486 CHAPTER 18. TOPOLOGICAL VECTOR SPACES

Theorem 18.2.5 Let K be closed and convex in a locally convex topological vector spaceand let p /∈ K. Then there exists a real number c, and f ∈ X ′ such that

Re f (p)> c > Re f (k)

for all k ∈ K.

Proof: Since K is closed, and p /∈ K, there exists a finite subset of Ψ,A, and a positiver > 0 such that

K∩BA (p,2r) = /0.

Pick k0 ∈ K and letU = K +BA (0,r)− k0, q = p− k0.

It follows that U is an open convex set containing 0 and q /∈ U . Therefore, by Lemma18.2.3, there exists f ∈ X ′ such that

Re f (p− k0) = Re f (q)> Re f (k+ e− k0) (18.2.7)

for all k ∈ K and e ∈ BA (0,r). If Re f (e) = 0 for all e ∈ BA (0,r), then Re f = 0 and 18.2.7could not hold. Therefore, Re f (e)> 0 for some e ∈ BA (0,r) and so,

Re f (p)> Re f (k)+Re f (e)

for all k ∈ K. Let c1 ≡ sup{Re f (k) : k ∈ K}. Then for all k ∈ K,

Re f (p)≥ c1 +Re f (e)> c1 +Re f (e)

2> Re f (k).

Let c = c1 +Re f (e)

2 .

K

{x : Re f (x) = c} p

Corollary 18.2.6 In the situation of the above theorem, there exist real numbers c,d suchthat Re f (p)> d > c > Re f (k) for all k ∈ K.

Proof: From the theorem, there exists ĉ such that Re f (p) > ĉ > Re f (k) for all k ∈K. Thus Re f (p) > ĉ ≥ supk∈K Re f (k). Now choose d,c such that f (p) > d > c > ĉ ≥supk∈K Re f (k)> f (k).

Note that if the field of scalars comes from R rather than C there is no essential changeto the above conclusions. Just eliminate all references to the real part.