Math 752
Spring 2015
Class 4/20 - 4/27
1
Factorization
Let G be a region in C, let {ak } be a sequence that has no limit point in
G, and let {mk } be a sequence of positive integers. Is there an analytic
function of G that has zeros precisely at the points z = ak with multiplicity
mk ? The answer, due to Weierstrass, is affirmative.
Before we can give the construction, we have to consider infinite products since the solution to this problem will involve infinitely many factors
containing (z − ak )mk . Is a product of such terms always defined? The
answer is no.
Definition 1. For given zk ∈ C we define
∞
Y
k=1
zk := lim
n→∞
n
Y
zk .
k=1
Q
Assume that
zk 6= 0. (Limit zero needs to be handled separately.)
Necessary for the existence of this limit if it is not zero: zk → 1 as k → ∞.
We would like to define the product through an exponential of a sum. For
this purpose, need to be able to define log zn . This can be done by noting
that if the product converges to a non-zero limit, then for sufficiently large
zn we have <zn > 0 (since zn → 1). Hence for sufficiently large n the
principal value log zn is defined and we can set


∞
Y
X
zn = z1 · ... · zn0 −1 · exp 
log zk 
k=n0
which reduces questions of convergence of the product to a series investigation. We note that the convergences are equivalent: if the product converges,
then the series of logarithms does as well:
Define only for this proof log z = log |z| + iθ with −π < θ ≤ π. Let
<zn > 0. Set pn = z1 · ... · zn .
1
Assume that pn → z = reiθ . Define `(z) = log |z| + iη where θ − π <
η ≤ θ + π. Note that `(pn ) = log rn + iθn with rn → r and θn → θ. (Make
a sketch!)
Set sn = log z1 + ... + log zn . Then sn = `(pn ) + 2πikn with kn ∈ Z.
We note that sn − sn−1 = log zn → 0 since zn → 1 and <zn > 0 for all
n. We have also that
`(pn ) − `(pn−1 ) = log rn − log rn−1 + i(θn − θn−1 ) → 0
since pn converges. Hence
2πi(kn − kn−1 ) = (sn − `(pn )) − (sn−1 − `(pn−1 )) → 0
as n → ∞, and since kn ∈ Z for all n, the integer kn is constant from some
n on.
Proposition 1. Let <z
Pn > −1. Then the series log(1 + zn ) converges
absolutely iff the series
zn converges absolutely.
Proof. The main tool is the inequality
1
3
1
|z| ≤ | log(1 + z)| ≤ |z| for |z| < .
2
2
2
P
Assuming this, absolute
convergence of
zn immediately implies absoP
lute convergence
of
log(1
+
z
).
In
the
other
direction, absolute convern
P
gence of
| log(1 + zn )| implies in particular that zn → 0, so for sufficiently
large n we have |zn | < 21 .
It remains to establish the inequality. Starting point is
log(1 + z) =
∞
X
(−1)n−1
n=1
hence
zn
,
n
1 − log(1 + z) ≤ 1 (|z| + |z|2 + ...)
2
z
Factor out |z| and calculate the remaining geometric series. In |z| < 12
we get an absolute upper bound 1/2 for the difference above. Multiply by
z, and use triangle inequality in both directions.
Q
zn to be given
P We define therefore absolute convergence of
P when
log zn converges absolutely, which in turn is equivalent to
(zn − 1)
being absolutely convergent.
2
We need to know when an infinite product of analytic functions converges
to an analytic function. By the previous theorems this can be done by
investigating for a convergent sequence of functions fn → f under which
conditions the exponentials exp(fn ) converge to exp(f ).
Lemma 1. Let G ⊆ C and let fi : G → C be such that fn → f uniformly
on G. If there is a constant a so that <f (z) ≤ a for all z ∈ G, then
exp(fn (z)) → exp(f (z)) uniformly on G.
Proof. The function exp f (z) is not zero on G, hence
exp fn (z)
exp fn (z)
| exp(fn (z)) − exp(f (z))| = − 1 | exp(f (z))| ≤ − 1 ea .
exp f (z)
exp f (z)
Note that fn (z) − f (z) → 0, hence the quotient of exponents converges
to 1. The estimate needs to be uniform on G. Hence, let ε > 0. Then
there exists δ > 0 so that |z| < δ implies |ez − 1| < ε (continuity of ez at
z = 1). The uniform convergence of fn implies that there exists n0 so that
|fn (z) − f (z)| < δ for all n ≥ n0 and z ∈ G. Hence for all n ≥ n0 and z ∈ G
exp fn (z)
exp f (z) − 1 ≤ ε.
hence | exp fn (z) − exp f (z)| ≤ εea uniformly in G.
