Chapter 18

Topological Vector Spaces18.1 Fundamental Considerations

The right context to consider certain topics like separation theorems is in locally convextopological vector spaces, a generalization of normed linear spaces. Let X be a vectorspace and let Ψ be a collection of functions defined on X such that if ρ ∈ Ψ,

ρ(x+ y)≤ ρ(x)+ρ(y),

ρ(ax) = |a|ρ(x) if a ∈ F,

ρ(x)≥ 0,

where F denotes the field of scalars, either R or C, assumed to be C unless otherwisespecified. These functions are called seminorms because it is not necessarily true that x = 0when ρ (x) = 0. A basis for a topology, B, is defined as follows.

Definition 18.1.1 For A a finite subset of Ψ and r > 0,

BA(x,r)≡ {y ∈ X : ρ(x− y)< r for all ρ ∈ A}.

ThenB ≡ {BA(x,r) : x ∈ X ,r > 0,and A⊆Ψ,A finite}.

That this really is a basis is the content of the next theorem.

Theorem 18.1.2 B is the basis for a topology.

Proof: I need to show that if BA(x,r1) and BB(y,r2) are two elements of B and if z ∈BA(x,r1)∩BB(y,r2), then there exists U ∈B such that

z ∈U ⊆ BA(x,r1)∩BB(y,r2).

Letr = min(min{(r1−ρ(z− x)) : ρ ∈ A},

min{(r2−ρ(z− y)) : ρ ∈ B})

and consider BA∪B(z,r). If w belongs to this set, then for ρ ∈ A,

ρ(w− z)< r1−ρ(z− x).

Henceρ (w− x)≤ ρ(w− z)+ρ(z− x)< r1

for each ρ ∈ A and so BA∪B(z,r)⊆ BA(x,r1). Similarly, BA∪B(z,r)⊆ BB(y,r2). This provesthe theorem.

Let τ be the topology consisting of unions of all subsets of B. Then (X ,τ) is a locallyconvex topological vector space.

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