4ℝn as a normed space
IB Analysis II
4.4 Mappings between normed spaces
We are now going to look at functions between normed spaces, and see if they
are continuous.
Let (
V, ∥ · ∥
), (
V
′
, ∥ · ∥
′
) be normed spaces, and let
E ⊆ K
be a subset,
and
f
:
E → V
′
a mapping (which is just a function, although we reserve the
terminology “function” or “functional” for when V
′
= R).
Definition (Continuity of mapping). Let y
∈ E
. We say
f
:
E → V
′
is
continuous at y if for all ε > 0, there is δ > 0 such that the following holds:
(∀x ∈ E) ∥x − y∥
V
< δ ⇒ ∥f(x) − f(y)∥
V
′
< ε.
Note that x
∈ E
and
∥
x
−
y
∥ < δ
is equivalent to saying x
∈ B
δ
(y)
∩ E
.
Similarly,
∥f
(x)
− f
(y)
∥ < ε
is equivalent to
f
(x)
∈ B
ε
(
f
(y)). In other words,
x
∈ f
−1
(
B
ε
(
f
(y))). So we can rewrite this statement as there is some
δ >
0
such that
E ∩ B
δ
(y) ⊆ f
−1
(B
ε
(f(y))).
We can use this to provide an alternative characterization of continuity.
Theorem. Let (
V, ∥ · ∥
), (
V
′
, ∥ · ∥
′
) be normed spaces,
E ⊆ V
,
f
:
E →
V
′
.
Then
f
is continuous at y
∈ E
if and only if for any sequence y
k
→
y in
E
, we
have f(y
k
) → f(y ).
Proof.
(
⇒
) Suppose
f
is continuous at y
∈ E
, and that y
k
→
y. Given
ε >
0,
by continuity, there is some δ > 0 such that
B
δ
(y) ∩ E ⊆ f
−1
(B
ε
(f(y))).
For sufficiently large k, y
k
∈ B
δ
(y) ∩ E. So f(y
k
) ∈ B
ε
(f(y)), or equivalently,
|f(y
k
) − f(y)| < ε.
So done.
(
⇐
) If
f
is not continuous at
y
, then there is some
ε >
0 such that for any
k
,
we have
B
1
k
(y) ⊆ f
−1
(B
ε
(f(y))).
Choose y
k
∈ B
1
k
(y)
\ f
−1
(
B
ε
(
f
(y))). Then y
k
→
y, y
k
∈ E
, but
∥f
(y
k
)
−
f(y)∥ ≥ ε, contrary to the hypothesis.
Definition (Continuous function).
f
:
E → V
′
is continuous if
f
is continuous
at every point y ∈ E.
Theorem. Let (
V, ∥ · ∥
) and (
V
′
, ∥ · ∥
′
) be normed spaces, and
K
a compact
subset of V , and f : V → V
′
a continuous function. Then
(i) f(K) is compact in V
′
(ii) f(K) is closed and bounded
(iii)
If
V
′
=
R
, then the function attains its supremum and infimum, i.e. there
is some y
1
, y
2
∈ K such that
f(y
1
) = sup{f(y) : y ∈ K}, f(y
2
) = inf{f(y) : y ∈ K}.
Proof.
(i)
Let (x
k
) be a sequence in
f
(
K
) with x
k
=
f
(y
k
) for some y
k
∈ K
. By
compactness of
K
, there is a subsequence (y
k
j
) such that y
k
j
→
y. By the
previous theorem, we know that
f
(y
j
k
)
→ f
(y). So x
k
j
→ f
(y)
∈ f
(
K
).
So f(K) is compact.
(ii)
This follows directly from (
i
), since every compact space is closed and
bounded.
(iii)
If
F
is any bounded subset of
R
, then either
sup F ∈ F
or
sup F
is a limit
point of
F
(or both), by definition of the supremum. If
F
is closed and
bounded, then any limit point must be in
F
. So
sup F ∈ F
. Applying this
fact to F = f(K) gives the desired result, and similarly for infimum.
Finally, we will end the chapter by proving that any two norms on a finite
dimensional space are Lipschitz equivalent. The key lemma is the following:
Lemma. Let
V
be an
n
-dimensional vector space with a basis
{
v
1
, ··· ,
v
n
}
.
Then for any x
∈ V
, write x =
P
n
j=1
x
j
v
j
, with
x
j
∈ R
. We define the Euclidean
norm by
∥x∥
2
=
X
x
2
j
1
2
.
Then this is a norm, and S = {x ∈ V : ∥x∥
2
= 1} is compact in (V, ∥ · ∥
2
).
After we show this, we can easily show that every other norm is equivalent
to this norm.
This is not hard to prove, since we know that the unit sphere in
R
n
is
compact, and we can just pass our things on to R
n
.
