Locally Convex Spaces and Continuous Separation: Examples

The function $i\mapsto 0$ is a specified element of ${\mathbb{K}}^I$, and $(x+y)(i)=x(i)+y(i)$, $(ax)(i)=ax(i)$ and $(-x)(i)=-x(i)$ define functions on $I$. Associativity, commutativity and the zero/inverse laws follow at each coordinate from the scalar field; the two distributive laws, associativity of scalar action and the unit action also follow at each coordinate. Function equality is coordinatewise, so these give every vector-space axiom.

The $i$-th component of addition is $(x,y)\mapsto x_i+y_i$, a composite of the continuous coordinate maps and scalar addition. The $i$-th component of scalar multiplication is $(a,x)\mapsto a{x}_i$, similarly continuous by joint scalar multiplication. Therefore both vector operations are jointly continuous by product universality. This uses neither projection surjectivity nor nonemptiness of an arbitrary product of unrelated factors.

Each $N(F,\epsilon )$ is open, being a finite intersection of inverse images of open scalar disks, and contains zero. If $x,y$ are in it and $0\le t\le 1$, then $\mathrm{\mid }(1-t)x_i+t{y}_i\mathrm{\mid }\le (1-t)\mathrm{\mid }{x}_i\mathrm{\mid }+t\mathrm{\mid }{y}_i\mathrm{\mid }<{\epsilon }_i$ for $0<t<1$, with $t=0,1$ immediate. If $\mathrm{\mid }a\mathrm{\mid }\le 1$, then $\mathrm{\mid }a{x}_i\mathrm{\mid }\le \mathrm{\mid }{x}_i\mathrm{\mid }<{\epsilon }_i$, including $a=0$. Thus these neighborhoods are convex and balanced. Given any basic zero-neighborhood, choose a positive radius inside each of its finitely many coordinate neighborhoods, by finite choice after listing those coordinates. The resulting $N(F,\epsilon )$ is contained in it. Hence these sets form a base and the TVS is locally convex. For $F=\mathrm{\varnothing }$, the set is the whole space.

If $x\mathrm{\ne }y$, some coordinate $i$ has $d=\mathrm{\mid }{x}_i-y_i\mathrm{\mid }>0$. The inverse images of the disks of radius $d\mathrm{/}3$ about $x_i,y_i$ are open neighborhoods of $x,y$. They are disjoint, since a common scalar value would imply $d<2d\mathrm{/}3$ by the triangle inequality. Thus the space is Hausdorff. For $I=\mathrm{\varnothing }$, its only element is the empty function; its only zero-neighborhood is the whole singleton, the vector operations are constant, and the Hausdorff assertion is vacuous.

Pointwise operations give $f(x+y)=f(x)+f(y)$ and $f(cx)=cf(x)$, so the projection $f$ is scalar-linear; it is continuous. Thus $u=\mathrm{Re}f$ is continuous and real-linear. The rays $(-\mathrm{\infty },a]$ and $[b,\mathrm{\infty })$ are closed: their complements are unions of open intervals. Their inverse images under $u$ are therefore closed, since inverse images of their open complements are open. They are convex because real-linear maps preserve real convex combinations and each ray is convex.

The constant functions with values $a$ and $b$ belong to $A$ and $B$, respectively, so both sets are nonempty. Their intersection is empty since $b>a$. Set $\alpha =(a+b)\mathrm{/}2$ and $\epsilon =(b-a)\mathrm{/}2>0$. Then for every $x\in A,y\in B$, $\mathrm{Re}f(x)\le a=\alpha -\epsilon <\alpha +\epsilon =b\le \mathrm{Re}f(y)\mathrm{.}$ The constant-one function has $f(1)=1$, so $f$ is nonzero.

For distinct $x,y$, fix one index $i$ with $z=x_i-y_i\mathrm{\ne }0$. Over $\mathbb{C}$ put $c=\bar{z}\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$; then $\mathrm{\mid }c\mathrm{\mid }=1$ and $cz=\mathrm{\mid }z\mathrm{\mid }>0$. Over $\mathbb{R}$ put $c=z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, which gives the same identities. Thus $g(w)=c{w}_i$ is continuous and scalar-linear, and $\mathrm{Re}g(x)-\mathrm{Re}g(y)=\mathrm{\mid }z\mathrm{\mid }>0$. The index and unit scalar are chosen for this one supplied pair; there is no simultaneous choice. If $I$ is empty there is no $i_0$ and no pair of distinct functions, so the respective hypotheses do not arise.

The interval $[-1,1]$ is convex since $-1\le s,t\le 1$ implies $-1\le (1-r)s+rt\le 1$ for $0\le r\le 1$. It is closed, its complement being the open rays $(-\mathrm{\infty },-1)$ and $(1,\mathrm{\infty })$, and bounded by one in absolute value. Hence it is nonempty compact convex. For $s,t\in K$, $\mathrm{\mid }q(s)-q(t)\mathrm{\mid }=\mathrm{\mid }s-t\mathrm{\mid }\mathrm{\mid }s+t\mathrm{\mid }\le 2\mathrm{\mid }s-t\mathrm{\mid }$, which proves continuity on $K$ directly.

For $s,t\in K$ and $0\le r\le 1$, $(1-r)s^2+r{t}^2-((1-r)s+rt{)}^2=r(1-r)(s-t{)}^2\ge 0.$ Thus $q((1-r)s+rt)\le (1-r)q(s)+rq(t)$, the defining convexity inequality for a function. It includes $r=0,1$ and $s=t$, where equality holds.

On $K$, $q(t)\le 1$ with equality exactly when $t=-1$ or $t=1$, because $1-t^2=(1-t)(1+t)$ and both factors are nonnegative. The maximum set is therefore $M=\{-1,1\}$. Its midpoint is zero, and $q(0)=0<1$, so $0\mathrm{\notin }M$. Thus $M$ is not convex and cannot be a face. This refutes the claim with a continuous convex function on a compact convex set.

Nevertheless $M$ is extremal. If $(1-r)s+rt=1$ with $s,t\in K$ and $0<r<1$, then $(1-r)(1-s)+r(1-t)=0$. Each summand is nonnegative and each coefficient is positive, so $s=t=1$. Similarly a combination equal to $-1$ gives $(1-r)(s+1)+r(t+1)=0$ and forces $s=t=-1$. Thus the endpoint property holds even though convexity fails. All witnesses and computations are explicit and choice-free.