Banach–Dieudonné linear-subspace criterion

If $E$ is weak-star closed, then so is $E\cap B_{X^{\ast }}$, because $B_{X^{\ast }}={\bigcap }_{x\in B_X}\{f:\mathrm{\mid }f(x)\mathrm{\mid }\le 1\}$ is an intersection of closed evaluation constraints.

For the reverse implication first suppose $\mathbb{K}=\mathbb{R}$ and $A:=E\cap B_{X^{\ast }}$ is weak-star closed. If $f_n\in E$ and $f_n\to f$ in norm, boundedness of the convergent sequence gives $R>0$ with $R^{-1}{f}_n\in A$; norm convergence implies weak-star convergence, so closedness of $A$ gives $R^{-1}f\in A$ and $f\in E$. If some $f_0\mathrm{\notin }E$ had ${\inf }_{f\in E}\parallel f-f_0\parallel =0$, DC could select $f_n\in E$ with $\parallel f_n-f_0\parallel <1\mathrm{/}(n+1)$, contradicting this sequential norm-closedness. Hence $d:=\mathrm{dist}(f_0,E)>0$; fix $0<\delta <d$.

For finite sets $S_k\subseteq B_X$, let $P_n(S_1,\dots ,S_{n-1})$ mean: every $f\in E$ with $\parallel f-f_0\parallel \le n\delta$ violates at least one earlier test, so $\mathrm{\mid }(f-f_0)(x)\mathrm{\mid }>k\delta$ for some $1\le k<n$ and $x\in S_k$. The assertion $P_1$ is vacuous because $\delta <d$.

Suppose $P_n$ holds. For finite $S\subseteq B_X$, let $E(S)$ consist of those $f\in E$ with $\parallel f-f_0\parallel \le (n+1)\delta$, all earlier tests at most $k\delta$, and the $S$-test at most $n\delta$. Put $R=\parallel f_0\parallel +(n+1)\delta$. The set $K=E\cap R{B}_{X^{\ast }}=RA$ is weak-star compact: $R{B}_{X^{\ast }}$ is compact by scaling [F1], $RA$ is weak-star closed, and [F1] uses the ultrafilter lemma through [F2]. Each $E(S)$ is weak-star closed in $K$, because each norm bound is the intersection over $x\in B_X$ of closed evaluation constraints. If every $E(S)$ were nonempty, the identity ${\bigcap }_{i=1}^{q}E(S_i)=E({\bigcup }_i{S}_i)$ would give the finite-intersection property; compactness would produce $f$ in every $E(S)$. Taking singleton $S=\{x\}$ for every $x\in B_X$ would give $\parallel f-f_0\parallel \le n\delta$, while all earlier tests hold, contradicting $P_n$. Thus some finite listed $S_n$ has $E(S_n)=\mathrm{\varnothing }$, and that emptiness is exactly $P_{n+1}$.

Apply DC to the relation that extends a finite list $(S_1,\dots ,S_{n-1})$ satisfying $P_n$ by a finite listed $S_n$ supplied in step 3.1. Starting from the empty list, it yields finite listed sets $S_n\subseteq B_X$ for all $n\ge 1$ with every $P_n$ true. Recording the finite listing as part of each state avoids a later countable choice of enumerations.

Concatenate, for $n=1,2,\dots$, the finite list $n^{-1}{S}_n$ followed by one zero padding term, obtaining a sequence $(x_i)$ in $B_X$. If a term lies in the $n$th block its norm is at most $1\mathrm{/}n$; because each block is finite and nonempty after padding, the block number tends to infinity with $i$. Hence $\parallel x_i\parallel \to 0$.

For every $f\in E$, choose an integer $n\ge \max (2,{\delta }^{-1}\parallel f-f_0\parallel )$. Property $P_n$ gives $k<n$ and $x\in S_k$ with $\mathrm{\mid }(f-f_0)(x)\mathrm{\mid }>k\delta$; the coordinate $x\mathrm{/}k$ occurs in $(x_i)$, so ${\sup }_i\mathrm{\mid }(f-f_0)(x_i)\mathrm{\mid }>\delta$.

