Norming and Separation under Hahn–Banach

All four objects have the types required by HB, and the given inequality holds for every $m\in M$. Applying HB to this quadruple yields real linear $F$ with $F{\mathrm{\mid }}_M=g$ and $F(x)\le p(x)$ for every $x\in X$.

For each $x\in X$, also $-x\in X$, so $F(-x)\le p(-x)$. Since $F(-x)=-F(x)$, multiplication by $-1$ gives $-p(-x)\le F(x)$. Together with the upper bound this proves the claim. At zero, $p(0)=F(0)=0$, so both inequalities are equalities.

Put $C=\mathrm{\parallel }g\mathrm{\parallel }\ge 0$. For $m\mathrm{\ne }0$, the vector $m\mathrm{/}\mathrm{\parallel }m\mathrm{\parallel }$ is in the unit ball of $M$, so $\mathrm{\mid }g(m)\mathrm{\mid }=\mathrm{\parallel }m\mathrm{\parallel }\mathrm{\mid }g(m\mathrm{/}\mathrm{\parallel }m\mathrm{\parallel })\mathrm{\mid }\le C\mathrm{\parallel }m\mathrm{\parallel }$; for $m=0$ the same inequality holds because $g(0)=0$. Set $p(x)=C\mathrm{\parallel }x\mathrm{\parallel }$. Then $p(tx)=tp(x)$ for $t\ge 0$ and $p(x+y)\le p(x)+p(y)$ by the norm axioms.

Over $\mathbb{R}$, $g\le p{\mathrm{\mid }}_M$ by the preceding estimate. The real extension theorem gives $F{\mathrm{\mid }}_M=g$ and $-C\mathrm{\parallel }x\mathrm{\parallel }\le F(x)\le C\mathrm{\parallel }x\mathrm{\parallel }$. Thus $\mathrm{\mid }F(x)\mathrm{\mid }\le C\mathrm{\parallel }x\mathrm{\parallel }$, so $F\in X^{\ast }$ and $\mathrm{\parallel }F\mathrm{\parallel }\le C$.

Over $\mathbb{C}$, the underlying real space of $M$ is a real linear subspace of the underlying real $X$, because closure under complex scalars includes closure under real scalars. Let $u=\mathrm{Re}g$. It is real linear and $u(m)\le \mathrm{\mid }g(m)\mathrm{\mid }\le p(m)$. The real extension theorem gives real-linear $U:X\to \mathbb{R}$ with $U{\mathrm{\mid }}_M=u$ and $U\le p$.

Define $F(x)=U(x)-iU(ix)$. The reconstruction lemma gives complex linearity and $\mathrm{Re}F=U$. For $m\in M$, also $im\in M$, whence $F(m)=u(m)-iu(im)=g(m)$ by the same lemma applied to $g$.

If $F(x)=0$ then $\mathrm{\mid }F(x)\mathrm{\mid }\le C\mathrm{\parallel }x\mathrm{\parallel }$. Otherwise set $a=\overline{F(x)}\mathrm{/}\mathrm{\mid }F(x)\mathrm{\mid }$. Then $\mathrm{\mid }a\mathrm{\mid }=1$ and $F(ax)=aF(x)=\mathrm{\mid }F(x)\mathrm{\mid }$ is real, so $\mathrm{\mid }F(x)\mathrm{\mid }=U(ax)\le C\mathrm{\parallel }ax\mathrm{\parallel }=C\mathrm{\parallel }x\mathrm{\parallel }$. Consequently the complex extension is bounded and $\mathrm{\parallel }F\mathrm{\parallel }\le C$.

In either field, $F$ extends $g$. For every $m\in M$ with $\mathrm{\parallel }m\mathrm{\parallel }\le 1$, $\mathrm{\mid }g(m)\mathrm{\mid }=\mathrm{\mid }F(m)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }$; taking the supremum gives $C\le \mathrm{\parallel }F\mathrm{\parallel }$. Together with the upper bounds this yields $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }g\mathrm{\parallel }$. If $C=0$, the bound forces $F=0$; in particular this covers $M=\{0\}$ and the zero space.

