Basic Analysis – Gently Done Topological Vector Spaces

7: Locally Convex Topological Vector Spaces 73
Definition 7.11 A subset A of a vector space over K is said to be convex if
sx+(1−s)y ∈ A whenever x,y ∈ A and 0 ≤ s ≤ 1. In other words, A is convex
if the line segment between any two points in A also lies in A.
Suppose that p is a seminorm on a vector space X, r > 0 and let A = {x ∈
X : p(x) < r}. Then A is balanced and convex. Indeed, for any x ∈ A, if |t| ≤ 1,
then p(tx) = |t|p(x) < r. Also, if 0 < s < 1 and x,y ∈ A, then
p(sx+(1−s)y) ≤ p(sx)+p((1−s)y)
= sp(x)+(1−s)p(y)
< sr +(1−s)r = r
so that sp(x) + (1 − s)p(y) ∈ A. Furthermore, for any x ∈ X and t > 0,
p(tx) = tp(x) < r for all sufficiently small t. It follows that A is absorbing.
To summarise, we have shown that A is a convex, balanced absorbing set. Since
these properties are each preserved under (finite) intersections, it follows that the
basic neighbourhoods V(0,p 1 ,...,p n ;r) of 0 determined by a family of seminorms
on X are convex, balanced and absorbing.
Definition 7.12 A topological vector space is said to be locally convex if there is
a neighbourhood base at 0 consisting of convex sets.
Thus, a topological vector space given by a family of seminorms is a locally
convextopologicalvectorspace. Infact, weshallshowthattheconverseholds,that
is, any locally convex topological vector space is a topological vector space whose
topology is determined by a family of seminorms. Such a family is given by certain
seminorms associated with the collection of convex and balanced neighbourhoods
of 0.
Proposition 7.13 Any convex neighbourhood of 0 contains a balanced convex
neighbourhood of 0. In particular, in any locally convex topological vector space,
there is a neighbourhood base at 0 consisting of convex, balanced (and absorbing)
sets.
Proof Let U be any convex neighbourhood of 0. Then U contains a balanced
neighbourhood W, say. For any s ∈ K with |s| = 1, we have sW ⊆ W and
s −1 W ⊆ W, so that sW = W = s −1 W. In particular, s −1 W = W ⊆ U so
that W ⊆ sU. Set V = �
|s|=1sU. Then V ⊆ U and W ⊆ V so that V is
a neighbourhood of 0 contained in U. Furthermore, each sU is convex, and so,
therefore, is their intersection, V. We claim that V is balanced. To see this, let
t ∈ K with |t| ≤ 1. For any v ∈ V, we have v ∈ sU for all s with |s| = 1. Hence,
for any s with |s| = 1, tv ∈ s|t|U ⊆ sU, since U is convex and contains 0. It
follows that tv ∈ V and so V is balanced.