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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.

Mazur theorem says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number $\alpha$ such thatThis weakly open separation shows that $x$ is outside the weak closure.

By definition,For each $n$, regard $x_n$ as the bounded functional $J{x}_n:X^{\ast }\to \mathbb{F}$ given by $J{x}_n(f)=f(x_n)$. Pointwise convergence makes the family $(J{x}_n)$ pointwise bounded on the Banach space $X^{\ast }$. The Uniform boundedness principle gives

Suppose first that $x_n\rightharpoonup 0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\epsilon$.

Conversely, if weak convergence fails, some $f\in X^{\ast }$ and an infinite subsequence satisfy either $\mathrm{Re}f(x_n)\ge \epsilon$ throughout or $\mathrm{Re}f(x_n)\le -\epsilon$ throughout. Every convex combination from that subsequence then has norm at least $\epsilon /\Vert f\Vert$, contradicting the stated property.

Now let the $x_n\in {\ell }^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\Vert x_n{\Vert }_p\le C$. For $N$ distinct terms,The convex-combination criterion therefore proves $x_n\rightharpoonup 0$. This is weak convergence of bounded disjointly supported sequences in lp.

The final implication for a commutative unital C*-algebra is true. By the Commutative Gelfand--Naimark theorem, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the Riesz-Markov-Kakutani representation theorem; the dominated convergence theorem givesHence $a_n\rightharpoonup 0$.

The useful forms of the Hahn-Banach separation theorem in a real locally convex space are:

The Banach-Alaoglu theorem says that the closed unit ball of $E^{\ast }$ is compact in $\sigma (E^{\ast },E)$. Goldstine theorem says that the canonical image of the closed unit ball of a normed space $E$ is weak-star dense in the closed unit ball of $E^{\ast \ast }$.

Give $Z$ its norm inherited from $X^{\ast }$ and defineIt is linear and contractive, and it is injective because $Z$ separates points. Goldstine's theorem followed by restriction from $X^{\ast }$ to $Z$ shows that $J(B_X)$ is weak-star dense in $B_{Z^{\ast }}$: a functional on $Z$ first extends norm-preservingly to $X^{\ast }$, and elements of $B_X$ approximate that extension on every finite subset of $Z$.

The topology induced by $\sigma (Z^{\ast },Z)$ on $J(B_X)$ is exactly the given topology $\sigma (X,Z)$. By hypothesis $B_X$ is compact, so $J(B_X)$ is weak-star compact and therefore closed in the Hausdorff space $B_{Z^{\ast }}$. Density now givesThus $J$ is surjective and maps closed unit ball onto closed unit ball, so it is an isometric isomorphism. Hence $X$ is a dual space. This is the compact norming dual-pair criterion.

Put $z_n=x_n-x$, so $z_n\rightharpoonup 0$. We show that the norm closure of the convex hull of every tailcontains zero. If it did not, the Hahn-Banach separation theorem would give a bounded linear functional $\varphi$, a real number $c>0$, and, after multiplying $\varphi$ by a complex scalar if necessary,for every $z$ in that closed convex hull. In particular $\mathrm{Re}\varphi (z_n)\ge c$ for all $n\ge k$. By the Riesz representation theorem, $\varphi (z)=⟨z,y⟩$ for some $y\in H$, contradicting $z_n\rightharpoonup 0$.

Consequently, for every $k$ there is a finite convex combinationwith $\Vert w_k{\Vert }_H<1/k$. DefineThen every ${\tilde{x}}_k$ is a convex combination of terms of the original sequence andThis is Mazur lemma in the present Hilbert-space setting.

For a real locally convex space $(X,\mathcal{P})$, the continuous dual space $X^{\ast }$ consists of all continuous real-linear maps $X\to \mathbb{R}$. If $g\in Y^{\ast }$ on a subspace $Y$, continuity gives seminorms $p_1,\dots ,p_k\in \mathcal{P}$ and $C>0$ such thatThe right side is a continuous sublinear functional on $X$. The dominated Hahn-Banach theorem extends $g$ to a linear $f:X\to \mathbb{R}$ satisfying the same bound, so $f\in X^{\ast }$.

