Reflexivity and Eberlein Smulian

For every $f\in X^{\ast }$ and $x\in X$, $(x^{\ast \ast }\mapsto x^{\ast \ast }(f))(J_{X}x)=f(x)$. Consequently the pullback by $J_X$ of the weak-star initial topology on $X^{\ast \ast }$ is exactly the weak initial topology on $X$: both are generated by the same family $\{f:f\in X^{\ast }\}$. Since [F2] makes $J_X$ injective, it is a homeomorphism from weak $X$ onto $J_X(X)$ with the relative weak-star topology. This also covers $X=\{0\}$, when both spaces are singletons.

Isometry gives $J_X(B_X)=J_X(X)\cap B_{X^{\ast \ast }}$. Indeed $\parallel J_{X}x\parallel =\parallel x\parallel$, so both inclusions include the closed boundary $\parallel x\parallel =1$; if $X=0$, both sides are the singleton $\{0\}$.

Suppose first that $X$ is reflexive. By [F1], $J_X(X)=X^{\ast \ast }$, so step 1.2 gives $J_X(B_X)=B_{X^{\ast \ast }}$. Apply Banach–Alaoglu to the normed space $X^{\ast }$: under the ultrafilter lemma its dual ball $B_{X^{\ast \ast }}$ is weak-star compact. The homeomorphism in step 1.1 therefore transfers this compactness to weak $B_X$. This is the only direction, and the only step, that spends the ultrafilter lemma; in the selected Alaoglu proof it enters through [F9].

Conversely, suppose that $B_X$ is weakly compact. Step 1.1 makes the restriction $J_X{\mathrm{\mid }}_{B_X}$ weak-to-weak-star continuous, so [F6] makes $J_X(B_X)$ weak-star compact. The weak-star topology on $X^{\ast \ast }$ is Hausdorff by [F4], and hence [F7] makes $J_X(B_X)$ weak-star closed in $X^{\ast \ast }$. No compactness choice principle is used in this reverse implication: its compact set is the one in the hypothesis.

Goldstine [F8] says that $J_X(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$. It is contained in that ball by step 1.2 and is weak-star closed by step 2.2. Therefore $J_X(B_X)=B_{X^{\ast \ast }}$. This uses topological closure, not merely sequential closure.

Let $x^{\ast \ast }\in X^{\ast \ast }$. If $x^{\ast \ast }=0$, then $x^{\ast \ast }=J_{X}0$. If $x^{\ast \ast }\mathrm{\ne }0$, put $u^{\ast \ast }=x^{\ast \ast }\mathrm{/}\parallel x^{\ast \ast }\parallel \in B_{X^{\ast \ast }}$. Step 3.1 supplies $u\in B_X$ with $J_{X}u=u^{\ast \ast }$; scalar linearity then gives $x^{\ast \ast }=J_X(\parallel x^{\ast \ast }\parallel u)$. Thus $J_X$ is onto, so $X$ is reflexive by [F1]. This normalization treats the zero and norm-one endpoints separately and selects only one witness for the supplied $x^{\ast \ast }$, not a family of witnesses.

Steps 2.1 and 4.1 prove the two implications. HB is used through the canonical isometry [F2] and Goldstine [F8], and [F10] records exactly which additional principle that name denotes. The ultrafilter lemma is used only through Alaoglu in step 2.1; the reverse implication is choice-free once its weak compactness hypothesis and HB-backed Goldstine are supplied.

If $X=\{0\}$, every scalar-linear functional on $X$ is zero, so $X^{\ast }=X^{\ast \ast }=X^{\ast \ast \ast }=\{0\}$ and both canonical maps are surjective. The equivalence therefore holds in the zero-space case.

Suppose first that $X$ is reflexive. Let $\mathrm{\Lambda }\in X^{\ast \ast \ast }$ and define $f=\mathrm{\Lambda }\circ J_X$. Scalar linearity of the two maps makes $f$ scalar-linear, and [F2] gives $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }{J}_{X}x\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$, so $f\in X^{\ast }$.