Fix $x\mathrm{\ne }0$. The set $M=\mathbb{K}x$ contains zero and is closed under addition and scalar multiplication, so is a linear subspace. The coefficient of $x$ is unique: $(a-b)x=0$ with $a\mathrm{\ne }b$ would imply $x=0$ on multiplying by $(a-b{)}^{-1}$. Thus $g(ax)=a\mathrm{\parallel }x\mathrm{\parallel }$ is well-defined and scalar-linear. Moreover $\mathrm{\mid }g(ax)\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }ax\mathrm{\parallel }$ and $g(x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel })=1$, so $\mathrm{\parallel }g\mathrm{\parallel }=1$.

For the final assertion alone, suppose $h(x)=0$ for all $h\in H$ and the span of $H$ is norm dense. Every finite combination $g={\sum }_{j<n}{a}_j{h}_j$ satisfies $g(x)={\sum }_{j<n}{a}_j{h}_j(x)=0$, including $n=0$, so every element of the span vanishes at $x$.

Apply norm-preserving extension to this $M$ and $g$. Its hypotheses were checked in step 1.1, so it gives $f\in X^{\ast }$ with $\mathrm{\parallel }f\mathrm{\parallel }=1$ and $f(x)=g(x)=\mathrm{\parallel }x\mathrm{\parallel }$. This is an existence statement for the fixed $x$.

For $v\mathrm{\ne }w$, apply step 2.1 to $x=v-w\mathrm{\ne }0$. The resulting functional satisfies $f(v)-f(w)=f(v-w)=\mathrm{\parallel }v-w\mathrm{\parallel }>0$, hence separates these points.

For any $h\in X^{\ast }$ and $x\mathrm{\ne }0$, normalization gives $\mathrm{\mid }h(x)\mathrm{\mid }=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\mid }h(x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel })\mathrm{\mid }\le \mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$; for $x=0$ both sides vanish. Thus every unit-ball value is at most $\mathrm{\parallel }x\mathrm{\parallel }$. For nonzero $x$ step 2.1 attains this upper bound; for $x=0$ the zero functional has norm zero and attains value zero. This proves the maximum formula even if $X=\{0\}$.

Fix $f\in X^{\ast }$ and $\epsilon >0$. By density a ball of radius $\epsilon$ about $f$ meets the span, so there is $g$ in the span with $\mathrm{\parallel }f-g\mathrm{\parallel }<\epsilon$. Thus $\mathrm{\mid }f(x)\mathrm{\mid }=\mathrm{\mid }(f-g)(x)\mathrm{\mid }\le \mathrm{\parallel }f-g\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le \epsilon \mathrm{\parallel }x\mathrm{\parallel }$. If $\mathrm{\parallel }x\mathrm{\parallel }>0$ and $\mathrm{\mid }f(x)\mathrm{\mid }>0$, taking $\epsilon =\mathrm{\mid }f(x)\mathrm{\mid }\mathrm{/}(2\mathrm{\parallel }x\mathrm{\parallel })$ is impossible; if $\mathrm{\parallel }x\mathrm{\parallel }=0$, then $x=0$ already. Therefore all $f$ vanish at $x$, and the maximum formula gives $\mathrm{\parallel }x\mathrm{\parallel }=0$, hence $x=0$.

By the operator-norm lemma, the space $X^{\ast }$ of bounded scalar-linear maps $X\to \mathbb{K}$ is itself a normed vector space. Therefore its dual $(X^{\ast }{)}^{\ast }$ is defined; this is the meaning of $X^{\ast \ast }$.

Fix $x\in X$. For $a,b\in \mathbb{K}$ and $f,g\in X^{\ast }$, evaluation gives $(af+bg)(x)=af(x)+bg(x)$, so the map $E_x:f\mapsto f(x)$ is scalar-linear. If $x\mathrm{\ne }0$, $\mathrm{\mid }f(x)\mathrm{\mid }=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\mid }f(x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel })\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }f\mathrm{\parallel }$; if $x=0$, $f(x)=0$. Thus $E_x$ is bounded and belongs to $X^{\ast \ast }$.

Define $J_X(x)=E_x$. For $x,y\in X$ and $a,b\in \mathbb{K}$, evaluating at every $f\in X^{\ast }$ gives $J_X(ax+by)(f)=f(ax+by)=a{J}_X(x)(f)+b{J}_X(y)(f)$. Equality at all arguments is equality of functions, hence $J_X$ is scalar-linear.