If $Y$ is closed and $x_0\notin Y$, the Hausdorff locally convex quotient $X/Y$ has a continuous seminorm $p$ with $p(x_0+Y)>0$. On $Y+\mathbb{R}{x}_0$, define $g(y+t{x}_0)=t$ and rescale $p$ so that $|g|\le p$. Hahn--Banach extends it to $f\in X^{\ast }$ with $f{|}_Y=0$ and $f(x_0)=1$.

The separation of a point and an open convex set says that if $C\subseteq X$ is nonempty, open, and convex and $x_0\notin C$, there is $f\in X^{\ast }$ such thatTo prove it, choose $a\in C$, put $U=C-a$, and let $p_U$ be its Minkowski functional. For $z=x_0-a\notin U$, $p_U(z)\ge 1$. Define $h(tz)=t{p}_U(z)$ on $\mathbb{R}z$; then $h\le p_U$. Hahn--Banach extends $h$ to $f\le p_U$. Since $p_U(u)<1$ for $u\in U$, one has $f(c-a)<1\le f(x_0-a)$, proving the claim.

If $K$ is closed and convex and $x_0\notin K$, a Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number separating $x_0$ from $K$. The corresponding inverse image of an open interval is a weak neighbourhood of $x_0$ disjoint from $K$, so $K$ is closed in $\sigma (X,X^{\ast })$.

The unit sphere $S_X$ of a normed space is norm closed. If $X$ is infinite-dimensional, every basic weak neighbourhood of an interior point $x\in B_X$ constrains only finitely many functionals $f_1,\dots ,f_n$. Their common kernel contains a nonzero $y$. Continuity of $t\mapsto \Vert x+ty\Vert$, together with its value below one at $t=0$ and divergence as $|t|\to \infty$, supplies $t$ with $\Vert x+ty\Vert =1$. Since all $f_i$ have the same values at $x+ty$ and $x$, every weak neighbourhood of $x$ meets $S_X$. Points outside $B_X$ are separated from it by Hahn--Banach, so the weak closure of the unit sphere is exactly $B_X$. In particular, $S_X$ is not weakly closed.

A linear map $T:X\to Y$ is a weakly compact operator when $T(B_X)$ is relatively weakly compact. Every weakly compact subset of a Banach space is norm bounded, soThus even without assuming continuity initially, a weakly compact linear map is bounded.

The key bidual criterion isFor the forward implication, approximate $x^{\ast \ast }\in B_{X^{\ast \ast }}$ weak-star by points $J_X{x}_{\alpha }$ using the Goldstine theorem. Weak compactness supplies a subnet for which $T{x}_{\alpha }$ converges weakly to some $y\in Y$, while weak-star continuity of $T^{\ast \ast }$ makes $J_{Y}T{x}_{\alpha }$ converge to $T^{\ast \ast }{x}^{\ast \ast }$. Hence $T^{\ast \ast }{x}^{\ast \ast }=J_{Y}y$. Conversely, if the displayed inclusion holds, the weak-star compact set $T^{\ast \ast }(B_{X^{\ast \ast }})$ lies in $J_Y(Y)$, where the inherited weak-star topology is the weak topology of $Y$. It is a weakly compact set containing $J_{Y}T(B_X)$.