Conversely, suppose that $X^{\ast }$ is reflexive and put $W=J_X(X)\subseteq X^{\ast \ast }$. The image $W$ is a scalar-linear subspace by [F2]. It is complete: if $(J_X{x}_n)$ is Cauchy in its restricted norm, then $\mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }=\mathrm{\parallel }{J}_X{x}_n-J_X{x}_m\mathrm{\parallel }$ makes $(x_n)$ Cauchy in the Banach space $X$; for its limit $x$, the same equality gives $J_X{x}_n\to J_{X}x$ in $W$.

For arbitrary $x^{\ast \ast }\in X^{\ast \ast }$, reflexivity of $X$ supplies an $x\in X$ with $x^{\ast \ast }=J_{X}x$. Then $J_{X^{\ast }}(f)(x^{\ast \ast })=x^{\ast \ast }(f)=J_{X}x(f)=f(x)=\mathrm{\Lambda }(J_{X}x)=\mathrm{\Lambda }(x^{\ast \ast })$. Thus $J_{X^{\ast }}(f)=\mathrm{\Lambda }$. Since $\mathrm{\Lambda }$ was arbitrary, $J_{X^{\ast }}$ is surjective and $X^{\ast }$ is reflexive. This chooses only one representing vector for one arbitrary $x^{\ast \ast }$ at a time.

Apply [F3] under the assumed ${\mathrm{A}\mathrm{C}}_{\omega }$. The complete subspace $W$ is closed in $X^{\ast \ast }$. It is also nonempty and convex because it is a linear subspace.

Suppose for contradiction that some $z\in X^{\ast \ast }\setminus W$ exists. By [F4] there is a nonzero $\mathrm{\Gamma }\in X^{\ast \ast \ast }$ that strictly separates the point $z$ from $W$. In particular, $\mathrm{Re}\mathrm{\Gamma }$ is bounded above on $W$. For $w\in W$ and every real $t$, also $tw\in W$; boundedness of $t\mathrm{Re}\mathrm{\Gamma }(w)$ for all $t\in \mathbb{R}$ forces $\mathrm{Re}\mathrm{\Gamma }(w)=0$. In the complex case $iw\in W$ as well, so $0=\mathrm{Re}\mathrm{\Gamma }(iw)=-\mathrm{Im}\mathrm{\Gamma }(w)$; hence in either field $\mathrm{\Gamma }{\mathrm{\mid }}_W=0$.

Reflexivity of $X^{\ast }$ supplies $f\in X^{\ast }$ with $\mathrm{\Gamma }=J_{X^{\ast }}(f)$. For every $x\in X$, step 3.1 yields $0=\mathrm{\Gamma }(J_{X}x)=J_{X}x(f)=f(x)$. Thus $f=0$, whence $\mathrm{\Gamma }=J_{X^{\ast }}(0)=0$, contradicting the nonzero separator in step 3.1.

No point of $X^{\ast \ast }$ lies outside $W$, so $J_X(X)=X^{\ast \ast }$ and [F1] says that $X$ is reflexive. This proves the reverse implication and hence the equivalence. HB is used only in the isometry [F2] and separation [F4]; ${\mathrm{A}\mathrm{C}}_{\omega }$ is used only in step 2.2 through [F3].

Let $R:X^{\ast }\to Y^{\ast }$ be restriction, $R(f)=f{\mathrm{\mid }}_Y$. It is scalar-linear and bounded with $\parallel Rf\parallel \le \parallel f\parallel$. Hence for a supplied $y^{\ast \ast }\in Y^{\ast \ast }$ the composite $x^{\ast \ast }=y^{\ast \ast }\circ R$ belongs to $X^{\ast \ast }$. Since $X$ is reflexive, [F1] supplies $x\in X$ such that $J_{X}x=x^{\ast \ast }$, meaning $f(x)=y^{\ast \ast }(f{\mathrm{\mid }}_Y)$ for every $f\in X^{\ast }$.

If $f\in Y^{\perp }$, then $f{\mathrm{\mid }}_Y=0$, so step 1.1 gives $f(x)=y^{\ast \ast }(0)=0$. Thus every functional annihilating $Y$ also annihilates $x$.

