1Normed vector spaces
II Linear Analysis
1 Normed vector spaces
In IB Linear Algebra, we have studied vector spaces in quite a lot of detail.
However, just knowing something is a vector space usually isn’t too helpful.
Often, we would want the vector space to have some additional structure. The
first structure we will study is a norm.
Definition
(Normed vector space)
.
A normed vector space is a pair (
V, k · k
),
where
V
is a vector space over a field
F
and
k · k
is a function
k · k
:
V 7→ R
,
known as the norm, satisfying
(i) kvk ≥ 0 for all v ∈ V , with equality iff v = 0.
(ii) kλvk = |λ|kvk for all λ ∈ F, v ∈ V .
(iii) kv + wk ≤ kvk + kwk for all v, w ∈ V .
Intuitively, we think of kvk as the “length” or “magnitude” of the vector.
Example.
Let
V
be a finite dimensional vector space, and
{e
1
, ··· , e
n
}
a basis.
Then, for any v =
P
n
i=1
v
i
e
i
, we can define a norm as
kvk =
v
u
u
t
n
X
i=1
v
2
i
.
If we are given a norm of a vector space
V
, we immediately obtain two more
structures on V for free, namely a metric and a topology.
Recall from IB Metric and Topological Spaces that (
V, d
) is a metric space if
the metric d : V × V → R satisfies
(i) d(x, x) = 0 for all x ∈ V .
(ii) d(x, y) = d(y, x) for all x, y ∈ V .
(iii) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ V .
Also, a topological spaces is a set
V
together with a topology (a collection of
open subsets) such that
(i) ∅ and V are open subsets.
(ii) The union of open subsets is open.
(iii) The finite intersection of open subsets is open.
As we have seen in IB Metric and Topological Spaces, a norm on a vector space
induces a metric by
d
(
v, w
) =
kv −wk
. This metric in terms defines a topology
on
V
where the open sets are given by “
U ⊆ V
is open iff for any
x ∈ U
, there
exists some ε > 0 such that B(x, ε) = {y ∈ V : d(x, y) < ε} ⊆ U”.
This induced topology is not just a random topology on the vector space.
They have the nice property that the vector space operators behave well under
this topology.
Proposition.
Addition + :
V ×V → V
, and scalar multiplication
·
:
F×V → V
are continuous with respect to the topology induced by the norm (and the usual
product topology).
Proof.
Let
U
be open in
V
. We want to show that (+)
−1
(
U
) is open. Let
(
v
1
, v
2
)
∈
(+)
−1
(
U
), i.e.
v
1
+
v
2
∈ U
. Since
v
1
+
v
2
∈ U
, there exists
ε
such that
B
(
v
1
+
v
2
, ε
)
⊆ U
. By the triangle inequality, we know that
B
(
v
1
,
ε
2
)+
B
(
v
2
,
ε
2
)
⊆
U. Hence we have (v
1
, v
2
) ∈ B
(v
1
, v
2
),
ε
2
⊆ (+)
−1
(U). So (+)
−1
(U) is open.
Scalar multiplication can be done in a very similar way.
This motivates the following definition — we can do without the norm, and
just require a topology in which addition and scalar multiplication are continuous.
Definition
(Topological vector space)
.
A topological vector space (
V, U
) is
a vector space
V
together with a topology
U
such that addition and scalar
multiplication are continuous maps, and moreover singleton points
{v}
are
closed sets.
The requirement that points are closed is just a rather technical requirement
needed in certain proofs. We should, however, not pay too much attention to
this when trying to understand it intuitively.
A natural question to ask is: when is a topological vector space normable? i.e.
Given a topological vector space, can we find a norm that induces the topology?
To answer this question, we will first need a few definitions.
Definition
(Absolute convexity)
.
Let
V
be a vector space. Then
C ⊆ V
is
absolutely convex (or balanced convex ) if for any
λ, µ ∈ F
such that
|λ|
+
|µ| ≤
1,
we have λC + µC ⊆ C. In other words, if c
1
, c
2
∈ C, we have λc
1
+ µc
2
∈ C.
Proposition.