Taking the supremum of $\mathrm{\mid }{E}_x(f)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }f\mathrm{\parallel }$ over $\mathrm{\parallel }f\mathrm{\parallel }\le 1$ yields $\mathrm{\parallel }{J}_{X}x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$. This also proves boundedness of $J_X$ with constant one. At $x=0$, $E_0$ is the zero functional and has norm zero. No step required HB or completeness.

By the evaluation construction, $J_X$ is scalar-linear and $\mathrm{\parallel }{J}_{X}x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$ for every $x$. In particular $J_{X}0=0$ and equality of the norms holds at zero.

For $x\mathrm{\ne }0$, HB norming gives $f\in X^{\ast }$ with $\mathrm{\parallel }f\mathrm{\parallel }=1$ and $f(x)=\mathrm{\parallel }x\mathrm{\parallel }$. The bidual norm is the supremum over the dual unit ball, which contains this $f$, so $\mathrm{\parallel }{J}_{X}x\mathrm{\parallel }\ge \mathrm{\mid }{J}_{X}x(f)\mathrm{\mid }=\mathrm{\mid }f(x)\mathrm{\mid }=\mathrm{\parallel }x\mathrm{\parallel }$. Combining with step 1.1 proves equality at every $x$.

For $x,y\in X$, linearity and the established equality give $\mathrm{\parallel }{J}_{X}x-J_{X}y\mathrm{\parallel }=\mathrm{\parallel }{J}_X(x-y)\mathrm{\parallel }=\mathrm{\parallel }x-y\mathrm{\parallel }$. If $J_{X}x=J_{X}y$, the left side is zero, so definiteness of the norm gives $x=y$.

Write $p=p_U$. The gauge definition gives $p(x)\ge 0$. For every $t>\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$ one has $x\mathrm{/}t\in B(0,r)\subseteq U$, so $p(x)\le t$. If $p(x)>\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$, their midpoint is such a $t$ smaller than $p(x)$, impossible. Hence $p(x)\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$.

For $a>0$, the condition $s\in S_{ax}$ is equivalent to $s\mathrm{/}a\in S_x$, so $S_{ax}=a{S}_x$. The number $ap(x)$ is a lower bound of this set. Conversely, for every $\epsilon >0$, choose $t\in S_x$ with $t<p(x)+\epsilon \mathrm{/}a$; then $at\in S_{ax}$ and $at<ap(x)+\epsilon$. The infimum criterion gives $p(ax)=ap(x)$. For $a=0$, both sides are zero by $p(0)=0$.

If $s>p(x)$, choose $s^{\mathrm{\prime }}\in S_x$ with $s^{\mathrm{\prime }}<s$ using the infimum criterion with $\epsilon =s-p(x)$. Since $0<s^{\mathrm{\prime }}\mathrm{/}s<1$, convexity and $0\in U$ give $x\mathrm{/}s=(s^{\mathrm{\prime }}\mathrm{/}s)(x\mathrm{/}{s}^{\mathrm{\prime }})+(1-s^{\mathrm{\prime }}\mathrm{/}s)0\in U$. Thus every $s>p(x)$ is admissible.

If $p(x)<1$, choose $s\in S_x$ with $s<1$ using the infimum criterion with $\epsilon =1-p(x)$. Then $0<s<1$ and $x=s(x\mathrm{/}s)+(1-s)0\in U$ by convexity.

Given $\epsilon >0$, set $s=p(x)+\epsilon$ and $t=p(y)+\epsilon$. Both are positive and admissible by step 1.3. Convexity gives $(x+y)\mathrm{/}(s+t)=(s\mathrm{/}(s+t))(x\mathrm{/}s)+(t\mathrm{/}(s+t))(y\mathrm{/}t)\in U$. Hence $p(x+y)\le s+t=p(x)+p(y)+2\epsilon$. If the desired inequality failed with positive difference $d$, taking $\epsilon =d\mathrm{/}4$ would give $d\le d\mathrm{/}2$. Therefore $p(x+y)\le p(x)+p(y)$. Together with step 1.2, this is sublinearity on the real space.

