Convex sets and continuous real-hyperplane separation in a normed space

Let $X$ be a normed space over $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$ with the metric and scalar convention of Real and complex scalar conventions for normed spaces. A subset $C\subseteq X$ is convex when $x,y\in C,0\le t\le 1⟹(1-t)x+ty\in C,$ where $t$ is real, also when $\mathbb{K}=\mathbb{C}$. The empty set and every singleton are convex: the former has no pair of points to test, and $(1-t)x+tx=x$ for the latter. At $t=0,1$ the convex combination is one of its endpoints.

For a nonzero $f\in X^{\ast }$ (the bounded scalar-linear dual of The dual space X^* of a normed space and its dual norm) put $u=\mathrm{Re}f$, with $u=f$ over $\mathbb{R}$. A continuous real affine hyperplane is a set $\{x:u(x)=a\}$ for $a\in \mathbb{R}$. For subsets $A,B\subseteq X$, this hyperplane gives:

• weak separation if $u(x)\le a\le u(y)$ for all $x\in A,y\in B$;
• open-side strict separation in the indicated orientation if $u(x)<a\le u(y)$ for all such $x,y$;
• uniform strict separation if there is $\epsilon >0$ with $u(x)\le a-\epsilon <a+\epsilon \le u(y)$ for all such $x,y$.

Only real numbers are ordered in these formulas. The last condition requires one positive margin that works for all pairs, rather than merely pointwise strict inequalities.

Here $u$ is a nonzero bounded real-linear functional. Indeed, if $f(w)\mathrm{\ne }0$ in the complex case, put $b=\overline{f(w)}\mathrm{/}\mathrm{\mid }f(w)\mathrm{\mid }$. Then $u(bw)=\mathrm{Re}(bf(w))=\mathrm{\mid }f(w)\mathrm{\mid }>0$; over the real field, $u=f\mathrm{\ne }0$. Normalizing a nonzero vector $v$ in the dual-norm definition gives $\mathrm{\mid }f(v)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$, also true at zero. Hence $\mathrm{\mid }u(x)-u(y)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }x-y\mathrm{\parallel }$ and $\mathrm{\parallel }f\mathrm{\parallel }>0$. If $u(x)\mathrm{\ne }a$, every $y$ with $\mathrm{\parallel }y-x\mathrm{\parallel }<\mathrm{\mid }u(x)-a\mathrm{\mid }\mathrm{/}(2\mathrm{\parallel }f\mathrm{\parallel })$ still has $u(y)\mathrm{\ne }a$. Thus the complement of the level set is open by The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, so the level set is closed. If $u(v)\mathrm{\ne }0$, the point $av\mathrm{/}u(v)$ is in the level set, and the set is its translate of $\ker u$; every $x$ decomposes as $x-u(x)v\mathrm{/}u(v)+u(x)v\mathrm{/}u(v)$ with the first term in $\ker u$. Thus it is an affine hyperplane of the underlying real space.

Brezis §1.2 definitions, pp.4–5; Teschl Theorems 5.2–5.3, pp.138–139.