 Solution

The Hahn-Banach separation theorem for two convex sets states that if $A$ and $B$ are disjoint nonempty convex subsets of a real locally convex space $X$ and $A$ is open, then there are a continuous linear functional $f\in X^{\ast }$ and a real number $c$ such thatTo prove it, formThis is open and convex and does not contain zero. The separation of a point and an open convex set, proved from the Minkowski functional and the Hahn-Banach theorem, gives a nonzero continuous $f$ with $f(c)<0$ for every $c\in C$. Hence $f(a)<f(b)$ for all $a,b$. Taking $c$ between ${\sup }_{A}f$ and ${\inf }_{B}f$ gives the stated form.

For a dual pair $(E,F)$, the topology $\sigma (E,F)$ has a neighbourhood base at zero consisting ofEvery $f\in F$ is continuous by definition. Conversely, if a linear functional $g$ is continuous, some such neighbourhood lies in $\{|g|<1\}$. Therefore ${\bigcap }_j\ker f_j\subseteq \ker g$. Elementary linear algebra then gives $g\in \mathrm{span}\{f_1,\dots ,f_n\}\subseteq F$. Thus

For a normed space $X$, the weak topology is $\sigma (X,X^{\ast })$; on $X^{\ast }$ the weak-star topology is $\sigma (X^{\ast },X)$, using the canonical image of $X$ in $X^{\ast \ast }$. If $X$ is reflexive, $X^{\ast \ast }=J_X(X)$, so the weak and weak-star topologies on $X^{\ast }$ coincide. Conversely, if they coincide, every $x^{\ast \ast }\in X^{\ast \ast }$ is weak-star continuous. The dual-pair result says that every such functional is evaluation at some $x\in X$, so $J_X$ is onto and $X$ is reflexive.

For $A\subseteq E$, every $a\in A$ and zero satisfy every inequality defining $A^{\circ \circ }$, so $A\cup \{0\}\subseteq A^{\circ \circ }$. The latter is an intersection of weakly closed convex half-spaces, hence containsIf $x\notin C$, choose an open convex neighbourhood $V$ of zero with $(x+V)\cap C=\varnothing$. Applying the separation theorem to $x+V$ and $C$ gives $f\in F$ withBecause $0\in C$, the supremum is nonnegative. After multiplying $f$ by a positive scalar, $f\le 1$ on $C$ while $f(x)>1$. Thus $f\in A^{\circ }$ but $x\notin A^{\circ \circ }$. We conclude with the Bipolar theorem for a dual pair: