Geometric Hahn Banach and Convex Separation

For $r>0$, $rx\in tC$ holds exactly when $x\in (t\mathrm{/}r)C$; taking infima gives $p_C(rx)=r{p}_C(x)$, while $r=0$ gives $p_C(0)=0$.

Absorption and convexity first give $0\in C$. Given $a>p_C(x)$ and $b>p_C(y)$, choose $s<a$, $t<b$ with $x\in sC$, $y\in tC$; convexity with $0$ enlarges these to $x=ac$, $y=bd$ for some $c,d\in C$. Then $(ac+bd)\mathrm{/}(a+b)\in C$, hence $x+y\in (a+b)C$.

Therefore $p_C(x+y)\le a+b$ for every such $a,b$; letting them decrease to the two infima proves subadditivity.

If $\lambda =0$ the assertion follows from [F1]. For $\lambda \mathrm{\ne }0$, balancedness gives $(\lambda \mathrm{/}\mathrm{\mid }\lambda \mathrm{\mid })C=C$: one inclusion is balancedness and the reverse follows by applying it to the inverse scalar.

Hence $\lambda x\in tC$ exactly when $x\in (t\mathrm{/}\mathrm{\mid }\lambda \mathrm{\mid })C$, so taking infima yields $p_C(\lambda x)=\mathrm{\mid }\lambda \mathrm{\mid }{p}_C(x)$. Together with [F1] this is precisely the seminorm axioms.

Openness gives $B(0,r)\subseteq U$ for some $r>0$. For any $x$, $x\in (\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r+1)U$, so $U$ is absorbing and [F1] applies.

If $x\in U$, openness gives $\epsilon >0$ with $(1+\epsilon )x\in U$; hence $p_U(x)\le (1+\epsilon {)}^{-1}<1$.

If $p_U(x)<1$, choose $t<1$ with $x\in tU$, say $x=tu$. Convexity and $0\in U$ imply $tu\in U$. Thus both inclusions hold.

Choose $u_0\in U$ and put $V=U-u_0$, $y=x_0-u_0$. Then $y\mathrm{\ne }0$ and $V$ is open, convex, contains $0$, and $y\mathrm{\notin }V$; by [F1], $p_V(y)\ge 1$ and $p_V(v)<1$ for $v\in V$.

By [F1], $V$ is absorbing, and [F4] makes $p_V$ sublinear. On the real line $\mathbb{R}y$ define $g(ty)=t$. For $t\ge 0$, $g(ty)=t\le t{p}_V(y)=p_V(ty)$; for $t<0$, $g(ty)<0\le p_V(ty)$. Thus [F2] gives a real linear $h$ on $X$ with $h\le p_V$ and $h(y)=1$.

Choose $r>0$ with $B(0,r)\subseteq V$. For $t>\mathrm{\parallel }z\mathrm{\parallel }\mathrm{/}r$ one has $z\mathrm{/}t\in V$ and $-z\mathrm{/}t\in V$, so taking infima gives $p_V(\pm z)\le \mathrm{\parallel }z\mathrm{\parallel }\mathrm{/}r$. Domination applied to both $z$ and $-z$ gives $\mathrm{\mid }h(z)\mathrm{\mid }\le \mathrm{\parallel }z\mathrm{\parallel }\mathrm{/}r$. In the real case take $f=h$; in the complex case take $f(z)=h(z)-ih(iz)$, which is continuous by this estimate and has real part $h$ by [F3].

For $u=u_0+v\in U$, [F1] and $h\le p_V$ give $\mathrm{Re}f(u)-\mathrm{Re}f(u_0)=h(v)<1=h(y)=\mathrm{Re}f(x_0)-\mathrm{Re}f(u_0)$. Since $h(y)=1$, $f$ is nonzero. Hence the stated strict separation holds.

The difference $D=U-V$ is open and convex, and $0\mathrm{\notin }D$ because $U\cap V=\mathrm{\varnothing }$.

Apply [F1] to $D$ and $0$. It supplies nonzero $f\in X^{\ast }$ with $\mathrm{Re}f(d)<0$ for every $d\in D$.

For $u\in U$ and $v\in V$, $u-v\in D$, so $\mathrm{Re}f(u)-\mathrm{Re}f(v)<0$. This is the required strict separation.

By [F1]--[F2], $d(k,C)$ has a minimum $\delta$ on $K$. It is positive: a zero minimum would put some $k\in K$ in the closed set $C$. Choose $0<r<\delta$.

The thickening $W=C+B(0,r)$ is open and convex and is disjoint from $K$. Apply [F3] to $W,K$ to obtain nonzero $f$ with $\mathrm{Re}f(w)<\mathrm{Re}f(k)$ for $w\in W,k\in K$.

For $c\in C$, take the supremum over $b\in B(0,r)$ in the inequalities from step 2.1. Since ${\sup }_{\mathrm{\parallel }b\mathrm{\parallel }<r}\mathrm{Re}f(b)=r\mathrm{\parallel }f\mathrm{\parallel }$, this gives $\mathrm{Re}f(c)+r\mathrm{\parallel }f\mathrm{\parallel }\le \mathrm{Re}f(k)$ for every $k\in K$. Hence ${\sup }_C\mathrm{Re}f+r\mathrm{\parallel }f\mathrm{\parallel }\le {\inf }_K\mathrm{Re}f$, a positive gap.