If (
V, k · k
) is a normed vector space, then
B
(
t
) =
B
(
0, t
) =
{v
:
kvk < t} is absolutely convex.
Proof. By triangle inequality.
Definition
(Bounded subset)
.
Let
V
be a topological vector space. Then
B ⊆ V
is bounded if for every open neighbourhood
U ⊆ V
of
0
, there is some
s >
0 such
that B ⊆ tU for all t > s.
At first sight, this might seem like a rather weird definition. Intuitively, this
just means that
B
is bounded if, whenever we take any open set
U
, by enlarging
it by a scalar multiple, we can make it fully contain B.
Example. B(t) in a normed vector space is bounded.
Proposition.
A topological vector space (
V, U
) is normable if and only if there
exists an absolutely convex, bounded open neighbourhood of 0.
Proof.
One direction is obvious — if
V
is normable, then
B
(
t
) is an absolutely
convex, bounded open neighbourhood of 0.
The other direction is not too difficult as well. We define the Minkowski
functional µ : V → R by
µ
C
(v) = inf{t > 0 : v ∈ tC},
where C is our absolutely convex, bounded open neighbourhood.
Note that by definition, for any
t < µ
C
(
v
),
v 6∈ tC
. On the other hand, by
absolute convexity, for any t > µ
C
(v), we have v ∈ tC.
We now show that this is a norm on V :
(i) If v = 0, then v ∈ 0C. So µ
C
(0) = 0. On the other hand, suppose v 6= 0.
Since a singleton point is closed,
U
=
V \{v}
is an open neighbourhood of
0. Hence there is some
t
such that
C ⊆ tU
. Alternatively,
1
t
C ⊆ U
. Hence,
v 6∈
1
t
C. So µ
C
(v) ≥
1
t
> 0. So µ
C
(v) = 0 iff v = 0.
(ii) We have
µ
C
(λv) = inf{t > 0 : λv ∈ tC} = λ inf{t > 0 : v ∈ tC} = λµ
C
(v).
(iii) We want to show that
µ
C
(v + w) ≤ µ
C
(v) + µ
C
(w).
This is equivalent to showing that
inf{t > 0 : v + w ∈ tC} ≤ inf{t > 0 : v ∈ tC} + inf{r > 0 : w ∈ rC}.
This is, in turn equivalent to proving that if
v ∈ tC
and
w ∈ rC
, then
(v + w) ∈ (t + r)C.
Let
v
0
=
v/t, w
0
=
w/r
. Then we want to show that if
v
0
∈ C
and
w
0
∈ C
,
then
1
(t+r)
(
tv
0
+
rw
0
)
∈ C
. This is exactly what is required by convexity.
So done.
In fact, the condition of absolute convexity can be replaced “convex”, where
“convex” means for every
t ∈
[0
,
1],
tC
+ (1
− t
)
C ⊆ C
. This is since for every
convex bounded
C
, we can find always find a absolutely convex bounded
˜
C ⊆ C
,
which is something not hard to prove.
Among all normed spaces, some are particularly nice, known as Banach
spaces.
Definition
(Banach spaces)
.
A normed vector space is a Banach space if it is
complete as a metric space, i.e. every Cauchy sequence converges.
Example.
(i)
A finite dimensional vector space (which is isomorphic to
F
n
for some
n
)
is Banach.
(ii) Let X be a compact Hausdorff space. Then let
B(X) = {f : X → R such that f is bounded}.
This is obviously a vector space, and we can define the norm be
kfk
=
sup
x∈X
f
(
x
). It is easy to show that this is a norm. It is less trivial to
show that this is a Banach space.
Let
{f
n
} ⊆ B
(
X
) be a Cauchy sequence. Then for any
x
,
{f
n
(
x
)
} ⊆ R
is
also Cauchy. So we can define f(x) = lim
n→∞
f
n
(x).
To show that
f
n
→ f
, let
ε >
0. By definition of
f
n
being Cauchy,
there is some
N
such that for any
n, m > N
and any fixed
x
, we have
|f
n
(
x
)
− f
m
(
x
)
| < ε
. Take the limit as
m → ∞
. Then
f
m
(
x
)
→ f
(
x
). So
|f
n
(
x
)
− f
(
x
)
| ≤ ε
. Since this is true for all
x
, for any
n > N
, we must
have kf
n
− fk ≤ ε. So f
n
→ f.
