Math 318 HW #1 Solutions
1. Exercise 6.2.3. Consider the sequence of functions
h
n
(
x
) =
x
1 +
x
n
over the domain [0
,
∞
).
(a) Find the pointwise limit of (
h
n
) on [0
,
∞
).
Answer.
Let
h
be the function
h
(
x
) =
x
if 0
≤
x <
1
1
/
2
if
x
= 1
0
if
x >
1
.
Then the claim is that
h
n
→
h
pointwise. To see this, let
± >
0. There are three cases
to consider:
0
≤
x <
1
:
If
x
= 0, then
h
n
(
x
) = 0 for all
n
. Otherwise, let
N >
ln
±
ln
x
, which is equivalent
to assuming that
x
N
+1
< ±
. Then, for any
n
≥
N
, we have that

h
n
(
x
)

h
(
x
)

=
±
±
±
±
x
1 +
x
n

x
±
±
±
±
=
±
±
±
±

x
n
+1
1 +
x
n
±
±
±
±
=
x
n
+1
1 +
x
n
< x
n
+1
≤
x
N
+1
< ±.
x
= 1
:
Clearly,
h
n
(1) = 1
/
2 for all
n
, so

h
n
(1)

h
(1)

= 0
< ±
for any
n
.
x >
1
:
Let
N >
ln(1
/±
)
ln
x
+1, which is equivalent to having
1
x
N

1
< ±
. Then for any
n
≥
N
,

h
n
(
x
)

h
(
x
)

=
±
±
±
±
x
1 +
x
n

0
±
±
±
±
=
x
1 +
x
n
<
x
x
n
=
1
x
n

1
≤
1
x
N

1
< ±.
In all three cases, there exists
N
∈
N
such that
n
≥
N
implies that

h
n
(
x
)

h
(
x
)

< ±.
Since the choice of
± >
0 was arbitrary, we conclude that
h
n
→
h
pointwise.
(b) Explain how we know that the convergence
cannot
be uniform on [0
,
∞
).
Answer.
Each
h
n
is continuous, whereas the function
h
is clearly not continuous at
x
= 1.
If the convergence
were
uniform, then the function
h
would have to be continuous by
Theorem 6.2.6. Since that isn’t the case, we know the convergence
cannot
be uniform.
(c) Choose a smaller set over which the convergence is uniform and supply an argument to
show that this is indeed the case.
Answer.
Pick
r >
1. Then I claim that
h
n
→
h
uniformly on [
r,
∞
). To see this, ﬁrst
note that
h
(
x
) = 0 on [
r,
∞
). Now, let
± >
0 and choose
N >
ln(1
/±
)
ln
r
+ 1. Then for all
n
≥
N
we know that

h
n
(
x
)

0

=
x
1 +
x
n
<
1
x
n

1
≤
1
r
n

1
≤
1
r
N

1
< ±.
Since the choice of
± >
0 was arbitrary, we conclude that
h
n
→
0 uniformly on [
r,
∞
).
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