3Uniform continuity and integration

IB Analysis II



3.1 Uniform continuity
Recall that we had a rather weak notion of convergence, known as pointwise
convergence, and then promoted it to uniform convergence. The process of this
promotion is to replace the condition “for each
x
, we can find an
ε
to “we can
find an
ε
that works for each
x
”. We are going to do the same for continuity to
obtain uniform continuity.
Definition (Uniform continuity). Let
E R
and
f
:
E R
. We say that
f
is
uniformly continuous on E if
(ε)(δ > 0)(x)(y) |x y| < δ |f(x) f(y)| < ε.
Compare this to the definition of continuity:
(ε)(x)(δ > 0)(y) |x y| < δ |f(x) f(y)| < ε.
Again, we have shifted the (
x
) out of the (
δ
) quantifier. The difference is
that in regular continuity,
δ
can depend on our choice of
x
, but in uniform
continuity, it only depends on
y
. Again, clearly a uniformly continuous function
is continuous.
In general, the converse is not true, as we will soon see in two examples.
However, the converse is true in a lot of cases.
Theorem. Any continuous function on a closed, bounded interval is uniformly
continuous.
Proof.
We are going to prove by contradiction. Suppose
f
: [
a, b
]
R
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
|x
n
y
n
| <
1
n
but
|f(x
n
) f(y
n
)| > ε.
Since we are on a closed, bounded interval, by Bolzano-Weierstrass, (
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
|f
(
x
n
i
)
f
(
y
n
i
)
| > ε
for
all n
i
. This is a contradiction.
Note that we proved this in the special case where the domain is [
a, b
] and
the image is
R
. In fact, [
a, b
] can be replaced by any compact metric space;
R
by any metric space. This is since all we need is for Bolzano-Weierstrass to hold
in the domain, i.e. the domain is sequentially compact (ignore this comment if
you have not taken IB Metric and Topological Spaces).
Instead of a contradiction, we can also do a direct proof of this statement,
using the Heine-Borel theorem which says that [0, 1] is compact.
While this is a nice theorem, in general a continuous function need not be
uniformly continuous.
Example. Consider
f
: (0
,
1]
R
given by
f
(
x
) =
1
x
. This is not uniformly
continuous, since when we get very close to 0, a small change in
x
produces a
large change in
1
x
.
In particular, for any
δ <
1 and
x < δ
,
y
=
x
2
, then
|x y|
=
x
2
< δ
but
|f(x) f(y)| =
1
x
> 1.
In this example, the function is unbounded. However, even bounded functions
can be not uniformly continuous.
Example. Let f : (0, 1] R, f(x) = sin
1
x
. We let
x
n
=
1
2
, y
n
=
1
(2n +
1
2
)π
.
Then we have
|f(x
n
) f(y
n
)| = |0 1| = 1,
while
|x
n
y
n
| =
π
2n(4n + 1)
0.