and you want to find an analytic function,
f such that these numbers are the zeros of f. How can you do it? The problem is easy if
there are only finitely many of these zeros,
{z1,z2,⋅⋅⋅,zm}
. You just write
(z − z1)
(z − z2)
⋅⋅⋅
(z − zm )
. Now if none of the z_{k} = 0 you could also write it at
∏_{k=1}^{m}
( z)
1 − zk
and this might have a better chance of success in the case of
infinitely many prescribed zeros. However, you would need to verify something like
∑_{n=1}^{∞}
| |
||zzn||
< ∞ which might not be so. The way around this is to adjust the product,
making it ∏_{k=1}^{∞}
( -z)
1− zk
e^{gk(z)
} where g_{
k}
(z)
is some analytic function. Recall also that
for
|x|
< 1, ln
( )
(1− x)−1
= ∑_{n=1}^{∞}
xnn-
. If you had x∕x_{n} small and real, then
1 =
(1− x∕xn)
exp
( ( −1))
ln (1− x∕xn)
and ∏_{k=1}^{∞}1 of course converges but loses all
the information about zeros. However, this is why it is not too unreasonable to consider
factors of the form
( ) ∑p ( )k
1− -z e kk=1 zzk 1k
zk
where p_{k} is suitably chosen.
First here are some estimates.
Lemma 56.1.1For z ∈ ℂ,
z |z|
|e − 1| ≤ |z|e , (56.1.3)
(56.1.3)
and if
|z|
≤ 1∕2,
||∑∞ k|| m
|| z-||≤ -1-|z|--≤ 2-|z|m ≤ 1---1--. (56.1.4)
|k=m k | m 1 − |z| m m 2m −1
With this estimate, it is easy to prove the Weierstrass product formula.
Theorem 56.1.4Let
{zn}
be a sequence of nonzero complex numbers which haveno limit point in ℂ and let
{pn}
be a sequence of nonnegative integers suchthat
∑∞ ( )pn+1
-R- < ∞ (56.1.5)
n=1 |zn|
(56.1.5)
for all R ∈ ℝ. Then
( )
∞∏ z-
P (z) ≡ Epn zn
n=1
is analytic on ℂ and has a zero at each point, z_{n}and at no others. If w occurs m timesin
{zn}
, then P has a zero of order m at w.
Proof: Since
{z }
n
has no limit point, it follows lim_{n→∞}
|z |
n
= ∞. Therefore, if
p_{n} = n− 1 the condition, 56.1.5 holds for this choice of p_{n}. Now by Theorem 56.0.2, the
infinite product in this theorem will converge uniformly on
|z|
≤ R if the same is true of
the sum,
∞ | ( ) |
∑ ||Ep z- − 1||. (56.1.6)
n=1 | n zn |
(56.1.6)
But by Corollary 56.1.3 the n^{th} term of this sum satisfies
|| ( z) || ||z ||pn+1
||Epn z- − 1|| ≤ 3 ||z || .
n n
By the Weierstrass M test, the series in 56.1.6 converges uniformly for
|z|
< R and so the
same is true of the infinite product. It follows from Lemma 50.3.13 on Page
5538 that P
(z)
is analytic on
|z|
< R because it is a uniform limit of analytic
functions.
Also by Theorem 56.0.2 the zeros of the analytic P
(z)
are exactly the points,
{zn}
,
listed according to multiplicity. That is, if z_{n} is a zero of order m, then if it is listed m
times in the formula for P
(z)
, then it is a zero of order m for P. This proves the
theorem.
The following corollary is an easy consequence and includes the case where there is a
zero at 0.
Corollary 56.1.5Let
{zn}
be a sequence of nonzero complex numbers whichhave no limit point and let
{pn}
be a sequence of nonnegative integers suchthat
( )
∑∞ -r- 1+pn
|zn| < ∞ (56.1.7)
n=1
(56.1.7)
for all r ∈ ℝ. Then
( )
m ∏∞ -z
P (z) ≡ z Epn zn
n=1
is analytic Ω and has a zero at each point, z_{n}and at no others along with a zero oforder m at 0. If w occurs m times in
{zn}
, then P has a zero of order m atw.
The above theory can be generalized to include the case of an arbitrary open set.
First, here is a lemma.
Lemma 56.1.6Let Ω be an open set. Also let
{zn}
be a sequence of points in Ω
which is bounded and which has no point repeated more than finitely many timessuch that
{zn}
has no limit point in Ω. Then there exist
{wn }
⊆ ∂Ω such that
lim_{n→∞}
|zn − wn |
= 0.
Proof:Since ∂Ω is closed, there exists w_{n}∈ ∂Ω such that dist
(zn,∂Ω)
=
|zn − wn|
.