We claim $x\in Y$. This is immediate if $Y=X$. Otherwise suppose $x\mathrm{\notin }Y$; then $Y$ is a nonempty closed convex set and [F3] supplies $0\mathrm{\ne }f\in X^{\ast }$, $a\in \mathbb{R}$, and $\epsilon >0$ with $\mathrm{Re}f(y)\le a-\epsilon <a+\epsilon \le \mathrm{Re}f(x)$ for every $y\in Y$. Since $ty\in Y$ for every real $t$, the real-linear function $\mathrm{Re}f$ can be bounded above on the line $\mathbb{R}y$ only when $\mathrm{Re}f(y)=0$. In the complex case applying this also to $iy\in Y$ gives $\mathrm{Re}f(iy)=-\mathrm{Im}f(y)=0$, so in either field $f{\mathrm{\mid }}_Y=0$. Taking $y=0$ in the separation inequality gives $0\le a-\epsilon$ and hence $\mathrm{Re}f(x)\ge a+\epsilon \ge 2\epsilon >0$, contradicting step 2.1. Therefore $x\in Y$.

Let $g\in Y^{\ast }$ be arbitrary. By norm-preserving Hahn–Banach [F6], one functional $f\in X^{\ast }$ extends $g$; this also covers $g=0$ and $Y=\{0\}$. Steps 1.1 and 3.1 then give $y^{\ast \ast }(g)=y^{\ast \ast }(f{\mathrm{\mid }}_Y)=f(x)=g(x)=(J_{Y}x)(g)$. Hence $y^{\ast \ast }=J_{Y}x$.

The closed-subspace theorem [F4] makes $Y$ a Banach space. Since the arbitrary $y^{\ast \ast }\in Y^{\ast \ast }$ of step 1.1 lies in the range of $J_Y$ by step 4.1, that canonical map is surjective, and [F1] makes $Y$ reflexive. If $Y=0$, its bidual and all maps above are zero and the same argument gives the singleton range directly.

HB is used exactly twice: geometric separation in step 3.1 and the extension of one supplied $g$ in step 4.1; [F5] records the principle being assumed. No compactness principle or simultaneous family choice occurs.

Put $Z=X\mathrm{/}Y$ and write $q:X\to Z$ for the quotient map. By [F4], under the assumed ${\mathrm{A}\mathrm{C}}_{\omega }$ the normed quotient $Z$ is Banach. This includes $Y=X$, when $Z=\{0\}$, and $Y=\{0\}$, when the quotient norm is the original norm.

Let $z^{\ast \ast }\in Z^{\ast \ast }$ be arbitrary. The isometric bijection $Q:Z^{\ast }\to Y^{\perp }$ from [F2] has a scalar-linear isometric inverse. Define $g:Y^{\perp }\to \mathbb{K}$ by $g(u)=z^{\ast \ast }(Q^{-1}u)$. It is a bounded scalar-linear functional with $\mathrm{\parallel }g\mathrm{\parallel }=\mathrm{\parallel }{z}^{\ast \ast }\mathrm{\parallel }$; if $Y^{\perp }=\{0\}$, both sides are zero.

Apply [F3] under HB to the subspace $Y^{\perp }\subseteq X^{\ast }$. There is $x^{\ast \ast }\in X^{\ast \ast }$ with $x^{\ast \ast }{\mathrm{\mid }}_{Y^{\perp }}=g$ and $\mathrm{\parallel }{x}^{\ast \ast }\mathrm{\parallel }=\mathrm{\parallel }g\mathrm{\parallel }$. Only this one supplied functional is extended; no family of extensions is chosen.

Reflexivity of $X$ supplies an $x\in X$ with $x^{\ast \ast }=J_{X}x$. Put $z=q(x)\in Z$.

For every $h\in Z^{\ast }$, [F2] gives $Qh=h\circ q\in Y^{\perp }$, and therefore $J_{Z}z(h)=h(qx)=Qh(x)=J_{X}x(Qh)=x^{\ast \ast }(Qh)=g(Qh)=z^{\ast \ast }(h)$. Hence $J_{Z}z=z^{\ast \ast }$. The calculation is scalar-linear over both fields and uses the bilinear evaluation convention, with no conjugation.

