1Normed vector spaces
II Linear Analysis
1.2 Dual spaces
We will frequently be interested in one particular case of B(X, Y ).
Definition (Dual space). Let V be a normed vector space. The dual space is
V
∗
= B(V, F).
We call the elements of V
∗
functionals. The algebraic dual of V is
V
0
= L(V, F),
where we do not require boundedness.
One particularly nice property of the dual is that
V
∗
is always a Banach
space.
Proposition. Let V be a normed vector space. Then V
∗
is a Banach space.
Proof.
Suppose
{T
i
} ∈ V
∗
is a Cauchy sequence. We define
T
as follows: for
any
v ∈ V
,
{T
i
(
v
)
} ⊆ F
is Cauchy sequence. Since
F
is complete (it is either
R
or C), we can define T : V → R by
T (v) = lim
n→∞
T
n
(v).
Our objective is to show that
T
i
→ T
. The first step is to show that we indeed
have T ∈ V
∗
, i.e. T is a bounded map.
Let
kvk ≤
1. Pick
ε
= 1. Then there is some
N
such that for all
i > N
, we
have
|T
i
(v) − T (v)| < 1.
Then we have
|T (v)| ≤ |T
i
(v) − T (v)| + |T
i
(v)|
< 1 + kT
i
k
V
∗
kvk
V
≤ 1 + kT
i
k
V
∗
≤ 1 + sup
i
kT
i
k
V
∗
Since
T
i
is Cauchy,
sup
i
kT
i
k
V
∗
is bounded. Since this bound does not depend
on v (and N ), we get that T is bounded.
Now we want to show that kT
i
− T k
V
∗
→ 0 as n → ∞.
For arbitrary ε > 0, there is some N such that for all i, j > N, we have
kT
i
− T
j
k
V
∗
< ε.
In particular, for any v such that kvk ≤ 1, we have
|T
i
(v) − T
j
(v)| < ε.
Taking the limit as j → ∞, we obtain
|T
i
(v) − T (v)| ≤ ε.
Since this is true for any v, we have
kT
i
− T k
V
∗
≤ ε.
for all i > N. So T
i
→ T .
Exercise: in general, for
X, Y
normed vector spaces, what condition on
X
and Y guarantees that B(X, Y ) is a Banach space?