Since every closed affine half-space in the family contains $C$, their intersection contains $C$.

If $x\mathrm{\notin }C$, choose $r>0$ with $B(x,r)\cap C=\mathrm{\varnothing }$. The set $C+B(0,r\mathrm{/}2)$ is open and convex and still omits $x$; [F1] separates it from $\{x\}$.

The closed half-space $H_x=\{z:\mathrm{Re}f(z)\le \mathrm{Re}f(x)-\epsilon \}$ obtained by taking a positive fraction of the separation gap contains $C$ and excludes $x$. Thus every exterior point is excluded from the intersection, proving equality.

By [F1], $\delta =\mathrm{dist}(x,M)>0$. On $M+\mathbb{K}x$ define $g(m+\lambda x)=\lambda$; the representation is unique because $x\mathrm{\notin }M$.

For $\lambda \mathrm{\ne }0$, $\mathrm{\parallel }m+\lambda x\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }x+m\mathrm{/}\lambda \mathrm{\parallel }\ge \mathrm{\mid }\lambda \mathrm{\mid }\delta$, while the case $\lambda =0$ is immediate. Thus $\mathrm{\parallel }g\mathrm{\parallel }\le 1\mathrm{/}\delta$.

Extend $g$ by [F2] to $f\in X^{\ast }$. Then $f{\mathrm{\mid }}_M=0$ and $f(x)=1$, so $f\in M^{\perp }$ as required.

If $z\in \bar{M}$ and $f\in M^{\perp }$, continuity of $f$ and $f{\mathrm{\mid }}_M=0$ give $f(z)=0$. Thus $\bar{M}$ lies in the intersection.

If $z\mathrm{\notin }\bar{M}$, [F1] gives $f\in M^{\perp }$ with $f(z)=1$, so $z$ is absent from the intersection.

The two inclusions prove the formula.

If $M$ is dense, [F1] says every $f\in M^{\perp }$ vanishes on $X$, hence is zero.

If $M^{\perp }=\{0\}$, the intersection in [F1] is $X$, so $\bar{M}=X$ and $M$ is dense.

Choose a basis $e_1,\dots ,e_n$ of $M$, and let ${\varphi }_j:M\to \mathbb{K}$ be its coordinate maps. By [F1]--[F2], extend each ${\varphi }_j$ to $f_j\in X^{\ast }$.

Define $P:X\to X$ by $P(x)={\sum }_{j=1}^n{f}_j(x)e_j$. It is bounded, has range in $M$, and for $m=\sum a_j{e}_j\in M$ satisfies $P(m)=m$.

Thus $P^2=P$ and $\mathrm{ran}P=M$; [F3] makes $M$ complemented.

By [F1], $Q=X\mathrm{/}M$ is a normed finite-dimensional quotient. Choose a basis $q_1,\dots ,q_n$ of $Q$, representatives $x_j\in X$, and bounded coordinate maps ${\varphi }_j$ supplied by [F2].

The map $S:Q\to X$, $S(q)={\sum }_j{\varphi }_j(q)x_j$, is bounded and satisfies $\pi S=I_Q$. Hence $P=I_X-S\pi$ is bounded, $P^2=P$, and $\mathrm{ran}P=M$.

By [F3], $M$ is complemented.

Suppose $H$ is closed and choose $x\mathrm{\notin }H$. By [F1] there is $f\in X^{\ast }$ with $f{\mathrm{\mid }}_H=0$ and $f(x)=1$. Thus $H\subseteq \ker f$.

Since $X\mathrm{/}H$ has dimension one, a proper subspace containing $H$ cannot strictly contain $H$. As $f(x)=1$, $\ker f$ is proper; hence $\ker f=H$.

Conversely, if $H=\ker f$ with $f\in X^{\ast }$ nonzero, continuity makes $H$ closed. The induced nonzero map $X\mathrm{/}H\to \mathbb{K}$ is injective and onto, so $\dim (X\mathrm{/}H)=1$.

Every $f\in X^{\ast }$ is norm-continuous, so every displayed basic weak neighbourhood is norm-open. Thus every weak-open set is norm-open, and $D\subseteq {\bar{C}}^{\mathrm{w}\mathrm{e}\mathrm{a}\mathrm{k}}$.

If $C=\mathrm{\varnothing }$, both closures are empty. Otherwise $D$ is nonempty; if $x\mathrm{\notin }D$, choose $r>0$ with $x\mathrm{\notin }D+B(0,r)$. This latter set is open and convex, so [F1] yields $f\mathrm{\ne }0$ with $\mathrm{Re}f(d+b)<\mathrm{Re}f(x)$ for $d\in D$, $\mathrm{\parallel }b\mathrm{\parallel }<r$.

Taking suprema over $b\in B(0,r)$ gives ${\sup }_{d\in D}\mathrm{Re}f(d)+r\mathrm{\parallel }f\mathrm{\parallel }\le \mathrm{Re}f(x)$. The weak neighbourhood $\{y:\mathrm{\mid }f(y-x)\mathrm{\mid }<r\mathrm{\parallel }f\mathrm{\parallel }\mathrm{/}2\}$ therefore misses $D$, hence misses $C$.

Thus $x$ is not in the weak closure whenever it is not in $D$, proving the reverse inclusion and equality.