Since $z^{\ast \ast }\in Z^{\ast \ast }$ was arbitrary, $J_Z$ is surjective; together with the Banach conclusion in step 1.1, [F1] shows that $Z=X\mathrm{/}Y$ is reflexive. When $Y=X$, step 1.2 starts from the unique zero bidual functional and the same computation gives the zero representer; when $Y=0$, $Q$ is the usual identification and the computation reduces to ambient reflexivity. HB is spent only in step 2.1, and ${\mathrm{A}\mathrm{C}}_{\omega }$ only in step 1.1.

Regard real $L^p(\mu )$ as the real-valued subspace of complex $L^p(\mu ;\mathbb{C})$. The maps $A(u)=\mathrm{Re}\mathrm{\Lambda }(u)$ and $B(u)=\mathrm{Im}\mathrm{\Lambda }(u)$ are bounded real-linear functionals there, with $\mathrm{\mid }A(u)\mathrm{\mid },\mathrm{\mid }B(u)\mathrm{\mid }\le \parallel \mathrm{\Lambda }\parallel \parallel u{\parallel }_p$. Applying [F1] twice gives real $a,b\in L^q(\mu )$ such that $A(u)=\int ua$ and $B(u)=\int ub$ for every real $u\in L^p$. Put $h=a+ib\in L^q(\mu ;\mathbb{C})$; component inequalities in [F3] make its $q$-norm finite.

For real-valued $u$, componentwise complex integration gives $\mathrm{\Lambda }(u)=A(u)+iB(u)=\int u(a+ib)=\int uh$. If $f=u+iv$ is an arbitrary complex $L^p$ class, [F3] puts its real and imaginary parts in real $L^p$, and complex linearity gives $\mathrm{\Lambda }(f)=\mathrm{\Lambda }(u)+i\mathrm{\Lambda }(v)=\int uh+i\int vh=\int fh$. All identities depend only on a.e. classes by the quotient and integration conventions in [F2]–[F3].

Hölder [F3] gives $\mathrm{\mid }\int fh\mathrm{\mid }\le \parallel f{\parallel }_p\parallel h{\parallel }_q$, hence $\parallel \mathrm{\Lambda }\parallel \le \parallel h{\parallel }_q$. If $h=0$ a.e., step 2.1 gives $\mathrm{\Lambda }=0$ and equality follows. Otherwise define $v=0$ on $\{h=0\}$ and $v=\mathrm{\mid }h{\mathrm{\mid }}^{q-2}\bar{h}$ where $h\mathrm{\ne }0$. Then $\mathrm{\mid }v\mathrm{\mid }=\mathrm{\mid }h{\mathrm{\mid }}^{q-1}$ and $vh=\mathrm{\mid }h{\mathrm{\mid }}^q$ pointwise. Since $(q-1)p=q$, [F2]–[F3] give $v\in L^p$, $\parallel v{\parallel }_p=\parallel h{\parallel }_q^{q-1}$, and $\mathrm{\Lambda }(v)=\int \mathrm{\mid }h{\mathrm{\mid }}^q=\parallel h{\parallel }_q^q$. Testing on $v\mathrm{/}\parallel v{\parallel }_p$ proves $\parallel \mathrm{\Lambda }\parallel \ge \parallel h{\parallel }_q$, including the closed unit-norm endpoint.

If $k\in L^q(\mu ;\mathbb{C})$ gives the same pairing functional, put $d=h-k$. Then $\int fd=0$ for every $f\in L^p$. If $d$ were nonzero, the phase test of step 3.1 with $d$ in place of $h$ would produce $v\in L^p$ with $\int vd=\parallel d{\parallel }_q^q>0$, a contradiction. Thus $d=0$ in $L^q$, so the density is unique.

Steps 2.1–4.1 prove existence, equality of norms, and uniqueness. Countable Choice is used only inside the real arbitrary-measure representation [F1], whose construction makes countably many local choices; applying that theorem to $A$ and $B$ requires only two instances and no stronger choice principle. The empty and zero-measure spaces have only zero $L^p$ classes and are covered by the $h=0$ branch of step 3.1; the forbidden endpoints $p=1,\mathrm{\infty }$ never enter because $1<p,q<\mathrm{\infty }$.