Conversely, let $x\in U$. For $x=0$, $p(x)=0<1$. If $x\mathrm{\ne }0$, openness gives $\eta >0$ with $B(x,\eta )\subseteq U$. Put $d=\eta \mathrm{/}(2\mathrm{\parallel }x\mathrm{\parallel })>0$. Since $\mathrm{\parallel }(1+d)x-x\mathrm{\parallel }=\eta \mathrm{/}2$, $(1+d)x\in U$, so $1\mathrm{/}(1+d)\in S_x$ and $p(x)\le 1\mathrm{/}(1+d)<1$. This proves both inclusions in the asserted set equality.

Subadditivity gives $p(x)-p(y)\le p(x-y)\le \mathrm{\parallel }x-y\mathrm{\parallel }\mathrm{/}r$ and, with $x,y$ interchanged, $p(y)-p(x)\le p(y-x)\le \mathrm{\parallel }x-y\mathrm{\parallel }\mathrm{/}r$. These two real inequalities yield the Lipschitz bound. When $x=y$ both differences are zero; no use of $p(-v)=p(v)$ occurs.

Fix $c_0\in C$, set $U=C-c_0$ and $v=z-c_0$. Then $0\in U$ and $v\mathrm{\notin }U$, hence $v\mathrm{\ne }0$. If $u_j=c_j-c_0\in U$ and $0\le t\le 1$, then $(1-t)u_1+t{u}_2=((1-t)c_1+t{c}_2)-c_0\in U$. A ball in $C$ about $c$ translates to a ball of the same radius in $U$ about $c-c_0$, so $U$ is open. Fix $r>0$ with $B(0,r)\subseteq U$.

Let $p=p_U$. Its gauge is finite and sublinear on the underlying real space, nonnegative everywhere, and bounded above by $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$. Because $v\mathrm{\notin }U=\{p<1\}$, $p(v)\ge 1$.

The set $M=\mathbb{R}v$ is a real linear subspace. Since $v\mathrm{\ne }0$, $tv$ has a unique real coefficient $t$, and $g(tv)=t$ defines a real-linear functional. For $t\ge 0$, $g(tv)=t\le tp(v)=p(tv)$; for $t<0$, $g(tv)=t<0\le p(tv)$. This checks domination on every element of M, including zero.

Applying relative dominated extension on the underlying real $X$ gives real-linear $F$ extending $g$ and satisfying $-p(-x)\le F(x)\le p(x)$. The two gauge upper bounds imply $-\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r\le F(x)\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$, so $\mathrm{\mid }F(x)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$. Also $F(v)=1$.

Over $\mathbb{R}$ put $f=F$. Over $\mathbb{C}$ put $f(x)=F(x)-iF(ix)$; reconstruction gives complex linearity and real part $F$, and $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\mid }F(x)\mathrm{\mid }+\mathrm{\mid }F(ix)\mathrm{\mid }\le 2\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$. Thus in either field $f\in X^{\ast }$, and it is nonzero because $\mathrm{Re}f(v)=1$. The explicit bounds give continuity: $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le (2\mathrm{/}r)\mathrm{\parallel }x-y\mathrm{\parallel }$ in both cases.

For every $c\in C$, $c-c_0\in U$, so $F(c-c_0)\le p(c-c_0)<1=F(z-c_0)$. Adding $F(c_0)$ gives $\mathrm{Re}f(c)=F(c)<F(z)=\mathrm{Re}f(z)$, as required.

Form the set of all admissible pairs $T=\{(a,r)\in K\times (0,\mathrm{\infty }):B(a,3r)\cap C=\mathrm{\varnothing }\}$ and the family $\mathcal{U}=\{K\cap B(a,r):(a,r)\in T\}$. For each fixed $k\in K$, the open complement of $C$ contains $k$, so some $s>0$ has $B(k,s)\cap C=\mathrm{\varnothing }$. Then $(k,s\mathrm{/}3)\in T$ and $k\in K\cap B(k,s\mathrm{/}3)$. Thus $\mathcal{U}$ covers $K$ without selecting radii for all $k$ simultaneously.

