Basic Analysis – Gently Done Topological Vector Spaces

72 Basic Analysis
at 0 is to say that Λ is continuous at 0. Thus, by Theorem 7.8, (i) and (iii) are
equivalent.
Next we show that (iv) follows from (iii). Indeed, if (iii) holds, and if ℓ 1 ,...,ℓ m
are elements of F such that p i (x) = |ℓ i (x)| for x ∈ X and 1 ≤ i ≤ m, then it is
clear that
m�
kerℓi ⊆ kerΛ.
i=1
The statement (iv) now follows, by Corollary 5.17. Finally, it is clear that (iv)
implies (i), since each |ℓ i |, and hence each ℓ i , 1 ≤ i ≤ m, is continuous at 0, by
Theorem 7.1
Theorem 7.10 Let X and Y be topological vector spaces with topologies
determined by families P and Q of seminorms, respectively. For a linear map
T : X → Y, the following statements are equivalent.
(i) T is continuous at 0 in X.
(ii) T is continuous.
(iii) For any given seminorm q in Q, there is a finite set of seminorms p 1 ,...,p m
in P and a constant C > 0, possibly depending on q, such that
q(Tx) ≤ C(p 1 (x)+···+p m (x)) for x ∈ X.
If, in particular, X and Y are normed spaces, then (i), (ii) and (iii) are equivalent
to the following.
(iv) There is a constant C > 0 such that
�Tx� ≤ C�x� for x ∈ X.
Proof We have already shown that (i) and (ii) are equivalent—as a direct consequence
of the linearity of T (Proposition6.19). The map x ↦→ q(Tx) is a seminorm
on X and so (ii) implies (iii), by Theorem 7.8. Suppose that (iii) holds. Let x ν be
any net converging to 0 in X. Then p(x ν ) → 0, for any p in P, so that q(Tx ν ) → 0,
by (iii), for any q in Q. Hence Tx ν → 0 in Y and (i) holds.
If X and Y are normed, then the families P and Q can be taken to be singleton
sets consisting of just the norm. Clearly, (iv) follows from (iii) and (i) follows from
(iv).
Department of Mathematics King’s College, London