Local convexity, convex and balanced sets, and the continuous dual

Let $X$ be a real or complex TVS (Topological vector spaces over the real and complex fields). A subset $C$ is convex if $(1-t)x+ty\in C$ whenever $x,y\in C$ and $0\le t\le 1$, with real coefficients even when $X$ is complex. Its convex hull is $\mathrm{co}(S)=\{{\sum }_{j=1}^n{t}_j{x}_j:n\ge 1,x_j\in S,t_j\ge 0,{\sum }_{j=1}^n{t}_j=1\}\mathrm{.}$ In particular $\mathrm{co}(\mathrm{\varnothing })=\mathrm{\varnothing }$. This is the smallest convex superset of $S$: single-term sums contain $S$, and concatenating two weighted lists proves convexity. Every convex superset contains every finite convex combination: induct on list length, remove a zero coefficient, and otherwise group the first $n-1$ terms with weight $1-t_n$. If $t_n=1$ the value is $x_n$; if $t_n<1$, divide those first weights by $1-t_n$ and apply the induction hypothesis followed by binary convexity.

A set $C$ is balanced if $\lambda C\subseteq C$ for every scalar with $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$, and absolutely convex if it is convex and balanced. Empty sets satisfy both conditions; every nonempty balanced set contains zero, by taking $\lambda =0$, and is symmetric, by taking $\lambda =-1$ twice.

The TVS $X$ is locally convex if every zero-neighborhood contains a convex zero-neighborhood. Equivalently it has a base of open convex zero-neighborhoods. Indeed, if $C$ is a convex zero-neighborhood, its interior contains zero. For $x,y\in \mathrm{int}C$ and $0<t<1$, the set $(1-t)\mathrm{int}C+t\mathrm{int}C$ is open: it is a union of translates of the open set $(1-t)\mathrm{int}C$, using Translations, dilations and absorption in a topological vector space. It contains $(1-t)x+ty$ and lies in $C$, so this point lies in the interior. The cases $t=0,1$ are immediate. Conversely an open convex zero-neighborhood is a convex zero-neighborhood.

A seminorm is a finite-valued $p:X\to [0,\mathrm{\infty })$ satisfying $p(x+y)\le p(x)+p(y)$ and $p(\lambda x)=\mathrm{\mid }\lambda \mathrm{\mid }p(x)$ for all scalars. It is continuous if continuous for the given topology and the usual real topology. Thus $p(0)=0$, but $p(x)=0$ need not imply $x=0$. Over the underlying real vector space it is sublinear in the sense of A sublinear functional on a real vector space; restriction of scalars is justified by A field is a vector space over itself, and over any subfield $K\subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars.

The continuous dual $X^{\mathrm{\prime }}$ consists of all continuous $\mathbb{K}$-linear maps $f:X\to \mathbb{K}$. The scalar field is a vector space over itself by A field is a vector space over itself, and over any subfield $K\subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars, clause 1, so these are linear functionals as in Linear functionals and the algebraic dual $V^{\ast }=\mathcal{L}(V,F)$. Pointwise operations make $X^{\mathrm{\prime }}$ a vector subspace of the algebraic dual: zero is continuous, and sums and scalar multiples are continuous by the scalar-operation continuity proved in Translations, dilations and absorption in a topological vector space. Explicitly, continuity of $f,g$ at $x$ bounds their errors by $\epsilon \mathrm{/}2$ to control the sum, and by $\epsilon \mathrm{/}(\mathrm{\mid }a\mathrm{\mid }+1)$ to control $af$. All linear axioms are inherited pointwise.

For separation inequalities write $u=\mathrm{Re}f$, or $u=f$ over $\mathbb{R}$. The real part is continuous because $\mathrm{\mid }\mathrm{Re}z-\mathrm{Re}w\mathrm{\mid }\le \mathrm{\mid }z-w\mathrm{\mid }$. It is real-linear. No Hahn–Banach or choice principle is part of these definitions.