Basic Analysis – Gently Done Topological Vector Spaces

76 Basic Analysis
are equivalent; s ∈ A x , x ∈ sC, tx ∈ tsC, ts ∈ A tx , which completes the proof of
(i).
To prove (ii), let x and y be given elements of X and let a ∈ A x and b ∈ A y . Then
we have that x ∈ aC and y ∈ bC, and so x + y ∈ aC + bC ⊆ (a + b)C, by the
convexity of C. That is, a+b ∈ A x+y .
Now assume that C is also balanced and let t ∈ K with |t| > 0. Write t = α|t|,
where |α| = 1. Then αC ⊆ C and also α −1 C ⊆ C, so that αC = C. Hence, the
following statements are equivalent; s ∈ A |t|x , |t|x ∈ sC, α|t|x ∈ αsC, tx ∈ sC,
s ∈ A tx . We conclude that A |t|x = A tx .
Theorem 7.17 For any convex absorbing set C in a vector space X the map
p C : x ↦→ p C (x) = inf{s > 0 : x ∈ sC}, x ∈ X, is positively homogeneous and
subadditive on X.
If, in addition, C is balanced, then p C is a seminorm on X. Furthermore, {x :
p C (x) < 1} ⊆ C ⊆ {x : p C (x) ≤ 1}, and if B is any convex, absorbing and
balanced set such that {x : p C (x) < 1} ⊆ B ⊆ {x : p C (x) ≤ 1} then p B = p C .
Proof We use the notation of the previous theorem, Theorem 7.16. By part (i)
of Theorem 7.16, for any x ∈ X and t > 0, we have that p C (tx) = inf A tx =
inf tA x = t inf A x = tp C (x) which shows that p C is positively homogeneous.
Also, for any x,y ∈ X, we have (by part (ii) of Theorem 7.16), p C (x + y) =
infA x+y ≤ a+b for any a ∈ A x , b ∈ A y . Hence p C (x+y)−b ≤ infA x , b ∈ A y ,
and so p C (x+y)−infA x ≤ infA y which gives p C (x+y) ≤ p C (x)+p C (y). That
is, p C is subadditive.
Now suppose that C is also balanced. We wish to show that p C is a seminorm.
Clearly p C (x) ≥ 0 and p C (0) = infA 0 = 0. Now let t ∈ K. Then p C (tx) =
infA tx = infA |t|x = inf|t|A x = |t|p C (x) and it follows that p C is a seminorm on
X, as required.
If z ∈ {x : p C (x) < 1}, then infA z < 1 and so there is some s ∈ A z with s < 1.
Thus z ∈ sC and so (1/s)z ∈ C. It follows that z ∈ C since C is convex and
contains 0. If x ∈ C, then 1 ∈ A x and so infA x ≤ 1, i.e., p C (x) ≤ 1.
Suppose that D,D ′ are any two convex, absorbing, balanced sets such that
D ⊆ D ′ . Then, for any x ∈ X, {s : sx ∈ D} ⊆ {s : sx ∈ D ′ }. Taking the infima,
it follows that p D ′(x) ≤ p D (x). It follows that if B is any convex, absorbing and
balanced set such that D ⊆ B ⊆ D ′ then p D ′ ≤ p B ≤ p D .
In particular, this holds with D = {x : p C (x) < 1} and D ′ = {x : p C (x) ≤ 1}.
However, p D (x) = inf{s > 0 : x ∈ sD} = inf{s > 0 : x/s ∈ D} = inf{s > 0 :
p C (x) < s} = p C (x). Similarly, p D ′(x) = inf{s > 0 : x ∈ sD ′ } = inf{s > 0 : x/s ∈
D ′ } = inf{s > 0 : p C (x) ≤ s} = p C (x) and we have shown that p D = p D ′ = p C .
We conclude that p C = p D ′ ≤ p B ≤ p D = p C and the proof is complete.
Department of Mathematics King’s College, London