Now if there is a subsequence,
{zn }
k
such that
|zn − wn |
k k
≥ ε for all k, then
{zn }
k
must possess a limit point because it is a bounded infinite set of points.
However, this limit point can only be in Ω because
{zn}
k
is bounded away from
∂Ω. This is a contradiction. Therefore, lim_{n→∞}
|zn − wn |
= 0. This proves the
lemma.
Corollary 56.1.7Let
{zn}
be a sequence of complex numbers contained in Ω, anopen subset of ℂ which has no limit point in Ω. Suppose each z_{n}is repeated nomore than finitely many times. Then there exists a function f which is analytic on
Ω whose zeros are exactly
{zn }
. If w ∈
{zn}
and w is listed m times, then w is azero of order m of f.
Proof: There is nothing to prove if
{zn}
is finite. You just let f
(z)
= ∏_{j=1}^{m}
(z − zj)
where
{zn}
=
{z1,⋅⋅⋅,zm}
.
Pick w ∈ Ω ∖
{zn}
_{n=1}^{∞} and let h
(z)
≡
z1−w-
. Since w is not a limit point of
{zn}
,
there exists r > 0 such that B
(w, r)
contains no points of
{zn}
. Let Ω_{1}≡ Ω ∖
{w}
. Now
h is not constant and so h
(Ω1)
is an open set by the open mapping theorem. In fact, h
maps each component of Ω to a region.
converges uniformly for z ∈ K. This implies ∏_{n=1}^{∞}E_{n}
( )
h(zn)−wn
h(z)−wn
also converges
uniformly for z ∈ K by Theorem 56.0.2. Since K is arbitrary, this shows f defined in
56.1.8 is analytic on Ω_{1}.
Also if z_{n} is listed m times so it is a zero of multiplicity m and w_{n} is the point from
∂
(h (Ω1))
closest to h
(zn)
, then there are m factors in 56.1.8 which are of the form
( ) ( )
En h(zn)−-wn- = 1 − h(zn)−-wn- egn(z)
h(z)− wn ( h (z)− wn)
= h(z)−-h(zn)- egn(z)
h(z)− wn
zn − z ( 1 ) g (z)
= (z −-w-)(zn-−-w) h(z)−-wn- en
= (z − zn)Gn (z) (56.1.10)
where G_{n} is an analytic function which is not zero at and near z_{n}. Therefore, f has a
zero of order m at z_{n}. This proves the theorem except for the point, w which has been
left out of Ω_{1}. It is necessary to show f is analytic at this point also and right now, f is
not even defined at w.
The
{wn }
are bounded because
{h (zn )}
is bounded and lim_{n→∞}
|wn − h(zn)|
= 0
which implies
|wn − h(zn)|
≤ C for some constant, C. Therefore, there exists δ > 0 such
that if z ∈ B^{′}
and n so by Theorem 56.0.2, the infinite product in
56.1.8 converges uniformly on B^{′}
(w,δ)
. This implies f is bounded in B^{′}
(w, δ)
and so w
is a removable singularity and f can be extended to w such that the result is analytic. It
only remains to verify f
(w )
≠0. After all, this would not do because it would be another
zero other than those in the given list. By 56.1.10, a partial product is of the
form
∏N ( )
h(z)−-h(zn)- egn(z) (56.1.11)
n=1 h(z)− wn
(56.1.11)
where
( h(zn)− wn 1 (h (zn)− wn )2 1 ( h(zn)− wn)n )
gn(z) ≡ h-(z)−-w---+ 2 -h(z)−-w-- + ⋅⋅⋅+ n- h-(z)−-w---
n n n
Each of the quotients in the definition of g_{n}
(z)
converges to 0 as z → w and so the
partial product of 56.1.11 converges to 1 as z → w because
( )
h(z)−h(zn)
h(z)−wn
→ 1 as
z → w.
If f
(w )
= 0, then if z is close enough to w, it follows
|f (z)|
<
1
2
. Also, by the uniform
convergence on B^{′}
(w,δ)
, it follows that for some N, the partial product up to N must
also be less than 1∕2 in absolute value for all z close enough to w and as noted above,
this does not occur because such partial products converge to 1 as z → w. Hence
f
(w )
≠0. This proves the corollary.
Recall the definition of a meromorphic function on Page 5581. It was a function which
is analytic everywhere except at a countable set of isolated points at which the function
has a pole. It is clear that the quotient of two analytic functions yields a meromorphic
function but is this the only way it can happen?
Theorem 56.1.8Suppose Q is a meromorphic function on an open set, Ω. Thenthere exist analytic functions on Ω,f