Page 104 - Textos de Matemática Vol. 34

94 JOSE´ CARLOS PETRONILHO
Proof. The topology generated by ℘ is defined by the sets V℘(0,ε,{pf1,···,pfk}):={u∈P′ :pfν(u)<ε,ν=1,···,k}
(which constitute a basis of neighborhoods of the origin in P′ [18, p.129]), where {pf1,··· ,pfk} run over the set of all finite subsets of P and ε belongs to the set of all positive real numbers, and the topology generated by I is defined by the sets
VI(0,ε,{|·|n1,···,|·|nk}):={u∈P′ :|u|nν <ε,ν=1,···,k} ,
where {n1, · · · , nk} run over the set of all finite subsets of N0 and ε belongs to the set of all positive real numbers. Therefore, since for each n ∈ N0 there exists f ∈ P such that |u|n = pf (u) —it suffices to take as f the polynomial xm where m is an nonneg- ative integer number ≤ n for which sup0≤m≤n |⟨u, xm⟩| is attained—, then each set of the form VI (0,ε,{|·|n1,··· ,|·|nk}) is also of the form V℘ (0,ε,{pf1,··· ,pfk}). On the other hand, for a given f ∈ P, putting f(x) = nν=0 aνxν , we have
n (2.2) pf(u) = |⟨u,f⟩| =
ν=0
n aν⟨u,xν⟩ ≤ max |us|
0≤s≤n ν=0
|aν| = |u|n ∥f∥n ,
and so each set of the form V℘ (0,ε,{pf1,··· ,pfk}) contains some set of the form VI (0,ε,{|·|n1,··· ,|·|nk}). ThisshowsthatV℘ andVI generatethesametopology on P′. The last statement in the theorem follows, e.g., from [18, Theorem V.5].
At this point, the important fact asserts that for the topology considered in P every linear functional in P is continuous.
Theorem 2.2. The equality
(2.3) P∗ = P′
holds, P being carried with the topology of the hyper-strict inductive limit of the spaces Pn.
Proof. We just have to prove that any u ∈ P∗ is continuous for the topology in P. Since this topology is the hyper-strict inductive limit of the spaces Pn, it suffices to show that, for each n, the restriction u|Pn is continuous [18, p. 147]. Since Pn is a metric space and u is linear, we only need to prove that if {fk}k is an arbitrary sequence in Pn such that fk → 0 in Pn (for the norm ∥·∥n) then ⟨u,fk⟩ → 0 (in C). But this is an immediate consequence of the inequalities
|⟨u,fk⟩| ≤ |u|n∥fk∥n , k = 0,1,2,··· , which follow from (2.2).