Norming and Separation under Hahn–Banach: Examples

For $n\ge 1$, the finite list $\mathrm{\mid }{y}_1\mathrm{\mid },\dots ,\mathrm{\mid }{y}_n\mathrm{\mid }$ has a maximum. It is nonnegative and is zero exactly when each coordinate is zero. For any scalar $a$, ${\max }_k\mathrm{\mid }a{y}_k\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }{\max }_k\mathrm{\mid }{y}_k\mathrm{\mid }$, including $a=0$. Also each $\mathrm{\mid }{y}_k+z_k\mathrm{\mid }\le \mathrm{\mid }{y}_k\mathrm{\mid }+\mathrm{\mid }{z}_k\mathrm{\mid }\le \mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$, and taking the maximum gives the triangle inequality. These verify that $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$ is a norm over either field.

For $x\mathrm{\ne }0$, the set of maximizing indices in $\{1,\dots ,n\}$ is nonempty. Its least element $j$ exists by natural-number well-ordering. Then $\mathrm{\mid }{x}_j\mathrm{\mid }=\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}>0$, so $c=\overline{x_j}\mathrm{/}\mathrm{\mid }{x}_j\mathrm{\mid }$ is defined and $\mathrm{\mid }c\mathrm{\mid }=1$. The formula $f_x(y)=c{y}_j$ satisfies $f_x(ay+bz)=ac{y}_j+bc{z}_j=a{f}_x(y)+b{f}_x(z)$, and $\mathrm{\mid }{f}_x(y)\mathrm{\mid }=\mathrm{\mid }{y}_j\mathrm{\mid }\le \mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus $f_x$ is scalar-linear and bounded, with $\mathrm{\parallel }{f}_x\mathrm{\parallel }\le 1$.

Let $e_j$ have coordinate one in position $j$ and zero elsewhere, and put $y=(x_j\mathrm{/}\mathrm{\mid }{x}_j\mathrm{\mid })e_j$. Then $\mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and $f_x(y)=\overline{x_j}{x}_j\mathrm{/}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2=1$. Thus $\mathrm{\parallel }{f}_x\mathrm{\parallel }\ge 1$, proving $\mathrm{\parallel }{f}_x\mathrm{\parallel }=1$. Also $f_x(x)=\overline{x_j}{x}_j\mathrm{/}\mathrm{\mid }{x}_j\mathrm{\mid }=\mathrm{\mid }{x}_j\mathrm{\mid }=\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$.

For any bounded linear $f$ of norm at most one and nonzero $x$, $\mathrm{\mid }f(x)\mathrm{\mid }=\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }f(x\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }})\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$; step 3.1 attains equality. At $x=0$, every linear $f$ gives value zero and the zero functional attains the same maximum. For $n=0$ the vector space has just zero and every linear functional sends it to zero, so the dual has only the zero functional of norm zero; its unit ball is nonempty but it has no norm-one element.

At $x=(1,1)$, $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}=1$. Each coordinate functional satisfies $\mathrm{\mid }{f}_k(y)\mathrm{\mid }=\mathrm{\mid }{y}_k\mathrm{\mid }\le \mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $f_k(e_k)=1$, so $\mathrm{\parallel }{f}_k\mathrm{\parallel }=1$ for $k=1,2$. Both give $f_k(x)=1$, while $f_1(1,0)=1$ and $f_2(1,0)=0$, so they are distinct. In dimension one the same displayed construction is $f_x(y)=\bar{x}y\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, with its norm and value computed in steps 2.1 and 3.1. No extension or infinite selection was used.

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.