First verify the Banach condition. Real $L^p$ is complete by [F4]. If $(f_n)$ is Cauchy in complex $L^p$, [F5] makes $(\mathrm{Re}{f}_n)$ and $(\mathrm{Im}{f}_n)$ Cauchy in real $L^p$; [F4] gives limits $u,v\in L^p(\mu ;\mathbb{R})$. The upper component bound in [F5] gives $\parallel f_n-(u+iv){\parallel }_p\le \parallel \mathrm{Re}{f}_n-u{\parallel }_p+\parallel \mathrm{Im}{f}_n-v{\parallel }_p\to 0$. Thus complex $L^p$ is complete as well, including the zero and empty measure spaces.

Fix either scalar field $\mathbb{K}$ and write $E=L^p(\mu ;\mathbb{K})$ and $H=L^q(\mu ;\mathbb{K})$. By [F2] in the real case and [F3] in the complex case, the map $T_q:H\to E^{\ast }$ defined by $(T_{q}h)(f)=\int fhd\mu$ is a scalar-linear isometric bijection. The same theorem with $q$ in place of $p$ identifies $H^{\ast }$ isometrically with $L^p=E$ by the same bilinear formula.

Let $\mathrm{\Phi }\in E^{\ast \ast }$. Since $T_q$ is a bounded linear map, $\mathrm{\Phi }\circ T_q$ lies in $H^{\ast }$. The $q$-duality assertion in step 1.2 therefore supplies $u\in E$ such that $\mathrm{\Phi }(T_{q}h)=\int hud\mu$ for every $h\in H$. This includes $\mathrm{\Phi }=0$, for which uniqueness gives $u=0$.

Given any $\mathrm{\ell }\in E^{\ast }$, surjectivity of $T_q$ supplies $h\in H$ with $\mathrm{\ell }=T_{q}h$. Commutativity of scalar multiplication and the bilinear pairing then gives $\mathrm{\Phi }(\mathrm{\ell })=\mathrm{\Phi }(T_{q}h)=\int hu=\int uh=(T_{q}h)(u)=\mathrm{\ell }(u)=(J_{E}u)(\mathrm{\ell })$. Hence $\mathrm{\Phi }=J_{E}u$.

Every $\mathrm{\Phi }\in E^{\ast \ast }$ is therefore in the range of $J_E$. Step 1.1 makes $E$ Banach, so [F1] proves reflexivity in both scalar fields. The proof uses Countable Choice only through the completeness and arbitrary-measure duality suppliers cited in steps 1.1–1.2; [F6] records that exact assumption. No Hahn–Banach or compactness principle is additionally invoked. The argument requires both $p$ and $q$ to lie strictly between one and infinity, so it makes no endpoint claim.

Let ${\mathbb{Q}}_{\mathbb{K}}=\mathbb{Q}$ in the real case and $\mathbb{Q}+i\mathbb{Q}$ in the complex case, with the canonical embeddings into the scalar field understood. By [F2], ${\mathbb{Q}}_{\mathbb{K}}$ is at most countable: in the complex case it is the image of the countable product $\mathbb{Q}\times \mathbb{Q}$. It is nonempty, so fix one surjection $q:\mathbb{N}\to {\mathbb{Q}}_{\mathbb{K}}$. This is one instantiation of the countability theorem, not a countable family of choices.

The scalar set ${\mathbb{Q}}_{\mathbb{K}}$ is dense in $\mathbb{K}$. This is [F3] over $\mathbb{R}$. Over $\mathbb{C}$, approximate the real and imaginary parts separately and use $\mathrm{\mid }(a+ib)-(r+is)\mathrm{\mid }\le \mathrm{\mid }a-r\mathrm{\mid }+\mathrm{\mid }b-s\mathrm{\mid }$.

Every $F\in X^{\ast }$ restricts to a member of $Y^{\ast }$, so every ambient weak subbasic set has an intrinsically weak-open trace on $Y$. Conversely, given one $g\in Y^{\ast }$, [F4] supplies $F\in X^{\ast }$ with $F{\mathrm{\mid }}_Y=g$. Therefore the inverse image under $g$ of any scalar-open set is the trace on $Y$ of the corresponding ambient weak-open inverse image under $F$. Finite intersections behave the same way. The two topologies on $Y$ are equal.