(iii) Define X as before, and let
C(X) = {f : X → R such that f is continuous}.
Since any continuous
f
is bounded, so
C
(
X
)
⊆ B
(
X
). We define the norm
as before.
Since we know that
C
(
X
)
⊆ B
(
X
), to show that
C
(
X
) is Banach, it suffices
to show that
C
(
X
)
⊆ B
(
X
) is closed, i.e. if
f
n
→ f
,
f
n
∈ C
(
X
), then
f ∈ C
(
X
), i.e. the uniform limit of a continuous function is continuous.
Proof can be found in IB Analysis II.
(iv) For 1 ≤ p < ∞, define
ˆ
L
p
([0, 1]) = {f : [0, 1] → R such that f is continuous}.
We define the norm k · k
ˆ
L
p
by
kfk
ˆ
L
p
=
Z
1
0
|f|
p
dx
1/p
.
It is easy to show that
ˆ
L
p
is indeed a vector space, and we now check that
this is a norm.
(a) kfk
ˆ
L
p
≥
0 is obvious. Also, suppose that
kfk
ˆ
L
p
= 0. Then we must
have
f
= 0. Otherwise, if
f 6
= 0, say
f
(
x
) =
ε
for some
x
. Then there
is some
δ
such that for any
y ∈
(
x − δ, x
+
δ
), we have
kf
(
y
)
k ≥
ε
2
.
Hence
kfk
ˆ
L
p
=
Z
1
0
|f|
p
dx
1/p
≥
h
2δ
ε
2
p
i
1/p
> 0.
(b) kλfk = |λ|kf k is obvious
(c)
The triangle inequality is the exactly what the Minkowski inequality
says, which is in the example sheet.
It turns out that
ˆ
L
p
is not a Banach space. We can brute-force a hard
proof here, but we will later develop some tools that allow us to prove this
much more easily.
Hence, we define
L
p
([0
,
1]) to be the completion of
ˆ
L
p
([0
,
1]). In IID
Probability and Measure, we will show that L
p
([0, 1]) is in fact the space
L
p
([0, 1]) =
f : [0, 1] → R such that
Z
1
0
|f|
p
dx < ∞
/∼,
where the integral is the Lebesgue integral, and we are quotienting by the
relation
f ∼ g
if
f
=
g
Lebesgue almost everywhere. You will understand
what these terms mean in the IID Probability and Measure course.
(v) `
p
spaces: for p ∈ [1, ∞), define
`
p
(F) =
(
(x
1
, x
2
, ···) : x
i
∈ F,
∞
X
i=1
|x
i
|
p
< ∞
)
,
with the norm
kxk
`
p
=
∞
X
i=1
|x
i
|
p
!
1/p
.
It should be easy to check that this is a normed vector space. Moreover,
this is a Banach space. Proof is in example sheet.
(vi) `
∞
space: we define
`
∞
=
(x
1
, x
2
, ···) : x
i
∈ F, sup
i∈N
|x
i
| < ∞
with norm
kxk
`
∞
= sup
i∈N
|x
i
|.
Again, this is a Banach space.
(vii)
Let
B
=
B
(1) be the unit open ball in
R
n
. Define
C
(
B
) to be the set of
continuous functions
f
:
B → R
. Note that unlike in our previous example,
these functions need not be bounded. So our previous norm cannot be
applied. However, we can still define a topology as follows:
Let
{K
i
}
∞
i=1
be a sequence of compact subsets of
B
such that
K
i
⊆ K
i+1
and
S
∞
i=1
K
i
= B. We define the basis to include
f ∈ C(B) : sup
x∈K
i
|f(x)| <
1
m
for each m, i = 1, 2, ···, as well as the translations of these sets.
This weird basis is chosen such that
f
n
→ f
in this topology iff
f
n
→ f
uniformly in every compact set. It can be showed that this is not normable.