Solution.
The new series can be written in the form
z
∞
X
n
=0
a
n
(
z
2
)
n
. It clearly converges at
z
if and only if the original
series converges at
z
2
. In particular, it converges for

z
2

<
2; i.e., for

z

<
√
2, diverges for

z
2

>
2; i.e., for

z

>
√
2.
The radius is
√
2.
3. Let
{
a
n
}
be a sequence of positive numbers such that
∑
∞
n
=1
a
n
<
∞
. Prove: There exists a sequence
{
c
n
}
of positive
numbers diverging to
∞
such that
∑
∞
n
=1
a
n
c
n
<
∞
.
Solution.
This is too nice an exercise to be spoiled already
by a solution. Only two people selected it in the final and I am afraid that their “solutions” are totally wrong. Very
wrong. Fundamentally wrong. Everybody still has a chance to work on it.
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4. Let
F
be an equicontinuous set of real valued functions on the interval [
a, b
]. That is, each element of
F
is a function
from [
a, b
] to
R
and for every
² >
0 there is
δ
such that if
x, y
∈
[
a, b
] and

x

y

< δ
, then

f
(
x
)

f
(
y
)

< ²
. Assume
∞
< a < b <
∞
. Prove: If the sequence of real numbers
{
f
(
a
) :
f
∈ F}
is bounded, then the set
F
is uniformly
bounded.
Solution.
I did this in class! (Maybe slurring over some details.) The hypothesis is that
F
is equicontinuous and
{
f
(
a
) :
f
∈ F}
is bounded; let
M >
0 be such that

f
(
a
)
 ≤
M
for all
f
∈ F
. Let
δ
be the
δ
that works for
²
= 1 in
the definition of equicontinuity; that is, let
δ >
0 be such that if
x, y
∈
[
a, b
], then

f
(
x
)

f
(
y
)

<
1. Let
n
∈
N
be such
that (
b

a
)
/n < δ
and let
x
0
=
a, x
1
=
a
+
b

a
n
, . . . , x
k
=
a
+
k
b

a
n
, . . . , x
n
=
x
k
=
a
+
n
b

a
n
=
b.
Notice
x
k

x
k

1
= (
b

a
)
/n < δ
for
k
= 1
,
2
, . . . , n
.
Let
x
∈
[
a, b
],
f
∈ F
. There is
k
∈ {
1
, . . . , n
}
such that
x
∈
[
x
k

1
, x
k
], then

x

x
k

1

< δ
and we have:

f
(
x
)
 ≤ 
f
(
x
)

f
(
a
)

+

f
(
a
)

=

f
(
a
)

+
fl
fl
fl
fl
fl
fl
f
(
x
)

f
(
x
k

1
) +
k

1
X
j
=1
[
f
(
x
j
)

f
(
x
j

1
)]
fl
fl
fl
fl
fl
fl
.
If
k
= 1, we interpret
∑
0
j
=1
as 0; the sum does not appear.
The sum, of course, telescopes to
f
(
x
k

1
)

f
(
x
0
) =
f
(
x
k

1
)

f
(
a
). Thus

f
(
x
)
 ≤ 
f
(
a
)

+

f
(
x
)

f
(
x
k

1
)

+
k

1
X
j
=1

f
(
x
j
)

f
(
x
j

1
)
 ≤
M
+ 1 +
k

1
X
j
=1
1 =
M
+
k
≤
M
+
n.
Since
x
∈
[
a, b
] is arbitrary, we proved

f
(
x
)
 ≤
M
+
n
for all
x
∈
[
a, b
],
f
∈ F
. The sequence is uniformly bounded my
M
+
n
.
5. If
f, g
:
R
→
R
define
f
∼
g
if and only if
f
=
g
a.e. Prove:
∼
is an equivalence relation.
Solution.
We have to see
f
∼
f
,
f
∼
g
⇒
g
∼
f
and
f
∼
g, g
∼
h
⇒
f
∼
h
, for all functions
f, g, h
:
R
→
R
. Since
the empty set is a null set, it is clear that
f
∼
f
for all
f
:
R
→
R
. Assume
f, g
:
R
→
R
and
f
∼
g
. Then the set
E
=
{
x
:
f
(
x
)
6
=
g
(
x
)
}
is a null set; this is of course the same set as
{
x
:
g
(
x
)
6
=
f
(
x
)
}
, thus
g
=
f
a.e. and
g
∼
f
.
Alternatively, the definition of
f
=
g
a.e. is symmetric in
f, g
, thus
f
=
g
a.e. if and only if
g
=
f
a.e.. In any case
f
∼
g
⇒
g
∼
f
. Finally assume
f
∼
g, g
∼
h
. The sets
E
=
{
x
:
f
(
x
)
6
=
g
(
x
)
}
,
F
=
{
x
:
g
(
x
)
6
=
h
(
x
)
}
are then null
sets. Clearly
f
(
x
)
6
=
h
(
x
) implies that either
x
∈
E
or
x
∈
F
, thus
{
x
:
f
(
x
)
6
=
h
(
x
)
⊂
E
∪
F
, thus a null set. Thus
f
=
h
a,e. and
f
∼
h
.
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 Spring '11
 Speinklo
 Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Nonmeasurable set

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