Define $T:X^{\ast }\to c_0$ by $T(f)=(f(x_i){)}_i$. Step 5.1 makes every image a null sequence, and $\parallel T(f){\parallel }_{\mathrm{\infty }}\le \parallel f\parallel {\sup }_i\parallel x_i\parallel$, so $T$ is bounded and linear. With $y_0=T(f_0)$, step 6.1 gives $\parallel T(f)-y_0{\parallel }_{\mathrm{\infty }}>\delta$ for every $f\in E$; therefore the closed linear subspace $M=\overline{T(E)}$ has $\mathrm{dist}(y_0,M)\ge \delta$.

On $M+\mathbb{R}{y}_0$ define $g(m+a{y}_0)=a$. This is well defined because $y_0\mathrm{\notin }M$, and $\parallel m+a{y}_0\parallel \ge \mathrm{\mid }a\mathrm{\mid }\delta$ shows $\mathrm{\mid }g\mathrm{\mid }\le {\delta }^{-1}\parallel \cdot \parallel$. Applying HB to the sublinear function ${\delta }^{-1}\parallel \cdot \parallel$ extends $g$ to $\beta \in (c_0{)}^{\ast }$ with $\beta (y_0)=1$, $\beta {\mathrm{\mid }}_M=0$, and $\parallel \beta \parallel \le {\delta }^{-1}$. This is the sole HB use.

By [F5] there is $\alpha =({\alpha }_i)\in {\mathrm{\ell }}^1$ with $\beta (y)={\sum }_i{\alpha }_i{y}_i$ for $y\in c_0$ and ${\sum }_i\mathrm{\mid }{\alpha }_i\mathrm{\mid }=\parallel \beta \parallel \le {\delta }^{-1}$.

Since ${\sum }_i\parallel {\alpha }_i{x}_i\parallel \le {\sum }_i\mathrm{\mid }{\alpha }_i\mathrm{\mid }\le {\delta }^{-1}$ and $X$ is Banach, [F6] gives $x_0={\sum }_i{\alpha }_i{x}_i\in X$ with $\parallel x_0\parallel \le {\delta }^{-1}$.

Continuity of every $f\in X^{\ast }$ allows evaluation term by term: $f(x_0)={\sum }_i{\alpha }_{i}f(x_i)=\beta (Tf)$. Thus $f_0(x_0)=1$ and $f(x_0)=0$ for every $f\in E$. The weak-star neighborhood $\{h:\mathrm{\mid }(h-f_0)(x_0)\mathrm{\mid }<1\mathrm{/}2\}$ therefore misses $E$.

Every $f_0\mathrm{\notin }E$ has the weak-star neighborhood constructed in step 11.1 disjoint from $E$, so $E$ is weak-star closed in the real case. Together with step 1.1 this proves both directions there.

Now let $X$ be complex and write $X_{\mathbb{R}}$ for its realification. The map $\mathcal{R}:X^{\ast }\to (X_{\mathbb{R}}{)}^{\ast }$, $\mathcal{R}f=\mathrm{Re}f$, is a real-linear isometric bijection with inverse $g\mapsto (x\mapsto g(x)-ig(ix))$: complex linearity follows from the displayed formula, and rotating a vector shows norm equality. It is a weak-star homeomorphism because $g(x)=\mathrm{Re}f(x)$ and $\mathrm{Im}f(x)=-g(ix)$. For a complex-linear $E$, $\mathcal{R}(E)$ is real-linear and $\mathcal{R}(E\cap B_{X^{\ast }})=\mathcal{R}(E)\cap B_{(X_{\mathbb{R}}{)}^{\ast }}$. Hence closedness of the complex slice implies closedness of the real slice; step 12.1 makes $\mathcal{R}(E)$ real weak-star closed, and the homeomorphism makes $E$ complex weak-star closed.

Step 1.1 proves the forward implication over both scalar fields, step 12.1 proves the reverse implication over $\mathbb{R}$, and step 13.1 proves it over $\mathbb{C}$. Therefore the two weak-star closedness conditions are equivalent.