\mathbb{R}$-linear functionals.) By the [Dual of Weak Topology](/theorems/2649), the $w^*$-continuous $\mathbb{C}$-linear (resp.\ $\mathbb{R}$-linear) functionals on $X^{**}$ are precisely the evaluations $\hat{f} : \varphi \mapsto \varphi(f)$ for $f \in X^*$ (resp.\ their real parts in the complex case). Concretely, in the complex case, the $\mathbb{R}$-linear functional $\Psi$ extends uniquely to a $\mathbb{C}$-linear $\widetilde{\Psi}(\varphi) := \Psi(\varphi) - i \Psi(i\varphi)$, which is $w^*$-continuous and so equals $\hat{f}$ for some $f \in X^*$, with $\Psi = \operatorname{Re} \hat{f}$. In either case there exists $f \in X^*$ such that \begin{align*} \operatorname{Re} \varphi(f) > \alpha > \sup_{\psi \in \overline{\hat{B}_X}^{w^*}} \operatorname{Re} \psi(f). \tag{$*$} \end{align*} [guided] We need a $w^*$-continuous functional that separates $\varphi$ from $\overline{\hat{B}_X}^{w^*}$. The crucial property of locally convex spaces — and the reason we work in $(X^{**}, w^*)$ rather than the norm topology — is that the geometric Hahn-Banach theorem produces a separator in the **continuous dual of the topology in use**. Here that means a $w^*$-continuous functional, not a norm-continuous one. We apply the [Hahn-Banach for Locally Convex Spaces](/theorems/2638). Hypotheses: $(X^{**}, w^*)$ is a locally convex space (verified in the previous step), $\{\varphi\}$ is convex compact, $\overline{\hat{B}_X}^{w^*}$ is convex closed, the two sets are disjoint. Conclusion: there exist a $w^*$-continuous $\mathbb{R}$-linear $\Psi : X^{**} \to \mathbb{R}$ and $\alpha \in \mathbb{R}$ strictly separating the two sets. Now we identify $\Psi$ with an evaluation. By the [Dual of Weak Topology](/theorems/2649), the $w^*$-continuous linear functionals on $X^{**}$ are exactly $\{\hat{f} : f \in X^*\}$, where $\hat{f}(\varphi) = \varphi(f)$. (This is the defining property of $w^* = \sigma(X^{**}, X^*)$: the topology is engineered to make evaluations on $X^*$ continuous, and to make these the only continuous linear functionals.) In the real case, $\Psi = \hat{f}$ for some $f \in X^*$, so $(*)$ reads $\varphi(f) > \alpha > \sup_{\psi} \psi(f)$. In the complex case, the $\mathbb{R}$-linear $\Psi$ corresponds via the standard real-complex linearity bijection to a $\mathbb{C}$-linear $\widetilde{\Psi}$ with $\Psi = \operatorname{Re} \widetilde{\Psi}$, and $\widetilde{\Psi}$ is $w^*$-continuous, so $\widetilde{\Psi} = \hat{f}$ for some $f \in X^*$. In either case the separation reads as $(*)$. [/guided] [/step] [step:Compute the supremum of $\operatorname{Re} \hat{f}$ on $\hat{B}_X$ and derive the contradiction] Restricting the supremum in $(*)$ from $\overline{\hat{B}_X}^{w^*}$ to the smaller set $\hat{B}_X \subseteq \overline{\hat{B}_X}^{w^*}$ can only weaken or preserve the strict inequality: \begin{align*} \operatorname{Re} \varphi(f) > \alpha > \sup_{x \in B_X} \operatorname{Re} \hat{x}(f) = \sup_{x \in B_X} \operatorname{Re} f(x). \end{align*} By the [Hahn-Banach (Normed Space Version)](/theorems/2629), \begin{align*} \|f\|_{X^*} = \sup_{x \in B_X} |f(x)|. \end{align*} We claim further $\sup_{x \in B_X} |f(x)| = \sup_{x \in B_X} \operatorname{Re} f(x)$. Indeed, for any $x \in B_X$ and $\theta \in \mathbb{R}$, $e^{i\theta} x \in B_X$ (the unit ball is balanced), and $\operatorname{Re} f(e^{i\theta}x) = \operatorname{Re}(e^{i\theta} f(x))$ achieves $|f(x)|$ at $\theta = -\arg f(x)$. Hence the suprema over $B_X$ of $|f|$ and $\operatorname{Re} f$ coincide. (In the real case, $f(-x) = -f(x)$, so $\sup |f| = \sup f$ on the symmetric ball $B_X$.) Therefore $\sup_{x \in B_X} \operatorname{Re} f(x) = \|f\|_{X^*}$, and \begin{align*} \operatorname{Re} \varphi(f) > \alpha > \|f\|_{X^*}. \end{align*} On the other hand, by the operator norm inequality applied to $\varphi \in X^{**} = (X^*)^*$, \begin{align*} \operatorname{Re} \varphi(f) \leq |\varphi(f)| \leq \|\varphi\|_{X^{**}} \cdot \|f\|_{X^*} \leq 1 \cdot \|f\|_{X^*} = \|f\|_{X^*}, \end{align*} using $\|\varphi\| \leq 1$ since $\varphi \in B_{X^{**}}$. Combined with the previous strict inequality, $\|f\|_{X^*} \geq \operatorname{Re} \varphi(f) > \|f\|_{X^*}$, which is a contradiction. [guided] We use the strict separation $(*)$ together with the formula for the dual norm to derive a contradiction. First, we shrink the supremum domain. Since $\hat{B}_X \subseteq \overline{\hat{B}_X}^{w^*}$, the supremum over the smaller set is no larger: \begin{align*} \sup_{x \in B_X} \operatorname{Re} \hat{x}(f) = \sup_{x \in B_X} \operatorname{Re} f(x) \leq \sup_{\psi \in \overline{\hat{B}_X}^{w^*}} \operatorname{Re} \psi(f) < \alpha < \operatorname{Re} \varphi(f). \end{align*} Second, we identify $\sup_{x \in B_X} \operatorname{Re} f(x) = \|f\|$. By [Hahn-Banach (Normed Space Version)](/theorems/2629), $\|f\| = \sup_{x \in B_X} |f(x)|$. The trick to convert $|f|$ to $\operatorname{Re} f$ is the **balanced rotation argument**: for any $x \in B_X$, the rotated vector $e^{i\theta}x$ also lies in $B_X$ (since $|e^{i\theta}| = 1$), and choosing $\theta = -\arg f(x)$ aligns $f(e^{i\theta}x) = e^{i\theta} f(x) = |f(x)|$, so $\operatorname{Re} f(e^{i\theta}x) = |f(x)|$. Taking the supremum over $x \in B_X$, we recover $\sup_{x \in B_X} \operatorname{Re} f(x) \geq \sup_{x \in B_X} |f(x)| = \|f\|$. The reverse inequality $\sup \operatorname{Re} f \leq \sup |f|$ is automatic. So they are equal. Third, we estimate $\operatorname{Re} \varphi(f)$ from above. The operator norm definition gives $|\varphi(f)| \leq \|\varphi\|_{X^{**}} \|f\|_{X^*}$, and $\operatorname{Re} z \leq |z|$ always. Since $\varphi \in B_{X^{**}}$, $\|\varphi\| \leq 1$. So $\operatorname{Re} \varphi(f) \leq \|f\|$. Combining: $\|f\| \geq \operatorname{Re} \varphi(f) > \|f\|$, a contradiction. The contradiction shows our assumption $\varphi \in B_{X^{**}} \setminus \overline{\hat{B}_X}^{w^*}$ was false. Hence $B_{X^{**}} \subseteq \overline{\hat{B}_X}^{w^*}$. Combined with Step 1, $\overline{\hat{B}_X}^{w^*} = B_{X^{**}}$. [/guided] [/step] [step:Conclude] Steps 1 and 4 together prove $\overline{\hat{B}_X}^{w^*} = B_{X^{**}}$, completing the proof of Goldstein's Theorem. [/step]