Basic Analysis – Gently Done Topological Vector Spaces

7: Locally Convex Topological Vector Spaces 81
Corollary 7.28 The space of continuous linear functionals on a separated locally
convex topological vector space separates the points of X, that is, for any x 1 �= x 2
in X there is a continuous linear functional Λ on X such that Λ(x 1 ) �= Λ(x 2 ).
Proof Given x 1 �= x 2 in X, set z = x 1 −x 2 . Then z �= 0. Let M be the linear
subspace M = {tz : t ∈ K} and define λ : M → K by λ(tz) = t. Then λ is a linear
functional on M with kerλ = {0} which is closed in M, since X and therefore
M is Hausdorff, by hypothesis. It follows that λ is continuous on M. Hence,
by Theorem 7.27, there is a continuous linear functional Λ : X → K such that
Λ ↾ M = λ. Furthermore, Λ(x 1 )−Λ(x 2 ) = Λ(z) = λ(z) = 1 �= 0.
Remark 7.29 Itisworthwhileexplicitlyhighlightingthefollowingstatementwhich
has more or less been proved in the course of the above argument. For any nonzero
element z in a separated topological vector space X, the map tz ↦→ t is a
homeomorphism between M, the one-dimensional subspace spanned by z, and K.
It is clear that this map is one-one and onto. Indeed, its inverse is t ↦→ tz, which
is continuous due to the continuity of scalar multiplication. The somewhat less
trivialpartisinshowing thattz ↦→ tiscontinuous—which aswesaw above, follows
from the fact that a linear functional is continuous if (and only if) its kernel is
closed.
Corollary 7.30 Let M be a linear subspace of a locally convex topological vector
space X and suppose that x 0 is an element of X not belonging to the closure of
M. Then there is a continuous linear functional Λ on X such that Λ(x 0 ) = 1 and
Λ(x) = 0 for all x ∈ M.
Proof Let B be a Hamel basis for M, the closure of M. Since M is a linear
subspace and x 0 /∈ M, it follows that {x 0 }∪B is a linearly independent set. Let
M 1 be the linear span of M and x 0 , that is, M 1 = {tx 0 +u : t ∈ K,u ∈ M}. Any
x ∈ M 1 can be written as x = tx 0 +u for unique t ∈ K and u ∈ M. Thus we may
define λ : M 1 → K by
λ(x) = t, for x = tx 0 +u ∈ M 1 .
Now, M is closed and x 0 /∈ M so there is a neighbourhood of x 0 which does not
meet M. Thus,
V(x 0 ,p 1 ,...,p n ;r)∩M = ∅
for suitable seminorms p 1 ,...,p n and r > 0. Let s(x) = p 1 (x)+···+p n (x), x ∈ X.
Then, for any u ∈ M, p i (x 0 −u) ≥ r for some 1 ≤ i ≤ n so that s(x 0 −u) ≥ r. In
particular, for any t �= 0 and u ∈ M,
s(tx0 +u) = |t|s � x0 + u�
≥ |t|r.
t