Quantitative separation of a norm ball from an exterior point

Since $d=\mathrm{\parallel }z-a\mathrm{\parallel }>r\ge 0$, $z-a\mathrm{\ne }0$. Apply dual norming to this vector to get $\mathrm{\parallel }f\mathrm{\parallel }=1$ and $f(z-a)=d$, a positive real. Put $u=\mathrm{Re}f$. Linearity gives $u(z)=u(a)+d$.

If $\mathrm{\parallel }x-a\mathrm{\parallel }\le r$, then $u(x)-u(a)=\mathrm{Re}f(x-a)\le \mathrm{\mid }f(x-a)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }x-a\mathrm{\parallel }\le r$. Thus $u(x)\le u(a)+r<u(a)+d=u(z)$. If $\mathrm{\parallel }x-a\mathrm{\parallel }<r$ with $r>0$, the same chain gives $u(x)<u(a)+r$.

Set $b=u(a)+(r+d)\mathrm{/}2$ and $\epsilon =(d-r)\mathrm{/}2>0$. Direct subtraction and addition give $b-\epsilon =u(a)+r$ and $b+\epsilon =u(a)+d=u(z)$. Hence step 2.1 gives $u(x)\le b-\epsilon <b+\epsilon =u(z)$, the prescribed uniform margin. For $r=0$, the closed ball is $\{a\}$ by norm definiteness and these formulas give margin $d\mathrm{/}2>0$.

For a numerical instance, take $X=\mathbb{R}$, $a=0$, $r=1$, $z=3$, and $f(t)=t$. Then $\mathrm{\parallel }f\mathrm{\parallel }={\sup }_{\mathrm{\mid }t\mathrm{\mid }\le 1}\mathrm{\mid }t\mathrm{\mid }=1$, $d=3$, $b=2$, and $\epsilon =1$. Every $x\in [-1,1]$ satisfies $f(x)=x\le 1=b-\epsilon <3=b+\epsilon =f(3)$; at $x=1$ the left bound is attained. This explicitly realizes gap two and margin one.