To see that the bound for f is necessary, consider fn (z) = z + n1 on C,
which converges uniformly to f (z) = z, but exp(fn (z)) does not converge
uniformly to ez on C. (It does converge uniformly on every compact subset
of C, though.)
Lemma 2. Let G be compact. Let gn : G → C be continuous and such that
P
gn (x) converges absolutely and uniformly. Then the product
f (z) =
∞
Y
(1 + gn (z))
n=1
converges absolutely and uniformly. Also there exists n0 so that f (z) = 0 if
and only if gn (z) = −1 for some n ≤ n0 .
Proof. Remove the terms
P for which |gn (z)| ≥ 1/2. There is only a finite
number of them since
gn converges uniformly. Hence may assume that
|gn (z)| < 1/2 for all n. Then <(1 + gn (z)) > 0 and | log(1 + gn (z))| ≤
3/2|gn (z)|. Hence
X
h(z) =
log(1 + gn (z))
3
converges uniformly. Since G is compact, h is bounded. It follows that
<h(z) < a for some a and all z ∈ G. By the previous lemma
Y
exp h(z) =
(1 + gn (z))
converges uniformly for z ∈ G, and the left hand side is never zero.
Q
It follows in particular that the multiplicity of a zero of f (z) = (1 +
gn (z)) is the sum of the multiplicities of 1 + gn (z) at that zero.
Back to the problem of finding an analytic function with prescribed zeros.
Definition 2. We define E0 (z) = 1 − z and
z2
zn
En (z) = (1 − z) exp z +
+ ... +
2
n
Lemma 3. If |z| ≤ 1 and n ≥ 0 then |1 − En (z)| ≤ |z|n+1 .
Proof. Consider the power series expansion
En (z) = 1 +
∞
X
ak z k .
k=1
We have
En0 (z)
zn
= − exp z + ... +
n
zn
+ (1 − z) exp z + ... +
(1 + z + ... + z n−1 )
n
zn
n
= −z exp z + ... +
,
n
P
of coefficients shows a1 = ... =
and this equals k≥1 ak kz k−1 . ComparisonP
aN = 0 and ak ≤ 0. Since 0 = En (1) = 1 + k>n ak , it follows that
X
X
|ak | = −
ak = 1.
k>n
k>n
Hence in |z| ≤ 1
∞
X
k
ak z ≤ |z|n+1 .
|En (z) − 1| = k=n+1
4
Lemma 4. Let {ak } ⊆ C with lim |ak | = ∞ and ak 6= 0 for all n. Then for
all r > 0
∞ X
r n
< ∞.
|ak |
k=1
Proof. Fix r > 0 and note
Pthatk for sufficiently large k we have r/|ak | < 1/2.
Hence comparison with
1/2 gives the result.
Theorem 1. Let {ak } ⊆ C with lim |ak | = ∞ and ak 6= 0 for all n. (Repetitions are allowed.) If {nk } has the property that
∞ X
r nk +1
< ∞,
|ak |
k=1
for all r > 0, then
f (z) =
∞
Y
Enk (z/ak )
k=1
converges in H(C). The function f is entire, and has zeros only at z = an
with multplicity equal to the number of times that the zero is repeated in
{an }.
Proof. By the previous lemma such a sequence nk exists. Fix r > 0 and
consider all z with |z| ≤ r. The estimate
|1 − Enk (z/ak )| ≤ |z/ak |nk +1 ≤ |r/ak |nk +1
P
gives an absolutely convergent series dominating
|1 − Enk (z/ak )|. Hence
the product converges absolutely (and uniformly) on |z| ≤ r, and hence f is
entire.
The previous lemma shows that nk = n − 1 will always make the series converge. However, in practice the exponents nk can be chosen much
smaller.
Theorem 2 (Weierstrass Factorization Theorem). Let f be an entire function and let {an } be the non-zero zeros of f repeated with multiplicity. Let
m ∈ N0 be the order of the zero of f (z) at z = 0. Then there exists entire g
and {pn } such that
f (z) = z m eg(z)
∞
Y
n=1
5
Epn (z/an ).
Proof. Most is shown: can find integers pn so that
h(z) = z m
∞
Y
Epn (z/an )
n=1
is an entire function that has zeros exactly where f has, with the correct
multiplicty (by previous theorem). Hence g1 = f /h is entire and non-zero
on the simply connected domain C and can therefore be written as g1 = eg
with some entire function g.