The set ${\mathbb{N}}^{<\mathbb{N}}$ of finite strings of naturals has a choice-free enumeration: order strings first by $\mathrm{\mid }s\mathrm{\mid }+{\sum }_{j<\mathrm{\mid }s\mathrm{\mid }}s(j)$, then by length, and then lexicographically. Each fixed-value block is finite, and the displayed order lists every finite string. Map $s=(s(0),\dots ,s(m-1))$ to $d_s={\sum }_{j=0}^{m-1}q(s(j))x_j$, with empty sum $0$. Its image $D=\{d_s:s\in {\mathbb{N}}^{<\mathbb{N}}\}$ is nonempty and at most countable by [F2].

The set $D$ is norm dense in ${\mathrm{span}}_{\mathbb{K}}\{x_n:n\in \mathbb{N}\}$. Indeed, write a given vector there as $u={\sum }_{j=0}^{m-1}{\alpha }_j{x}_j$, padding with zero coefficients when necessary. If $m=0$, then $u=0=d_{\mathrm{\varnothing }}$. If $m>0$ and $\epsilon >0$, put $S={\sum }_{j<m}(1+\mathrm{\parallel }{x}_j\mathrm{\parallel })>0$. By step 1.2, make the finitely many choices $r_j\in {\mathbb{Q}}_{\mathbb{K}}$ with $\mathrm{\mid }{\alpha }_j-r_j\mathrm{\mid }<\epsilon \mathrm{/}S$. Choose indices $k_j$ with $q(k_j)=r_j$; only finitely many choices are involved. For $s=(k_0,\dots ,k_{m-1})$, the triangle inequality gives $\mathrm{\parallel }u-d_s\mathrm{\parallel }\le {\sum }_{j<m}\mathrm{\mid }{\alpha }_j-r_j\mathrm{\mid }\mathrm{\parallel }{x}_j\mathrm{\parallel }<(\epsilon \mathrm{/}S){\sum }_{j<m}\mathrm{\parallel }{x}_j\mathrm{\parallel }<\epsilon$. Thus ${\bar{D}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}=Y$.

By [F1] and step 3.1, $Y$ is separable. The norm closure of a linear subspace is again linear: approximating two vectors and using the triangle inequality proves closure under addition, and multiplying an approximating net by one fixed scalar proves closure under scalar multiplication, including the scalar zero. Hence $Y$ is a closed linear subspace of the Banach space $X$, so [F6] makes $Y$ Banach.

Finally take $z\in X\setminus Y$. The set $Y$ is nonempty, closed and convex, while $\{z\}$ is compact and disjoint from it. By [F5], there are $f\in X^{\ast }$, $a\in \mathbb{R}$ and $\delta >0$ such that $\mathrm{Re}f(y)\le a-\delta <a+\delta \le \mathrm{Re}f(z)$ for every $y\in Y$. The weakly open set $\{w\in X:\mathrm{Re}f(w)>a\}$ contains $z$ and misses $Y$. Every point of $X\setminus Y$ therefore has a weak neighborhood in the complement, so $Y$ is weakly closed. This uses HB only through [F4] and [F5]: the proof never selects extensions or separators simultaneously for a family.

If $Y=\{0\}$, then $K$ is either empty or the singleton $\{0\}$. In either case the zero function $d_K:K\times K\to \mathbb{R}$ is the unique metric and induces the only topology on $K$, which is its relative weak topology. Hence suppose below that $Y\mathrm{\ne }\{0\}$.

On $\mathbb{K}$ put $\delta (s,t)=\min \{1,\mathrm{\mid }s-t\mathrm{\mid }\}$. This is a metric bounded by $1$ and induces the usual scalar topology, because its balls of radius below $1$ are the usual metric balls. It is complete: a $\delta$-Cauchy sequence is eventually Cauchy for $\mathrm{\mid }\cdot \mathrm{\mid }$ at every tolerance below $1$, so [F5] gives a usual limit, and $\delta (s,t)\le \mathrm{\mid }s-t\mathrm{\mid }$ gives convergence in $\delta$.

