Basic Analysis – Gently Done Topological Vector Spaces

7: Locally Convex Topological Vector Spaces 79
Theorem 7.25 The topology of any locally convex topological vector space is
determined by a family of seminorms. Indeed, this family may be taken to be the
family P of Minkowski functionals associated with all convex, absorbing, balanced
neighbourhoods of 0.
Proof We have seen that in every locally convex topological vector space (X,T), 0
possesses a neighbourhood base of convex, balanced (absorbing) neighbourhoods.
Let P be the family of associated Minkowski functionals. Denote by T P the vector
topology on X determined by P. We must show that T P = T. Let p ∈ P. Then,
by definition, p is bounded on some neighbourhood of 0. Indeed, p = p C for some
(convex, balanced and absorbing) neighbourhood C of 0 and so C ⊆ {x ∈ X :
p C (x) ≤ 1}. That is to say, if x ∈ C, then p(x) = p C (x) ≤ 1. Thus p is bounded
on the neighbourhood C of 0 and so is continuous, by Theorem 7.8. Since T P is
the weakest vector topology on X such that each member of P is continuous, we
deduce that T P ⊆ T.
On the other hand, suppose that U is a convex, balanced neighbourhood of
0 with respect to the topology T. Then the interior of U is a convex, balanced
neighbourhood of 0, so we may assume that U is open. Let x ∈ U, so that U is
a neighbourhood of x. By continuity of scalar multiplication, there is δ > 0 such
that rx ∈ U whenever |r −1| < δ. In particular, there is r > 1 such that rx ∈ U,
(x) = inf{s > 0 : x ∈ sU} < 1 for any x ∈ U. It follows
i.e., x ∈ 1
rU. Hence pU from Theorem 7.17 that U = {x : pU (x) < 1}. This means that U = p −1
U ((−∞,1))
is in TP . Now, any non-empty T-open set is a union of translations of open convex,
balanced neighbourhoods of 0, so we conclude that any such set is also TP-open, i.e., T ⊆ TP , and we have equality, as required.
Proposition 7.26 A subset B of a locally convex topological vector space (X,T)
is bounded if and only if p(B) is a bounded subset of R for each continuous
seminorm p on X.
Proof Suppose that B is bounded and p is a continuous seminorm on X. Then
U = {x : p(x) < 1} is a neighbourhood of 0 in X. Since B is bounded, there is
s > 0 such that B ⊆ sU. But then if x ∈ B, we have (1/s)x ∈ U and therefore
p((1/s)x) < 1. Thus p(x) < s for all x ∈ B and p(B) is bounded.
Conversely, supposethatp(B)isboundedforeachcontinuousseminormponX
and let U be a neighbourhood of 0. Since (X,T) is locally convex, there is a family
of seminorms P on X which determine the topology T, by Theorem 7.25. Thus
there is some r > 0 and seminorms p 1 ,...,p n ∈ P such that V(0,p 1 ,...,p n ;r) ⊆
U. Now, each member of P is continuous and so, by hypothesis, each p i (B) is
bounded, so there is some d > 0 such that p i (x) ≤ d for x ∈ B and 1 ≤ i ≤ n.