Solution the new series can be written in the form z

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.

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 .