Each $V=K\cap B(a,r)$ in this family is open for the restricted metric on $K$. Indeed, if $k\in V$, then $r-\mathrm{\parallel }k-a\mathrm{\parallel }>0$, and every $y\in K$ with $\mathrm{\parallel }y-k\mathrm{\parallel }<r-\mathrm{\parallel }k-a\mathrm{\parallel }$ satisfies $\mathrm{\parallel }y-a\mathrm{\parallel }\le \mathrm{\parallel }y-k\mathrm{\parallel }+\mathrm{\parallel }k-a\mathrm{\parallel }<r$, so is in $V$. Thus $\mathcal{U}$ is an intrinsic open cover of $K$.

Compactness gives a finite subcover $V_0,\dots ,V_{n-1}$ with $n\ge 1$, since $K\mathrm{\ne }\mathrm{\varnothing }$. For each index $j<n$ define $W_j=\{(a,r)\in T:V_j=K\cap B(a,r)\}$. Each $W_j$ is nonempty by the definition of $\mathcal{U}$. Applying finite choice to the function $j\mapsto W_j$ supplies pairs $(a_j,r_j)\in W_j$ for these finitely many indices. Repeated $V_j$ or $W_j$ cause no problem: a choice function on the set of values can be evaluated at each $W_j$.

The finite list $r_0,\dots ,r_{n-1}$ consists of positive reals, so its minimum $\delta$ exists and is positive, since it equals one of those reals.

For any $k\in K$ choose an index $j<n$ with $k\in V_j$, possible because the finite family covers $K$. For every $c\in C$, admissibility gives $\mathrm{\parallel }c-a_j\mathrm{\parallel }\ge 3r_j$, whereas $\mathrm{\parallel }k-a_j\mathrm{\parallel }<r_j$. Hence $\mathrm{\parallel }c-k\mathrm{\parallel }\ge \mathrm{\parallel }c-a_j\mathrm{\parallel }-\mathrm{\parallel }k-a_j\mathrm{\parallel }>2r_j\ge \delta$. This proves the uniform bound for all $k,c$.

For (i), put $D=A-B=\{x-y:x\in A,y\in B\}$. It is nonempty. For $x_j-y_j\in D$ and $0\le t\le 1$, their convex combination equals $((1-t)x_1+t{x}_2)-((1-t)y_1+t{y}_2)\in D$. If $x-y\in D$, choose $r>0$ with $B(x,r)\subseteq A$; then $B(x-y,r)\subseteq D$ by keeping $y$ fixed. Thus $D$ is open and convex. If $0\in D$, then some $x\in A$ equals some $y\in B$, contrary to disjointness; hence $0\mathrm{\notin }D$.

For (ii), now suppose $A$ is closed and $B$ compact. The distance-gap lemma applied with $K=B,C=A$ gives $\delta >0$ with $\mathrm{\parallel }x-y\mathrm{\parallel }\ge \delta$ for all $x\in A,y\in B$. Put $\rho =\delta \mathrm{/}2$ and $O=A+B(0,\rho )$. The ball is convex by the triangle inequality, so for $x_j+h_j\in O$ each convex combination has its $A$ component in $A$ and its ball component of norm less than $\rho$, including the weights zero and one. Thus $O$ is convex. A ball about $x+h$ of radius $\rho -\mathrm{\parallel }h\mathrm{\parallel }$ stays in $O$, so $O$ is open; it contains nonempty $A$. If $x+h=y\in B$, then $\mathrm{\parallel }x-y\mathrm{\parallel }=\mathrm{\parallel }h\mathrm{\parallel }<\rho <\delta$, impossible. Therefore $O$ and $B$ are disjoint.

For (i), apply point separation to $D$ and $0$. It gives $0\mathrm{\ne }f\in X^{\ast }$ with $u(d)<u(0)=0$ for every $d\in D$, where $u=\mathrm{Re}f$ is nonzero bounded real-linear. Consequently $u(x)<u(y)$ for every $x\in A,y\in B$. Fix $y_0\in B$. The nonempty real set $u(A)$ is bounded above by $u(y_0)$, so it has a real supremum $a_0$. Explicitly, $a_0=-\inf (-u(A))$; reflection of lower bounds makes this the least upper bound. Since every $u(y)$ bounds $u(A)$, $a_0\le u(y)$ for all $y\in B$.