484 CHAPTER 18. TOPOLOGICAL VECTOR SPACES

Proof: Let x ∈ X be arbitrary. There exists A⊆Ψ such that

0 ∈ BA (0,r)⊆U.

Then rx2ρA (x)

∈ BA (0,r)⊆U

which implies2ρA (x)

r≥ m(x) . (18.2.6)

Thus m(x) is defined on X .Let x/t ∈U, y/s ∈U . Then since U is convex,

x+ yt + s

=

(t

t + s

)(xt

)+

(s

t + s

)(ys

)∈U.

It follows thatm(x+ y)≤ t + s.

Choosing s, t such that t− ε < m(x) and s− ε < m(y),

m(x+ y)≤ m(x)+m(y)+2ε.

Since ε is arbitrary, this shows 18.2.4. It remains to show 18.2.5. Let x/t ∈ U . Then ifλ > 0,

λxλ t∈U

and so m(λx)≤ λ t. Thus m(λx)≤ λm(x) for all λ > 0. Hence

m(x) = m(

λ−1

λx)≤ λ

−1m(λx)≤ λ−1

λm(x) = m(x)

and soλm(x) = m(λx) .

This proves the proposition.

Lemma 18.2.3 Let U be an open convex set containing 0 and let q /∈U. Then there existsf ∈ X ′ such that

Re f (q)> Re f (x)

for all x ∈U.

Proof: Let m be the Minkowski functional just defined and let

F (cq) = cm(q)

for c ∈ R. If c > 0 thenF (cq) = m(cq)