Basic Analysis – Gently Done Topological Vector Spaces

60 Basic Analysis
Corollary 6.20 Suppose that X and Y are normed spaces and T : X → Y is a
linear map. The following statements are equivalent.
(i) T is continuous at 0.
(ii) T is continuous.
(iii) There is C > 0 such that �Tx� ≤ C�x�, for all x ∈ X.
Proof By the proposition, (i) and (ii) are equivalent, and certainly (iii) implies
(i). We shall show that (i) implies (iii). Let V be the neighbourhood of 0 in Y
given by V = {y ∈ Y : �y� ≤ 1}. The continuity of T at 0 implies that there is a
neighbourhood U of 0 in X such that T(U) ⊆ V. But U contains a ball of radius
r, for some r > 0, so that �x� < r implies that Tx ∈ V, that is, �Tx� ≤ 1. Hence,
for any x ∈ X, with x �= 0, we have that �T(rx/2�x�)� ≤ 1. It follows that
�Tx� ≤ 2
�x� x ∈ X, x �= 0,
r
and this inequality persists for x = 0.
Theorem 6.21 Suppose that Λ is a linear functional on a topological vector
space X, and suppose that Λ(x) �= 0 for some x ∈ X. The following statements
are equivalent.
(i) Λ is continuous.
(ii) kerΛ is closed.
(iii) Λ is bounded on some neighbourhood of 0.
Proof Since kerΛ = Λ −1 ({0}) and {0} is closed in K, it is clear that (i) implies
(ii).
Suppose that kerΛ is closed. By hypothesis, kerΛ �= X, so the complement
of kerΛ is a non-empty open set in X. Hence there is a point x /∈ kerΛ and
V = X\kerΛ is a (open) neighbourhood of x. Let U = V −x. Then U is an open
neighbourhood of 0 and so contains a balanced neighbourhood W, say, of 0. Thus
x+W ⊆ V and so (x+W)∩kerΛ = ∅. The linearity of Λ implies that the set
Λ(W) is a balanced subset of K and so is either bounded or equal to the whole
of K. If Λ(W) = K, there is some w ∈ W such that Λ(w) = −Λ(x). This gives
Λ(x + w) = 0, which is impossible since x + W contains no points of the kernel
of Λ. It follows that Λ(W) is bounded, that is, there is some K > 0 such that
|Λ(w)| < K for all w ∈ W. Hence (ii) implies (iii).
If (iii) holds, there is some neighbourhood V of 0 and some K > 0 such that
V is a neighbourhood
K V we have ε x ∈ V so that |Λ(K ε x)| < K, that is, |Λ(x)| <
|Λ(v)| < K for all v ∈ V. Let ε > 0 be given. Then ε
K
of 0, and if x ∈ ε
K
ε. It follows that Λ is continuous at 0, and hence Λ is continuous on X, by
Proposition 6.19.
Department of Mathematics King’s College, London