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.