As an example we factor F (z) = sin πz. We require two auxiliary results.
P
Lemma 5. G open, fn ∈ H(G) so that n (fn (z) − 1)Qconverges uniformly
and absolutely in every compact subset of G. Let f = fn .
1. For all z ∈ G,
f 0 (z) =
∞
X
fk0 (z)
Y
fn (z).
n6=k
k=1
2. Let K be a compact subset of G with f (z) 6= 0 for all z ∈ K. Then
∞
f 0 (z) X fn0 (z)
=
.
f (z)
fn (z)
n=1
Proof. Let K be a compact subset of G. Consider the partial product
Fn (z) =
n
Y
fk (z).
k=1
We have Fn → f in H(G), and hence Fn0 → f 0 . Calculate this derivative:
Fn0 (z)
=
n
X
fk0 (z)
k=1
n
Y
fj (z).
j6=k,j=1
It remains to show that the sum on the right converges
P to the claimed series.
(This is not automatic since the sum is of the form k gn,k (z).)
Since fn (z) − 1 converges to zero uniformly on K by assumption, there
exists k0 so that |fk (z)| ≥ 1/2 for k ≥ k0 and all z ∈ K. Since the product
converges to f uniformly on K, we may write
n
Y
fj (z) = f (z)/fk (z) + hn (z)
j6=k,j=1
6
where |hn (z)| < ε for sufficiently large n and all z ∈ K. Hence


∞
n
n 0
X
X
X
f (z) 0
fk (z)hn (z) ≤ 2ε 
|fk0 (z)| + D
fk0 (z)
f (z) −
≤
fk (z) fk (z) k=1
k=k0
k=1
P
and the series on the right converges since (fk (z) − 1) converges uniformly
and absolutely, may therefore be differentiated termwise, and the series of
derivatives has the same convergence properties. This shows the first part.
The second part follows immediately by noting that the assumptions allow
us to divide by f (z).
Lemma 6. For all a ∈ Z\Z,
∞
π cot πa =
1 X 2a
+
.
a
a2 − n2
n=1
Proof. This is obtained by evaluating γn π(z 2 − a2 )−1 cot πzdz where γn is
the contour with corners ±(n+1/2)±in. We evaluate first the integral using
the residue theorem, and then show that the integral converges to zero as
n → ∞.
The residues at z = ±a are both π(2a)−1 cot πa. We need to find next
the residues at the integers. We have for n ∈ Z
R
sin πz = (−1)n sin π(z − n),
hence
π cot πz = π
cos πz
1
π(z − n) cos πz
=
(−1)n sin π(z − n)
z − n sin π(z − n) (−1)n
hence the residue of π cot πz at z = n ∈ Z is (−1)n cos πn = 1. The residues
of the integrand at the zeros of sin πz are therefore (n2 − a2 )−1 .
To estimate
Z
lim
π(z 2 − a2 )−1 cot πzdz
n→∞ γ
n
note that
| cos z|2 = cos2 x + sinh2 y,
hence
| cot z|2 =
| sin z|2 = sin2 x + sinh2 y,
cos2 x + sinh2 y
sin2 x + sinh2 y
7
which is uniformly bounded on γn with a constant independent of n.
Letting n → ∞ gives
0=
X
π cot πa
1
1
− 2+
,
2
a
a
n − a2
n6=0
and multiplication by a and combining the summands for n and −n gives
the result. Since the length of the integration path is ≤ 5n, the integral
converges to zero as n → ∞.
Back to factoring sin πz. Find the sequence pn first. The zeros of F are
all simple and located at the integers. Hence, need for fixed r > 0
X r pn +1
< ∞.
n
n∈Z
n6=0
This is satisfied for pn = 1, all n. (Note that pn = 0 does not work.) We
obtain with E1 (z) = (1 − z)ez
Y
z
sin πz = eg(z) z
1−
exp(z/n)
n
n6=0
∞ Y
z2
g(z)
=e z
1− 2
n
n=1
d
sin πz. We have (eg(z) )0 /eg(z) =
In order to find g(z), consider (sin πz)−1 dz
0
z /z = 1/z and
g 0 (z),
(1 − z 2 /n2 )0
2z
= 2
,
2
2
(1 − z /n )
z − n2
and hence
∞
π cot πz = g 0 (z) +
1 X 2z
+
.
z
z 2 − n2
n=1
Comparing this with the representation of π cot πz above, shows that
= 0. Evaluate the representation at z = 0 to get that eg(z) = π. Thus,
g 0 (z)
∞ Y
z2
sin πz = πz
1− 2
n
n=1
8