5Metric spaces
IB Analysis II
5.5 Continuous functions
We are going to look at continuous mappings between metric spaces.
Definition (Continuity). Let (
X, d
) and (
X
′
, d
′
) be metric spaces. A function
f : X → X
′
is continuous at y ∈ X if
(∀ε > 0)(∃δ > 0)(∀x) d(x, y) < δ ⇒ d
′
(f(x), f(y)) < ε.
This is true if and only if for every ε > 0, there is some δ > 0 such that
B
δ
(y) ⊆ f
−1
B
ε
(f(x)).
f is continuous if f is continuous at each y ∈ X.
Definition (Uniform continuity). f is uniformly continuous on X if
(∀ε > 0)(∃δ > 0)(∀x, y ∈ X) d(x, y) < δ ⇒ d(f(x), f(y)) < ε.
This is true if and only if for all ε, there is some δ such that for all y, we have
B
δ
(y) ⊆ f
−1
(B
ε
(f(y))).
Definition (Lipschitz function and Lipschitz constant).
f
is said to be Lipschitz
on X if there is some K ∈ [0, ∞) such that for all x, y ∈ X,
d
′
(f(x), f(y)) ≤ Kd(x, y)
Any such K is called a Lipschitz constant.
It is easy to show
Lipschitz ⇒ uniform continuity ⇒ continuity.
We have seen many examples that continuity does not imply uniform continuity.
To show that uniform continuity does not imply Lipschitz, take
X
=
X
′
=
R
.
We define the metrics as
d(x, y) = min{1, |x − y|}, d
′
(x, y) = |x − y|.
Now consider the function
f
: (
X, d
)
→
(
X
′
, d
′
) defined by
f
(
x
) =
x
. We can
then check that this is uniformly continuous but not Lipschitz.
Note that the statement that metrics
d
and
d
′
are Lipschitz equivalent is
equivalent to saying the two identity maps
i
: (
X, d
)
→
(
X, d
′
) and
i
′
: (
X, d
′
)
→
(X, d) are Lipschitz, hence the name.
Note also that the metric itself is also a Lipschitz map for any metric. Here
we are viewing the metric as a function
d
:
X × X → R
, with the metric on
X × X defined as
˜
d((x
1
, y
1
), (x
2
, y
2
)) = d(x
1
, x
2
) + d(y
1
, y
2
).
This is a consequence of the triangle inequality, since
d(x
1
, y
1
) ≤ d(x
1
, x
2
) + d(x
2
, y
2
) + d(y
1
, y
2
).
Moving the middle term to the left gives
d(x
1
, y
1
) − d(x
2
, y
2
) ≤
˜
d((x
1
, y
1
), (x
2
, y
2
))
Swapping the theorems around, we can put in the absolute value to obtain
|d(x
1
, y
1
) − d(x
2
, y
2
)| ≤
˜
d((x
1
, y
1
), (x
2
, y
2
))
Recall that at the very beginning, we proved that a continuous map from a
closed, bounded interval is automatically uniformly continuous. This is true
whenever the domain is compact.
Theorem. Let (
X, d
) be a compact metric space, and (
X
′
, d
′
) is any metric
space. If f : X → X
′
be continuous, then f is uniformly continuous.
This is exactly the same proof as what we had for the [0, 1] case.
Proof.
We are going to prove by contradiction. Suppose
f
:
X → X
′
is not
uniformly continuous. Since
f
is not uniformly continuous, there is some
ε >
0
such that for all
δ
=
1
n
, there is some
x
n
, y
n
such that
d
(
x
n
, y
n
)
<
1
n
but
d
′
(f(x
n
), f(y
n
)) > ε.
By compactness of
X
, (
x
n
) has a convergent subsequence (
x
n
i
)
→ x
. Then
we also have
y
n
i
→ x
. So by continuity, we must have
f
(
x
n
i
)
→ f
(
x
) and
f
(
y
n
i
)
→ f
(
x
). But
d
′
(
f
(
x
n
i
)
, f
(
y
n
i
))
> ε
for all
n
i
. This is a contradiction.
In the proof, we have secretly used (part of) the following characterization of
continuity:
Theorem. Let (
X, d
) and (
X
′
, d
′
) be metric spaces, and
f
:
X → X
′
. Then the
following are equivalent:
(i) f is continuous at y.
(ii) f(x
k
) → f (y) for every sequence (x
k
) in X with x
k
→ y.
(iii)
For every neighbourhood
V
of
f
(
y
), there is a neighbourhood
U
of
y
such
that U ⊆ f
−1
(V ).
Note that the definition of continuity says something like (iii), but with open
balls instead of open sets. So this should not be surprising.
Proof.
– (i) ⇔ (ii): The argument for this is the same as for normed spaces.
–
(i)
⇒
(iii): Let
V
be a neighbourhood of
f
(
y
). Then by definition there is
ε >
0 such that
B
ε
(
f
(
y
))
⊆ V
. By continuity of
f
, there is some
δ
such
that
B
δ
(y) ⊆ f
−1
(B
ε
(f(y))) ⊆ f
−1
(V ).
Set U = B
ε
(y) and done.
–
(iii)
⇒
(i): for any
ε
, use the hypothesis with
V
=
B
ε
(
f
(
y
)) to get a
neighbourhood U of y such that
U ⊆ f
−1
(V ) = f
−1
(B
ε
(f(y))).
Since U is open, there is some δ such that B
δ
(y) ⊆ U. So we get
B
δ
(y) ⊆ f
−1
(B
ε
(f(y))).
So we get continuity.
Corollary. A function
f
: (
X, d
)
→
(
X
′
, d
′
) is continuous if
f
−1
(
V
) is open in
X whenever V is open in X
′
.
Proof.
Follows directly from the equivalence of (i) and (iii) in the theorem
above.