Due date: Wed., Oct 5, 2011
5.
Let
f
j
:
D
(0
,
1)
→
C
be holomorphic and suppose that each
f
j
has
at least
k
roots in
D
(0
,
1), counting multiplicities. Suppose that
f
j
→
f
uniformly on compact subsets. Show by example that it does not follow
that
f
has at least
k
roots in
D
(0
,
1), counting multiplicities. In particular,
construct examples, for each ﬁxed
k
and each
l
, 0
≤
l
≤
k
, where
f
has
exactly
l
roots. What simple hypothesis can you add that will guarantee
that
f
does have at least
k
roots in
D
(0
,
1), counting multiplicities?
Solution
. Let
f
j
=
z
j
+
k

1
2
and
f
= 1
/
2. Then
f
j
has at least
k
roots in
D
(0
,
1). But
f
=
1
2
has no zeros on
D
(0
,
1).
Similarly, let
f
j
(
z
) =
z
j
+
k
+
l

1
2
z
l
. Then
f
j
has at least
k
roots in
D
(0
,
1).
But
f
=
1
2
z
l
has
l
zeros, counting multiplicities on
D
(0
,
1).
The simple hypothesis we can add is as follows: we assume that
f
j
be e
holomorphic on
D
(0
,
1)
⊂
U
and suppose that each
f
j
has at least
k
roots in
D
(0
,
1), counting multiplicities. Suppose that
f
j
→
f
uniformly on compact
subsets of
U
. Assume that
f
has no zeros on
∂D
(0
,
1). Then
f
has at
least
k
roots in
D
(0
,
1), counting multiplicities. This can be proved by using
Rouche’s theorem (similar to the proof of Problem 8 and extra problem 1
below): Since
f
has no no zeros on
∂D
(0
,
1), by the compactness of
∂D
(0
,
1),
δ
:=
min
z
∈
∂D
(0
,
1)

f
(
z
)

>
0
.
Since
f
j
→
f
uniformly on
∂D
(0
,
1), there exists
N >
0 such that for
j > N
,
for all
z
∈
∂D
(0
,
1),

f
j
(
z
)

f
(
z
)

< δ
≤ 
f
(
z
)

.
Rouch’s theorem implies that
f
and
f
n
have same number of zeros, counting
multiplicities on
D
(0
,
1).
Remark
: In above, we compare the zeros of
f
and
f
n
=
f

f
n
, so in order
to apply Rouche’s theorem, we need a lower bound of
f
and an upper bound
fo
f

f
n
.
1