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The dual test-function norm isIf $\mathcal{F}$ is closed, convex, and symmetric, the Bipolar theorem for a dual pair identifies its dual unit ball with $\mathcal{F}$. Submultiplicativity under pointwise products allows a polynomial in one test function to remain controlled in this norm.

Put $\delta =\exp (-{\epsilon }^{-C})$, with the absolute constant $C$ chosen sufficiently large below, and suppose for a contradiction that no $g_2$ with $0\le g_2\le 1$ satisfies $\Vert g_1-g_2{\Vert }_{\mathcal{F}}\le \epsilon$.

LetThe set $K$ is compact and convex, while $B$ is closed and convex, so $K+B$ is closed and convex. We use the following finite-dimensional form of the Hahn-Banach separation theorem: if a point lies outside a nonempty closed convex set, there is a linear functional whose value at the point is strictly greater than its supremum over that set. Identifying linear functionals on ${\mathcal{H}}_N$ through the inner product, there is therefore a function $\psi$ such thatThe separating functional cannot have zero dual norm, so rescale it to make $\Vert \psi {\Vert }_{\mathcal{F}}^{\ast }=1$. Because $\mathcal{F}$ is closed, convex, and symmetric, the finite-dimensional Bipolar theorem for a dual pair says that the unit ball of $\Vert \cdot {\Vert }_{\mathcal{F}}^{\ast }$ is precisely $\mathcal{F}$. Thus $\psi \in \mathcal{F}$ and $|\psi (n)|\le 1$.

The support function of $K$ is obtained by choosing $g(n)=1$ where $\psi (n)>0$ and $g(n)=0$ where $\psi (n)<0$. In terms of the positive part of a real-valued function ${\psi }_{+}=\max (0,\psi )$, the separating inequality becomesSince $0\le g_1\le \nu$, pointwise we have $g_1\psi \le \nu {\psi }_{+}$, and hence

Apply the supplied polynomial approximation of the positive part to $\psi$. If $P_{\epsilon }(t)={\sum }_{i=0}^m{a}_i{t}^i$, then its uniform approximation error and $\mathbb{E}\nu \le 1$ implyThe constant function $1$ and $\psi$ belong to the dual unit ball. By the assumed submultiplicativity, $\Vert {\psi }^i{\Vert }_{\mathcal{F}}^{\ast }\le 1$ for every $i\ge 0$. Dual seminorm therefore givesThe stated coefficient bound, with $C$ in $\delta$ chosen larger than the absolute constant in that bound, makes this last quantity at most $\epsilon /50$. Together with the polynomial-approximation error, this contradicts $⟨\nu -1,{\psi }_{+}⟩>\epsilon$. The required $g_2$ consequently exists. This proves the dense model theorem for a multiplicative test family.

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: