Basic Analysis – Gently Done Topological Vector Spaces

78 Basic Analysis
C = {f ∈ X : p 1 (f) < 1}. Then we see that the Minkowski functional p C is equal
to the norm p 1 . In particular, p C is a norm. However, we claim that C is not
bounded. To see this, let U be the neighbourhood of 0 given by
U = {f ∈ X : p 1 (f) < 1 and p 2 (f) < 1}.
IfC were bounded, thenC ⊆ sU forallsufficiently larges. However, for anys > 0,
the function g(x) = 1
2 sin4sx belongs to C but not to sU (since p2 (g) = 2s > s).
It follows that C is not bounded, as claimed.
Theorem 7.24 The topology on a separated topological vector space (X,T) is a
norm topology if and only if 0 possesses a bounded convex neighbourhood.
Proof Suppose that the topology on the topological vector space X is given by a
norm �·�. Then the set {x : �x� < 1} is an open bounded convex neighbourhood
of 0.
Conversely, suppose that U is a bounded convex neighbourhood of 0. Then, by
Proposition 7.13, there is a balanced, convex neighbourhood V of 0 with V ⊆ U.
Since U is bounded, so is V. By Proposition 7.22, µ V , the Minkowski functional
associated with V is a norm.
We shall show that �·� = µ V (·) induces the topology T on X. Since µ V (x) ≤ 1,
for x ∈ V, by Theorem 7.17, it follows from Theorem 7.8 that µ V is continuous.
On the other hand, suppose that �x ν � → 0. For any neighbourhood G of 0 there
is s > 0 such that V ⊆ sG, since V is bounded. Now, there is ν 0 such that
�x ν � < 1/s for all ν � ν 0 . For such ν, we have, �sx ν � < 1 and so sx ν ∈ V ⊆ sG,
by Theorem 7.17. Hence x ν ∈ G whenever ν � ν 0 . It follows that x ν → 0 with
respect to T. Hence, we have the following equivalent statements;
x ν → x with respect to T,
⇐⇒ x ν −x → 0 with respect to T,
⇐⇒ x ν −x → 0 with respect to �·�,
⇐⇒ x ν → x with respect to �·�.
It follows that T and the topology given by the norm �·� have the same convergent
nets, and hence the same closed sets and therefore the same open sets, i.e., they
are equal.
Department of Mathematics King’s College, London