Geometric Hahn Banach and Convex Separation Examples

For $x\in I$, recursively let $n_k(x)$ be the least unused index $n$ with $q_n\in (x-1\mathrm{/}k,x+1\mathrm{/}k)$, and put $A_x=\{n_k(x):k\ge 1\}$. Such an index exists by [F2] because every interval contains infinitely many rationals; [F1] makes the choice deterministic.

Each $A_x$ is infinite since its indices are chosen unused. If $x\mathrm{\ne }y$, then for sufficiently large $k,\mathrm{\ell }$ the intervals $(x-1\mathrm{/}k,x+1\mathrm{/}k)$ and $(y-1\mathrm{/}\mathrm{\ell },y+1\mathrm{/}\mathrm{\ell })$ are disjoint; any common selected index must therefore arise among finitely many early choices. Thus $A_x\cap A_y$ is finite.

The map $x\mapsto A_x$ is injective by the finite-intersection conclusion, so $\{A_x:x\in I\}$ is uncountable by [F2] and has the required property.

For $A\in \mathcal{A}$, let $u_A$ be its indicator sequence. Then $[u_A]\mathrm{\ne }0$ in $Q$, and for distinct $A_1,\dots ,A_m$ the quotient norm of $\sum c_j[u_{A_j}]$ is ${\max }_j\mathrm{\mid }{c}_j\mathrm{\mid }$, since the supports are disjoint after deleting finitely many coordinates.

For $g\in Q^{\ast }$ and $r>0$, the set $\{A:\mathrm{\mid }g([u_A])\mathrm{\mid }\ge r\}$ is finite: choose scalar phases on any finite subfamily and apply $\mathrm{\mid }g(\sum c_j[u_{A_j}])\mathrm{\mid }\le \mathrm{\parallel }g\mathrm{\parallel }$. Thus $\{A:g([u_A])\mathrm{\ne }0\}$ is countable as a union over positive reciprocal integers.

Given a countable family $(g_n)$ in $Q^{\ast }$, [F3] makes ${\bigcup }_n\{A:g_n([u_A])\mathrm{\ne }0\}$ countable. By [F2] choose $A$ outside it; then nonzero $[u_A]$ is annihilated by every $g_n$.

Suppose a bounded projection $P:{\mathrm{\ell }}^{\mathrm{\infty }}\to c_0$ exists. For each coordinate $n$, define ${\phi }_n(\pi x)=x_n-(Px{)}_n$. This is well defined because $Pz=z$ for $z\in c_0$, and it is a bounded functional on the quotient.

If ${\phi }_n(\pi x)=0$ for every $n$, then $(I-P)x=0$ coordinatewise, so $x=Px\in c_0$ and $\pi x=0$. Thus $({\phi }_n)$ is a countable separating family.

This contradicts [F1]. Hence no such projection exists, and $c_0$ is not complemented.

$x\in tB(0,r)$ exactly when $\mathrm{\parallel }x\mathrm{\parallel }<tr$, whose infimum over $t>0$ is $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$.

Likewise $x\in tE$ exactly when $q(x)<t^2$, so the infimum is $\sqrt{q(x)}$. Positive definiteness gives zero only at $x=0$, and the displayed formulas give norm homogeneity and triangle inequality.

From $x\in t(-1,2)⟺-t<x<2t$, [F1] gives $p_C(x)=x\mathrm{/}2$ for $x\ge 0$ and $p_C(x)=-x$ for $x<0$.

Hence $p_C(1)=1\mathrm{/}2$ while $p_C(-1)=1$, contradicting the required equality $p_C(-1)=\mathrm{\mid }-1\mathrm{\mid }{p}_C(1)$. The missing hypothesis is balancedness.

$A$ is closed convex. By [F1], $B$ is the closed epigraph of a convex function, hence closed and convex; it is disjoint from $A$ because $e^{-s}>0$.

The points $(n,0)\in A$ and $(n,e^{-n})\in B$ have distance $e^{-n}\to 0$. A strong gap $\beta -\alpha >0$ for $f$ would imply $\mathrm{\parallel }a-b\mathrm{\parallel }\ge (\beta -\alpha )\mathrm{/}\mathrm{\parallel }f\mathrm{\parallel }$ for all $a\in A,b\in B$, impossible.

For $f\in M^{\perp }$ with $\mathrm{\parallel }f\mathrm{\parallel }\le 1$, continuity gives $f{\mathrm{\mid }}_{\bar{M}}=0$. Hence, for every $m\in \bar{M}$, $\mathrm{\mid }f(x)\mathrm{\mid }=\mathrm{\mid }f(x-m)\mathrm{\mid }\le \mathrm{\parallel }x-m\mathrm{\parallel }$. Taking infima gives the $\le$ direction.

Put $\delta =\mathrm{dist}(x,\bar{M})$. If $\delta >0$, define $g:\bar{M}+\mathbb{K}x\to \mathbb{K}$ by $g(m+\lambda x)=\lambda \delta$. This is well defined, and $\mathrm{\mid }g(m+\lambda x)\mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }\delta \le \mathrm{\parallel }m+\lambda x\mathrm{\parallel },$ so $\mathrm{\parallel }g\mathrm{\parallel }\le 1$. In the real case apply [F1] with $p(y)=\mathrm{\parallel }y\mathrm{\parallel }$; domination at $y$ and $-y$ gives an extension $f$ with $\mathrm{\parallel }f\mathrm{\parallel }\le 1$. In the complex case apply [F1] to $\mathrm{Re}g$ on the underlying real subspace, obtaining a real linear $U$ with $\mathrm{\mid }U(y)\mathrm{\mid }\le \mathrm{\parallel }y\mathrm{\parallel }$. Then $f(y)=U(y)-iU(iy)$ is complex linear and extends $g$. For each $y$, choose a scalar $a$ of modulus one with $af(y)=\mathrm{\mid }f(y)\mathrm{\mid }$; thus $\mathrm{\mid }f(y)\mathrm{\mid }=U(ay)\le \mathrm{\parallel }y\mathrm{\parallel }$. This is the exact use of AC in both scalar cases; this extension lies in $M^{\perp }$ and satisfies $\mathrm{\mid }f(x)\mathrm{\mid }=\delta$. If $\delta =0$, then $x\in \bar{M}$, so continuity makes every annihilator vanish at $x$. Thus equality holds in both cases.

The inclusion in the statement is a closed-subspace inclusion by [F1].

It is uncomplemented by [F2], so it has both advertised properties.