Basic Analysis – Gently Done Topological Vector Spaces

46 Basic Analysis
Proposition 5.14 Suppose that ℓ is a linear functional on the vector space X
such that ℓ does not vanish on the whole of X. Then kerℓ is a maximal proper
subspace. Conversely, any maximal proper subspace of X is the kernel of a linear
functional.
Proof It is clear that the kernel of any linear functional ℓ is a linear subspace of
X. Moreover, kerℓ is a proper subspace if and only if ℓ is not identically zero.
Suppose that ℓ(z) �= 0 for some z ∈ X. For any x ∈ X, we have
x = ℓ(x)
�
z + x−
ℓ(z) ℓ(x)
ℓ(z) z
�
.
But x−(ℓ(x)/ℓ(z))z ∈ kerℓ, so it follows that kerℓ is maximal.
Now suppose that V is a maximal proper subspace of X, and let z be any
element of X such that z /∈ V. Since V is maximal, any x ∈ X can be written as
x = αz +v, for suitable α ∈ K and v ∈ V. Moreover, this decomposition of x is
unique; ifalsox = α ′ z+v ′ forsomeα ′ ∈ Kandv ′ ∈ V, then0 = (α−α ′ )z+(v−v ′ ).
Since z /∈ V, we must have α = α ′ and therefore v = v ′ . Define ℓ : X → K by
ℓ(x) = α, where x = αz+v, as above, Evidently, ℓ is well-defined and is linear. If
x = αz +v and ℓ(x) = 0, we have α = 0 and so x = v ∈ V. Clearly, ℓ(v) = 0 for
v ∈ V and so kerℓ = V.
Proposition 5.15 Let ℓ 1 and ℓ 2 be linear functionals on the vector space X,
neither being identically zero. Then ℓ 1 and ℓ 2 have the same kernel if and only if
they are proportional.
Proof Suppose that kerℓ 1 = kerℓ 2 . Let z ∈ X with ℓ 1 (z) �= 0. For any x ∈ X,
x−(ℓ 1 (x)/ℓ 1 (z))z ∈ kerℓ 1 = kerℓ 2 so that ℓ 2 (x) = ℓ 1 (x)ℓ 2 (z)/ℓ 1 (z). Thus ℓ 1 and
ℓ 2 are proportional.
Clearly, if ℓ 1 and ℓ 2 are proportional, then they have the same kernel since the
constant of proportionality is not zero, by hypothesis.
Proposition 5.16 Let ℓ and ℓ 1 ,...ℓ n be linear functionals on the vector space
X. Then either
(i) ℓ is a linear combination of ℓ 1 ,...ℓ n , or
(ii) there is z ∈ X such that ℓ(z) = 1 and ℓ 1 (z) = ··· = ℓ n (z) = 0.
Proof By considering a subset of the ℓ i if necessary, we may assume, without
loss of generality, that ℓ 1 ,...ℓ n are linearly independent. Define the map γ : X →
K n+1 by
γ(x) = (ℓ(x),ℓ 1 (x),...,ℓ n (x)), x ∈ X.
Department of Mathematics King’s College, London