Proof. ∥ · ∥
2
is well-defined since
x
1
, ··· , x
n
are uniquely determined by x (by
(a certain) definition of basis). It is easy to check that ∥ · ∥
2
is a norm.
Given a sequence x
(k)
in
S
, if we write x
(k)
=
P
n
j=1
x
(k)
j
v
j
. We define the
following sequence in R
n
:
˜
x
(k)
= (x
(k)
1
, ··· , x
(k)
n
) ∈
˜
S = {
˜
x ∈ R
n
: ∥
˜
x∥
Euclid
= 1}.
As
˜
S
is closed and bounded in
R
n
under the Euclidean norm, it is compact.
Hence there exists a subsequence
˜x
(k
j
)
and
˜x ∈
˜
S
such that
∥
˜
x
(k
j
)
−
˜
x∥
Euclid
→
0.
This says that x =
P
n
j=1
x
j
v
j
∈ S, and ∥x
k
j
− x∥
2
→ 0. So done.
Theorem. Any two norms on a finite dimensional vector space are Lipschitz
equivalent.
The idea is to pick a basis, and prove that any norm is equivalent to ∥ · ∥
2
.
To show that an arbitrary norm
∥ · ∥
is equivalent to
∥ · ∥
2
, we have to show
that for any ∥x∥, we have
a∥x∥
2
≤ ∥x∥ ≤ b∥x∥
2
.
We can divide by ∥x∥
2
and obtain an equivalent requirement:
a ≤
x
∥x∥
2
≤ b.
We know that any x
/∥
x
∥
2
lies in the unit sphere
S
=
{
x
∈ V
:
∥
x
∥
2
= 1
}
. So
we want to show that the image of
∥ · ∥
is bounded. But we know that
S
is
compact. So it suffices to show that ∥ · ∥ is continuous.
Proof.
Fix a basis
{
v
1
, ··· ,
v
n
}
for
V
, and define
∥ · ∥
2
as in the lemma above.
Then
∥ · ∥
2
is a norm on
V
, and
S
=
{
x
∈ V
:
∥
x
∥
2
= 1
}
, the unit sphere, is
compact by above.
To show that any two norms are equivalent, it suffices to show that if
∥ · ∥
is
any other norm, then it is equivalent to ∥ · ∥
2
, since equivalence is transitive.
For any
x =
n
X
j=1
x
j
v
j
,
we have
∥x∥ =
n
X
j=1
x
j
v
j
≤
X
|x
j
|∥v
j
∥
≤ ∥x∥
2
n
X
j=1
∥v
j
∥
2
1
2
by the Cauchy-Schwarz inequality. So ∥x∥ ≤ b∥x∥
2
for b =
P
∥v
j
∥
2
1
2
.
To find
a
such that
∥
x
∥ ≥ a∥
x
∥
2
, consider
∥ · ∥
: (
S, ∥ · ∥
2
)
→ R
. By above,
we know that
∥x − y∥ ≤ b∥x − y∥
2
By the triangle inequality, we know that
∥
x
∥ − ∥
y
∥
≤ ∥
x
−
y
∥
. So when x is
close to y under
∥ · ∥
2
, then
∥
x
∥
and
∥
y
∥
are close. So
∥ · ∥
: (
S, ∥ · ∥
2
)
→ R
is continuous. So there is some x
0
∈ S
such that
∥
x
0
∥
=
inf
x∈S
∥
x
∥
=
a
, say.
Since ∥x∥ > 0, we know that ∥x
0
∥ > 0. So ∥x∥ ≥ a∥x∥
2
for all x ∈ V .
The key to the proof is the compactness of the unit sphere of (
V, ∥ · ∥
).
On the other hand, compactness of the unit sphere also characterizes finite
dimensionality. As you will show in the example sheets, if the unit sphere of a
space is compact, then the space must be finite-dimensional.
Corollary. Let (V, ∥ · ∥) be a finite-dimensional normed space.
(i) The Bolzano-Weierstrass theorem holds for V , i.e. any bounded sequence
sequence in V has a convergent subsequence.
(ii) A subset of V is compact if and only if it is closed and bounded.
Proof.
If a subset is bounded in one norm, then it is bounded in any Lipschitz
equivalent norm. Similarly, if it converges to x in one norm, then it converges to
x in any Lipschitz equivalent norm.
Since these results hold for the Euclidean norm
∥ · ∥
2
, it follows that they
hold for arbitrary finite-dimensional vector spaces.
Corollary. Any finite-dimensional normed vector space (V, ∥ · ∥) is complete.
Proof.
This is true since if a space is complete in one norm, then it is complete
in any Lipschitz equivalent norm, and we know that
R
n
under the Euclidean
norm is complete.