Basic Analysis – Gently Done Topological Vector Spaces

7: Locally Convex Topological Vector Spaces 77
Definition 7.18 The map p C constructed in Theorem 7.17 is called the Minkowski
gauge or functional associated with the convex absorbing set C in the vector space
X.
Example 7.19 Let X be a normed space, and let C be the ball C = {x : �x� < r},
with r > 0. Then p C (x) = inf{s > 0 : x ∈ sC} = inf{s > 0 : �x/s� < r} =
inf{s > 0 : �x� < sr} = �x�/r.
Example 7.20 Let S be the strip S = {z ∈ C : |Imz| < 1} in the one-dimensional
complex vector space C. Then S is convex and absorbing but not balanced—for
example, 2 ∈ S but 2i /∈ S. Given s > 0, we see that z ∈ sS if and only if
|Imz| < s, so that p S (z) = |Imz|.
We see that p S (2) = 0 �= 2 = p S (2i), so that p S fails to be absolutely homogeneous
and so is not a seminorm. Notice that p S is positively homogeneous and
subadditive, as it should be.
Example 7.21 LetS bethestripS = {(x,y) ∈ R 2 : |y| < 1}inthetwo-dimensional
real vector space R 2 . We see that S is convex, absorbing and balanced. For given
s > 0, (x,y) ∈ sS if and only if |y| < s, which implies that p S ((x,y)) = |y|. In this
case, p S is a seminorm; clearly p S ((x,y)) ≥ 0, p S (t(x,y)) = |t||y| = |t|p ( S(x,y)),
and
p S ((x,y)+(x ′ ,y ′ )) = |y +y ′ | ≤ |y|+|y ′ | = p S ((x,y))+p S ((x ′ ,y ′ )).
Notice that S is an unbounded set in R 2 and that p S is not a norm. Indeed,
p S ((x,0)) = 0 for any x ∈ R.
Proposition 7.22 The Minkowski functional p V associated with a bounded,
balanced, convex neighbourhood of 0 in a topological vector space X is a norm.
Proof We recall that any neighbourhood of 0 is absorbing and so p V is welldefined.
Let x ∈ X with x �= 0. Then there is a neighbourhood W of 0 such that
x /∈ W. However, since V is bounded, V ⊆ sW for all sufficiently large s. But
x /∈ W implies that sx /∈ sW, so that sx /∈ V for some (sufficiently large) s. By
Theorem 7.17, it is false that p V (sx) < 1, and so we must have p V (sx) ≥ 1. In
particular, p V (x) = (1/s)p V (sx) �= 0, and we conclude that the seminorm p V is in
fact a norm.
Example 7.23 Let X be the real vector space of bounded, continuously differentiable
real-valued functions on (−1,1) which vanish at the origin and which
have bounded derivatives. Let p 1 and p 2 be the norms on X given by p 1 (f) =
sup{|f(x)| : x ∈ (−1,1)} and p 2 (f) = sup{|f ′ (x)| : x ∈ (−1,1)} for f ∈ X.
Equip X with the vector space topology T given by the pair P = {p 1 ,p 2 } and let