Relative dual norming, point separation, and recovery of the norm

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 $x=0$.