If $T$ is weakly compact and $y^{\ast \ast \ast }\in Y^{\ast \ast \ast }$, restriction of $y^{\ast \ast \ast }$ to $J_Y(Y)$ defines $y^{\ast }\in Y^{\ast }$. For $x^{\ast \ast }\in X^{\ast \ast }$,so $T^{\ast \ast \ast }{y}^{\ast \ast \ast }=J_{X^{\ast }}(T^{\ast }{y}^{\ast })$. The bidual criterion makes $T^{\ast }$ weakly compact. Conversely, if $T^{\ast }$ is weakly compact, then $T^{\ast \ast \ast }(Y^{\ast \ast \ast })\subseteq J_{X^{\ast }}(X^{\ast })$. Every $y^{\ast \ast \ast }$ annihilating $J_Y(Y)$ is sent to zero, so every $T^{\ast \ast }{x}^{\ast \ast }$ lies inThe criterion makes $T$ weakly compact. Hence $T$ is weakly compact exactly when $T^{\ast }$ is.

The bidual criterion also proves the structure of $\mathcal{W}(X,Y)$. It is closed under linear combinations. If $T_n\to T$ in operator norm and every $T_n$ is weakly compact, then $T_n^{\ast \ast }{x}^{\ast \ast }\to T^{\ast \ast }{x}^{\ast \ast }$ in norm; the canonical copy $J_Y(Y)$ is norm closed, so $T$ is weakly compact. For bounded composable maps $A$ and $B$, the image condition for $(ATB{)}^{\ast \ast }=A^{\ast \ast }{T}^{\ast \ast }{B}^{\ast \ast }$ proves the ideal property. Thus the weakly compact operators form a norm-closed operator ideal.

The Krein-Smulian theorem states that the norm-closed convex hull $\overline{\mathrm{co}}K$ of a weakly compact set $K$ is weakly compact. To prove it, view $K$ as a compact Hausdorff space in its weak topology. The probability measures on $K$ form a weak-star compact subset of $C(K{)}^{\ast }$ by the Banach-Alaoglu theorem. The barycentre mapis weak-star-to-weak continuous; scalar integration and weak compactness ensure that $b_{\mu }\in X$. Finitely supported probability measures are weak-star dense and their barycentres are precisely $\mathrm{co}K$. The image of all probability measures is therefore the weak closure of $\mathrm{co}K$, and it is weakly compact. A convex set has the same weak and norm closures by the Hahn-Banach separation theorem, proving the claim.

If $X$ is reflexive, $B_X$ is weakly compact and every bounded $T:X\to Y$ is weak-to-weak continuous, so $T$ is weakly compact. If $Y$ is reflexive, every bounded subset of $Y$ is relatively weakly compact, with the same conclusion. Thuswhenever either space is reflexive.

If $X={\ell }^1$ and $Y$ is nonreflexive, the Eberlein-Smulian theorem supplies a bounded sequence $(y_n)$ in $Y$ with no weakly convergent subsequence. The formuladefines a bounded operator $T:{\ell }^1\to Y$. Since $T{e}_n=y_n$, it cannot be weakly compact. Therefore $\mathcal{B}({\ell }^1,Y)\ne \mathcal{W}({\ell }^1,Y)$.

This conclusion does not hold for every pair of nonreflexive spaces. Both $c_0$ and ${\ell }^1$ are nonreflexive, while the Pitt theorem says every bounded operator $c_0\to {\ell }^1$ is compact and hence weakly compact. Thus

For a closed convex cone $C$ in a finite-dimensional real inner product space and $z\notin C$, there exists $y$ with $⟨y,z⟩<0$ and $⟨y,c⟩\ge 0$ for all $c\in C$. Here is a direct proof. A nearest point $c_0\in C$ exists by compactness after restricting to a sufficiently large ball. Differentiating squared distance along the segment towards $c\in C$ gives $⟨c_0-z,c-c_0⟩\ge 0$. Using $c=0$ and $c=2c_0$ shows $⟨c_0-z,c_0⟩=0$. Set $y=c_0-z\ne 0$: then $⟨y,c⟩\ge 0$ and $⟨y,z⟩=-\Vert y{\Vert }^2<0$. This is a conic form of the Hahn-Banach separation theorem.