An explicit norming functional for the finite-dimensional maximum norm

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.