By separability, take an at most countable norm-dense $D\subseteq Y$. It is nonempty because its closure is the nonempty space $Y$. The set $E=\{z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }:z\in D\setminus \{0\}\}$ is at most countable and nonempty. It is dense in the unit sphere: if $\mathrm{\parallel }u\mathrm{\parallel }=1$ and $\epsilon >0$, density gives $z\in D$ with $\mathrm{\parallel }z-u\mathrm{\parallel }<\min \{1\mathrm{/}2,\epsilon \mathrm{/}2\}$, so $z\mathrm{\ne }0$ and $\mathrm{\parallel }z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }-u\mathrm{\parallel }\le \mathrm{\mid }1-\mathrm{\parallel }z\mathrm{\parallel }\mathrm{\mid }+\mathrm{\parallel }z-u\mathrm{\parallel }\le 2\mathrm{\parallel }z-u\mathrm{\parallel }<\epsilon$. By [F1], enumerate $E$ as $(u_n{)}_{n\in \mathbb{N}}$, allowing repetitions.

Apply [F6] to countably many copies of $(\mathbb{K},\delta )$. The formula $D(a,b)={\sum }_{n=0}^{\mathrm{\infty }}{2}^{-(n+1)}\delta (a_n,b_n)$ is a metric on ${\mathbb{K}}^{\mathbb{N}}$ inducing its product topology. Its restriction to every subset is a metric inducing the subspace topology, and that metric topology is Hausdorff by [F7].

For each $n$, let $S_n=\{f\in Y^{\ast }:\mathrm{\parallel }f\mathrm{\parallel }=1,f(u_n)=1\}$. Each $S_n$ is nonempty by [F2], including in the complex case where the attained value is the positive real number $1$. Apply ${\mathrm{A}\mathrm{C}}_{\omega }$ once to the sequence $(S_n)$ and obtain $f_n\in S_n$ for every $n$.

The family $(f_n)$ separates points of $Y$. If $x\mathrm{\ne }y$, put $v=(x-y)\mathrm{/}\mathrm{\parallel }x-y\mathrm{\parallel }$ and choose $u_n$ with $\mathrm{\parallel }{u}_n-v\mathrm{\parallel }<1\mathrm{/}2$. Then $\mathrm{\mid }{f}_n(v)\mathrm{\mid }\ge \mathrm{\mid }{f}_n(u_n)\mathrm{\mid }-\mathrm{\mid }{f}_n(v-u_n)\mathrm{\mid }>1-1\mathrm{/}2>0$, since $\mathrm{\parallel }{f}_n\mathrm{\parallel }=1$. Therefore $f_n(x-y)=\mathrm{\parallel }x-y\mathrm{\parallel }{f}_n(v)\mathrm{\ne }0$.

Define $\mathrm{\Phi }:K\to {\mathbb{K}}^{\mathbb{N}}$ by $\mathrm{\Phi }(x)=(f_n(x){)}_n$. Every coordinate $f_n$ is weakly continuous, so the initial property of the product topology makes $\mathrm{\Phi }$ continuous. Step 4.1 makes it injective. Its corestriction ${\mathrm{\Phi }}_0:K\to \mathrm{\Phi }[K]$ is therefore a continuous bijection.

The weak space $K$ is compact by hypothesis, and $\mathrm{\Phi }[K]$ is Hausdorff by step 2.2. Hence [F8] makes ${\mathrm{\Phi }}_0$ a homeomorphism. Pulling the restricted product metric back along ${\mathrm{\Phi }}_0$ gives exactly $d_K(x,y)=D(\mathrm{\Phi }(x),\mathrm{\Phi }(y))$, the displayed metric, and its topology is precisely the relative weak topology on $K$. Together with step 1.1 this proves the claim for every $Y$ and for empty as well as nonempty $K$.

The only countable selection is step 3.1, where ${\mathrm{A}\mathrm{C}}_{\omega }$ selects the norming family. HB is used only inside the individual norming-functional supplier [F2]. Enumeration in step 2.1 is obtained from one at-most-countable set by its supplied surjection and uses no choice. The argument metrizes only the supplied weakly compact $K$ in a separable $Y$; it makes no metrizability claim for all of $Y$ or for nonseparable spaces.