1Metric spaces
IB Metric and Topological Spaces
1.3 Norms
There are several notions on vector spaces that are closely related to metrics.
We’ll look at norms and inner products of vector spaces, and show that they all
naturally induce a metric on the space.
First of all, we define the norm. This can be thought of as the “length” of a
vector in the vector space.
Definition (Norm). Let
V
be a real vector space. A norm on
V
is a function
k · k : V → R such that
– kvk ≥ 0 for all v ∈ V
– kvk = 0 if and only if v = 0.
– kλvk = |λ|kvk
– kv + wk ≤ kvk + kwk.
Example. Let
V
=
R
n
. There are several possible norms we can define on
R
n
:
kvk
1
=
n
X
i=1
|v
i
|
kvk
2
=
v
u
u
t
n
X
i=1
v
2
i
kvk
∞
= max{|v
i
| : 1 ≤ i ≤ n}.
In general, we can define the norm
kvk
p
=
n
X
i=1
|v
i
|
p
!
1/p
.
for any 1 ≤ p ≤ ∞, and kvk
∞
is the limit as p → ∞.
Proof that these are indeed norms is left as an exercise for the reader (in the
example sheets).
A norm naturally induces a metric on V :
Lemma. If k · k is a norm on V , then
d(v, w) = kv − wk
defines a metric on V .
Proof.
(i) d(v, w) = kv −wk ≥ 0 by the definition of the norm.
(ii) d(v, w) = 0 ⇔ kv − wk = 0 ⇔ v − w = 0 ⇔ v = w.
(iii) d(w, v) = kw −vk = k(−1)(v −w)k = | −1|kv − wk = d(v, w).
(iv) d(u, v) + d(v, w) = ku − vk + kv − wk ≥ ku − wk = d(u, w).
Example. We have the following norms on C[0, 1]:
kfk
1
=
Z
1
0
|f(x)| dx
kfk
2
=
s
Z
1
0
f(x)
2
dx
kfk
∞
= max
x∈[0,1]
|f(x)|
The first two are known as the
L
1
and
L
2
norms. The last is called the uniform
norm, since it induces the uniform metric.
It is easy to show that these are indeed norms. The only slightly tricky part
is to show that kfk = 0 iff f = 0, which we obtain via the following lemma.
Lemma. Let
f ∈ C
[0
,
1] satisfy
f
(
x
)
≥
0 for all
x ∈
[0
,
1]. If
f
(
x
) is not
constantly 0, then
R
1
0
f(x) dx > 0.
Proof.
Pick
x
0
∈
[0
,
1] with
f
(
x
0
) =
a >
0. Then since
f
is continuous, there is
a
δ
such that
|f
(
x
)
−f
(
x
0
)
| < a/
2 if
|x −x
0
| < δ
. So
|f
(
x
)
| > a/
2 in this region.
Take
g(x) =
(
a/2 |x − x
0
| < δ
0 otherwise
Then f(x) ≥ g(x) for all x ∈ [0, 1]. So
Z
1
0
f(x) dx ≥
Z
1
0
g(x) dx =
a
2
· (2δ) > 0.
Example. Let X = C[0, 1], and let
d
1
(f, g) = kf − gk
1
=
Z
1
0
|f(x) − g(x)| dx.
Define the sequence
f
n
=
(
1 − nx x ∈ [0,
1
n
]
0 x ≥
1
n
.
f
x
1
n
1
Then
kfk
1
=
1
2
·
1
n
· 1 =
1
2n
→ 0
as n → ∞. So f
n
→ 0 in (X, d
1
) where 0(x) = 0.
On the other hand,
kf
n
k
∞
= max
x∈[0,1]
kf(x)k = 1.
So f
n
6→ 0 in the uniform metric.
So the function (
C
[0
,
1]
, d
1
)
→
(
C
[0
,
1]
, d
∞
) that maps
f 7→ f
is not continu-
ous. This is similar to the case that the identity function from the usual metric
of
R
to the discrete metric of
R
is not continuous. However, the discrete metric
is a silly metric, but d
1
is a genuine useful metric here.
Using the same example, we can show that the function
G
: (
C
[0
,
1]
, d
1
)
→
(R, usual) with G(f) = f (0) is not continuous.
We’ll now define the inner product of a real vector space. This is a general-
ization of the notion of the “dot product”.
Definition (Inner product). Let
V
be a real vector space. An inner product on
V is a function h·, ·i : V × V → R such that
(i) hv, vi ≥ 0 for all v ∈ V
(ii) hv, vi = 0 if and only if v = 0.
(iii) hv, wi = hw, vi.
(iv) hv
1
+ λv
2
, w) = hv
1
, wi + λhv
2
, wi.
Example.
(i) Let V = R
n
. Then
hv, wi =
n
X
i=1
v
i
w
i
is an inner product.
(ii) Let V = C[0, 1]. Then
hf, gi =
Z
1
0
f(x)g(x) dx
is an inner product.
We just showed that norms induce metrics. The proof was completely trivial
as the definitions were almost the same. Now we want to prove that inner
products induce norms. However, this is slightly less trivial. To do so, we need
the Cauchy-Schwarz inequality.
Theorem (Cauchy-Schwarz inequality). If h·, ·i is an inner product, then
hv, wi
2
≤ hv, vihw, wi.
Proof. For any x, we have
hv + xw, v + xwi = hv, vi + 2xhv, wi + x
2
hw, wi ≥ 0.
Seen as a quadratic in
x
, since it is always non-negative, it can have at most one
real root. So
(2hv, wi)
2
− 4hv, vihw, wi ≤ 0.
So the result follows.
With this, we can show that inner products induce norms (and hence metrics).
Lemma. If h·, ·i is an inner product on V , then
kvk =
p
hv, vi
is a norm.
Proof.
(i) kvk =
p
hv, vi ≥ 0.
(ii) kvk = 0 ⇔ hv, vi = 0 ⇔ v = 0.
(iii) kλvk =
p
hλv, λvi =
p
λ
2
hv, vi = |λ|kvk.
(iv)
(kvk + kwk)
2
= kvk
2
+ 2kvkkwk + kwk
2
≥ hv, vi + 2hv, wi + hw, wi
= kv + wk
2