Basic Analysis – Gently Done Topological Vector Spaces

7: Locally Convex Topological Vector Spaces 75
out that discontinuous linear functionals cannot always be found. If the topology
of a topological vector space is sufficiently strong, then all linear functionals are
continuous.
This would certainly be the case if we could equip a vector space with the
discrete topology. However, this is not a vector topology so we must look further.
In fact, we will show that any vector space X can be equipped with a vector
topology T such that every linear functional on X is continuous. To construct
such a topology, let P denote the collection of all seminorms on X. P is not empty
because X can be equipped with a norm, and so this will appear in the family P.
Let T be the vector topology on X determined by P. We note that T is Hausdorff
because P is separating—after all, it contains a norm. By Theorem 7.1, every
member of P is continuous. Let ℓ : X → K be any linear functional on X. Then
|ℓ| is a seminorm on X and therefore |ℓ| belongs to P. It follows that |ℓ| is
continuous at 0 and hence ℓ is continuous at 0. By Corollary 7.9, we conclude that
ℓ is continuous.
Proposition 7.15 For any seminorm p on a vector space X over K, the sets
{x ∈ X : p(x) < 1} and {x ∈ X : p(x) ≤ 1} are convex, absorbing and balanced.
Proof Let x,y ∈ {x : p(x) < 1} and let 0 < s < 1. Then p(sx + (1 − s)y) ≤
p(sx) + p((1 − s)y) = sp(x) + (1 − s)p(y) < 1 which shows that {x : p(x) < 1}
is convex. For any z ∈ X, we have p(tz) = |t|p(z) < 1 for all t ∈ K with |t|
sufficiently large, so that {x : p(x) < 1} is absorbing. Now let t ∈ K with |t| ≤ 1.
Then, p(tx) = |t|p(x) ≤ p(x) < 1 and we see that {x : p(x) < 1} is balanced. An
almost identical argument applies to the set {x : p(x) ≤ 1}.
This rather easy result tells us how to get convex, absorbing and balanced sets
from seminorms. It turns out that one can go in the other direction—from convex,
absorbing and balanced sets we can construct seminorms, as we now show.
Theorem 7.16 Suppose that C is a convex absorbing set in a vector space
X. For each x ∈ X, let A x = {s > 0 : x ∈ sC}. Then A x enjoys the following
properties.
(i) A x �= ∅ for x ∈ X, A 0 = (0,∞), and A tx = tA x for any t > 0.
(ii) For any x,y ∈ X, A x +A y ⊆ A x+y .
(iii) If, in addition, C is balanced, then A tx = |t|A x for any t ∈ K with |t| > 0.
Proof A x is non-empty because C is absorbing—in fact, A x contains a set of
the form (s,∞) for some suitably large s. The set C contains 0 and so certainly
t0 = 0 ∈ C for all t > 0, i.e., A 0 = (0,∞). For any t > 0, the following statements