Basic Analysis – Gently Done Topological Vector Spaces

52 Basic Analysis
so that Λ ↾ M = f. Furthermore, for any given x ∈ X, let α ∈ C be such that
|α| = 1 and αΛ(x) = |Λ(x)|. Then
|Λ(x)| = αΛ(x) = Λ(αx)
= ReΛ(αx), since lhs is real,
= g(αx)
≤ p(αx)
= p(x), for x ∈ X.
For a normed space, with norm �·�, the corresponding result is as follows.
Corollary 5.23 Let M be a linear subspace of a normed space X over K and let
f be a linear functional on M such that |f(x)| ≤ C�x� for some constant C > 0
and for all x ∈ M. Then there is a linear functional Λ on X such that Λ ↾ M = f
and |Λ(x)| ≤ C�x� for all x ∈ X.
Proof This follows immediately from the observation that the mapping x ↦→
p(x) = C�x� is a seminorm on X.
Corollary 5.24 Suppose that X is a normed space over K and that x 0 ∈ X.
There is a linear functional Λ on X such that Λ(x 0 ) = �x 0 � and |Λ(x)| ≤ �x� for
all x ∈ X. In particular, for any pair of distinct points x �= y in X, there is a
bounded linear functional Λ on X such that Λ(x) �= Λ(y).
Proof If x 0 = 0, take Λ = 0 on X. If x 0 �= 0, let M be the one-dimensional linear
subspace of X spanned by x 0 . Set p(x) = �x�, for x ∈ X, and define f : M → K
by f(αx 0 ) = α�x 0 �, α ∈ K. Then, for x = αx 0 ∈ M,
|f(x)| = |α|�x 0 � = �αx 0 � = �x�.
By the last corollary, there is an extension Λ of f such that |Λ(x)| ≤ �x� for all
x ∈ X.
Finally, for x �= y, set x 0 = x−y and let Λ be as above. Then Λ(x)−Λ(y) =
Λ(x−y) = Λ(x 0 ) = �x 0 � = �x−y� �= 0.
Department of Mathematics King’s College, London