Basic Analysis – Gently Done Topological Vector Spaces

104 Basic Analysis
Now we consider c 0 , the linear space of all complex sequences which converge
to 0, equipped with the supremum norm
�x� ∞ = sup{|x n | : n ∈ N}, for x = (x n ) ∈ c 0 .
One checks that c 0 is a Banach space (a closed subspace of ℓ ∞ ). We shall show
that the dual of c 0 is ℓ 1 , that is, there is an isometric isomorphism between c ∗ 0
and ℓ 1 . To see this, suppose first that z = (z n ) ∈ ℓ 1 . Define ψ z : c 0 → C to be
ψz : x ↦→ �
nznxn , x = (xn ) ∈ c0 . It is clear that ψz and that
|ψz (x)| ≤ �
n
|z n ||x n | ≤ �z� 1 �x� ∞ .
is well-defined for any z ∈ ℓ1
Thus we see that ψ z is a bounded linear functional with norm �ψ z � ≤ �z� 1 . Let x
be the element of c 0 given by x = (u 1 ,...,u m ,0,0,...), where u k = sgnz k (with
the convention that sgn0 = 1). Then z k u k = |z k | and �x� ∞ = 1 and we see
that ψz (x) = �m k=1 |zk |. Therefore �ψz� ≥ �m k=1 |zk |, for any m. It follows that
�ψz� = �z�1 and therefore z ↦→ ψz is an isometric mapping of ℓ1 into c∗ 0 . We shall
show that every element of c∗ 0 is of this form and hence z ↦→ ψz is onto.
is the
To see this, let λ ∈ c∗ 0 , and, for n ∈ N, let zn = λ(en ), where en ∈ c0 sequence (δnm ) m∈N . For any given N ∈ N, let
Then v ∈ c 0 , �v� ∞ = 1 and
|λ(v)| =
v =
N�
k=1
N�
k=1
sgnz k e k .
|z k | ≤ �λ��v� ∞ = �λ�.
It follows that z = (z n ) ∈ ℓ 1 and that �z� 1 ≤ �λ�. Furthermore, for any element
x = (x n ) ∈ c 0 ,
λ(x) = λ � ∞ �
n=1
x n e n
�
= � ∞ �
xnλ(en ) �
n=1
= ψ z (x).
Hence λ = ψ z , and the proof is complete.
Department of Mathematics King’s College, London