1]?
(1)

f
n
(
y
)

=

1

y
n

=

1

y
 · 
1 +
y

<
1
2
, then it would
have to be true that

1

y

<
1
2

y
+ 1

or 1
<
1
2

1 +
y

+
y
This is not possible, so there is no way to choose a
δ
so that

1

y

< δ
, so
g
n
is not equicontinuos.
6.2.16
Given
f
n
:
[0
,
1]
→
R
, each
f
n
is bounded (by some
M
), and (
f
n
) is equicontinuous, show that
f
m
)
contains a uniformly convergence subsequence.
(a)
Since
Q
=
{
r
1
, r
2
, . . .
}
, and
f
n
is bounded, we can use problem 6.2.14 to find a subsequence
f
n
k
=
:
g
n
that
converges to each rational number in [0
,
1]. (See part (c) of that problem).
(b)
We are given a set Λ =
{
r
1
, r
2
, . . . , r
m
}
such that [0
,
1] =
[
r
∈
Λ
V
δ
(
r
), where

x

y

< δ
and

g
k
(
x
)

g
k
(
y
)

<
3
for some positive
.
4
We want to show that there is an
N
, so that

g
s
(
r
i
)

g
t
(
r
i
)

<
3
for all
s, t
≥
N
and
4
This is by equicontinuity
28
r
i
∈
Λ. Each
g
k
is Cauchy since it is uniformly continuous (by Theorem 6.2.5), so we can find an
N
for any
particular
r
i
in Λ. Because it is given that Λ is finite, we may choose the maximum of these
N
. If Λ was
infinite, we would not be be able to choose such an
N
.
(c) We can use the conclusion from part (b) with the fact that
g
n
is equicontinuous to conclude:

g
s
(
x
)

g
t
(
x
)

=

g
s
(
x
)

g
s
(
r
) +
g
s
(
r
)

g
t
(
r
) +
g
t
(
r
)

g
t
(
x
)

≤

g
s
(
x
)

g
s
(
r
)

+

g
s
(
r
)

g
t
(
r
)

+

g
t
(
r
)

g
t
(
x
)

<
3
3
=
Rate of convergence of lefthand Riemann Sums
Suppose that
f
:
[0
,
1]
→
R
has a continuous second
derivative. Prove that
Z
1
0
f
(
x
) d
x

1
n
n

1
X
k
=0
f
(
k/n
) =
f
(1)

f
(0)
2
n
+
O
1
n
2
Proof.
We can rewrite
Z
1
0
f
(
x
) d
x
as a sum of integrals,
Z
1
n
0
f
(
x
) d
x
+
Z
2
n
1
n
f
(
x
) d
x
+
· · ·
+
Z
1
n

1
n
f
(
x
) d
x
=
n

1
X
k
=0
Z
k
+1
n
k
n
f
(
x
) d
x
This is useful, becuase it allows to write the difference between
Z
1
0
f
(
x
)
d
x
and the lefthand Riemann sum as a
sum of integrals. To do this, consider some arbitrary term
1
n
f
(
k
/
n
)
from the lefthand sums, and note that
Z
k
+1
k
k
n
f
k
n
d
x
=
f
k
n
Z
k
+1
k
k
n
d
x
=
1
n
f
k
n
Using this fact, we can write
n

1
X
k
=0
f
(
k/n
) as
n

1
X
k
=0
Z
k
+1
n
k
n
f
(
k/n
)
d
x
. Then, multiplying the difference between the
integral and left hand sums by
n
, we see
n
Z
1
0
f
(
x
) d
x

1
n
n

1
X
k
=0
f
(
k/n
)
=
n
n

1
X
k
=0
Z
k
+1
n
k
n
f
(
x
) d
x

n

1
X
k
=0
Z
k
+1
n
k
n
f
(
k/n
) d
x
=
n
n

1
X
k
=0
Z
k
+1
n
k
n
f
(
x
)

f
(
k/n
) d
x
The first order Taylor approximation of some integrand about its lower limit is
f
(
x
)

f
k
n
=
"
f
k
n

f
k
n
#
+
29
f
0
k
n
x

k
n
+
R
(
x
) =
f
0
k
n
x

k
n
+
1
2
f
00
(
c
k
)(
x

k
n
)
2
Using this:
n
n

1
X
k
=0
Z
k
+1
n
k
n
f
(
x
)

f
(
k/n
) d
x
=
n
n

1
X
k
=0
Z
k
+1
n
k
n
f
0
k
n
x

k
n
d
x
+
n
n

1
X
k
=0
Z
k
+1
n
k
n
1
2
f
00
(
c
k
)(
x

k
n
)
2
d
x
=
n
n

1
X
k
=0
f
0
k
n
Z
1
n
0
u
d
u
+
n
n

1
X
k
=0
Z
k
+1
n
k
n
1
2
f
00
(
c
k
)(
x

k
n
)
2
d
x
=
n
n

1
X
k
=0
f
0
k
n
1
2
n
2
+
n
n

1
X
k
=0
Z
k
+1
n
k
n
1
2
f
00
(
c
k
)(
x

k
n
)
2
d
x
=
1
2
n

1
X
k
=0
f
0
k
n
1
n
+
n
n

1
X
k
=0
Z
k
+1
n
k
n
1
2
f
00
(
c
k
)(
x

k
n
)
2
d
x
=
1
2
Z
1
0
f
0
(
x
) d
x
+
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 Fall '15
 Topology, yn, Metric space, Xn, lim F