56.3 The Existence Of An Analytic Function With Given Values
The Weierstrass product formula, Theorem 56.1.4, along with the Mittag-Leffler theorem,
Theorem 55.2.1 can be used to obtain an analytic function which has given
values on a countable set of points, having no limit point. This is clearly an
amazing result and indicates how potent these theorems are. In fact, you can
show that it isn’t just the values of the function which may be specified at
the points in this countable set of points but the derivatives up to any finite
order.
Theorem 56.3.1Let P ≡
{zk}
_{k=1}^{∞}be a set of points in ℂ,which has no limit point.For each z_{k}, consider
∑mk
akj (z − zk)j . (56.3.18)
j=0
(56.3.18)
Then there exists an analytic function defined on ℂ such thatthe Taylor series of f at z_{k}has the first m_{k}terms given by56.3.18.^{1}
Proof:By the Weierstrass product theorem, Theorem 56.1.4, there exists an analytic
function, f defined on all of Ω such that f has a zero of order m_{k} + 1 at z_{k}. Consider this
z_{k} Thus for z near z_{k},
∑∞
f (z) = cj(z − zk)j
j=mk+1
where c_{mk+1}≠0. You choose b_{1},b_{2},
⋅⋅⋅
,b_{mk+1} such that
( )
mk∑+1 bl m∑k k j ∑∞ k j
f (z) -------k = aj (z − zk) + cj (z − zk) .
l=1 (z − zk) j=0 k=mk+1
Thus you need
mk+1 ∞ mk
∑ ∑ cb (z − z )j− l = ∑ ak(z − z )r + Higher order terms.
l=1 j=m +1 jl k r=0 r k
k
It follows you need to solve the following system of equations for b_{1},
Since c_{mk+1}≠0, it follows there exists a unique solution to the above system. You first
solve for b_{mk+1} in the top. Then, having found it, you go to the next and use c_{mk+1}≠0
again to find b_{mk} and continue in this manner. Let S_{k}
(z)
be determined in this manner
for each z_{k}. By the Mittag-Leffler theorem, there exists a Meromorphic function, g such
that g has exactly the singularities, S_{k}
(z)
. Therefore, f
(z)
g
(z)
has removable
singularities at each z_{k} and for z near z_{k}, the first m_{k} terms of fg are as prescribed. This
proves the theorem.
Corollary 56.3.2Let P ≡
{zk}
_{k=1}^{∞}be a set of points in Ω, an open set such that Phas no limit points in Ω. For each z_{k}, consider
∑mk
akj (z − zk)j . (56.3.19)
j=0
(56.3.19)
Then there exists an analytic function defined on Ω such that the Taylor series of f at z_{k}has the first m_{k}terms given by 56.3.19.
Proof: The proof is identical to the above except you use the versions of the
Mittag-Leffler theorem and Weierstrass product which pertain to open sets.
Definition 56.3.3Denote by H
(Ω)
the analytic functionsdefined on Ω, an open subset of ℂ. Then H
(Ω )
is a commutativering^{2}with the usual operations of addition and multiplication. A set, I ⊆ H
(Ω)
is called afinitely generated ideal of the ring if I is of the form
{ }
∑n
gkfk : fk ∈ H (Ω) for k = 1,2,⋅⋅⋅,n
k=1
where g_{1},
⋅⋅⋅
,g_{n}are given functions in H
(Ω)
. This ideal is also denoted as
[g1,⋅⋅⋅,gn]
and is called the ideal generated by the functions,
{g1,⋅⋅⋅,gn}
. Since there are finitelymany of these functions it is called a finitely generated ideal. A principal idealis one which is generated by a single function. An example of such a thing is
[1]
= H
(Ω)
.
Then there is the following interesting theorem.
Theorem 56.3.4Every finitely generated ideal in H
(Ω)
for Ω a connected openset (region) is a principal ideal.
Proof: Let I =
[g1,⋅⋅⋅,gn]
be a finitely generated ideal as described above.
Then if any of the functions has no zeros, this ideal would consist of H
(Ω )
because then g_{i}^{−1}∈ H
(Ω)
and so 1 ∈ I. It follows all the functions have zeros.
If any of the functions has a zero of infinite order, then the function equals
zero on Ω because Ω is connected and can be deleted from the list. Similarly, if
the zeros of any of these functions have a limit point in Ω, then the function
equals zero and can be deleted from the list. Thus, without loss of generality, all
zeros are of finite order and there are no limit points of the zeros in Ω. Let
m
(gi,z)
denote the order of the zero of g_{i} at z. If g_{i} has no zero at z, then
m
(gi,z)
= 0.
I claim that if no point of Ω is a zero of all the g_{i}, then the conclusion of the theorem
is true and in fact
[g1,⋅⋅⋅,gn ]
=
[1]
= H
(Ω )
. The claim is obvious if n = 1 because this
assumption that no point is a zero of all the functions implies g≠0 and so g^{−1} is analytic.
Hence 1 ∈
[g1]
. Suppose it is true for n− 1 and consider
[g1,⋅⋅⋅,gn]
where no point of
Ω is a zero of all the g_{i}. Even though this may be true of
{g1,⋅⋅⋅,gn}
, it may not be true
of
{g1,⋅⋅⋅,gn−1}
. By Corollary 56.1.7 there exists ϕ, a function analytic on Ω
such that m
(ϕ,z)
= min
{m (gi,z),i = 1,2,⋅⋅⋅,n− 1}
. Thus the functions
{g1∕ϕ,⋅⋅⋅,gn− 1∕ϕ }
.are all analytic. Could they all equal zero at some point, z? If so,
pick i where m
(ϕ,z)
= m
(gi,z)
. Thus g_{i}∕ϕ is not equal to zero at z after all and so these
functions are analytic there is no point of Ω which is a zero of all of them. By
induction,
[g1∕ϕ,⋅⋅⋅,gn−1∕ϕ]
= H
(Ω)
. (Also there are no new zeros obtained in this
way.)
Now this means there exist functions f_{i}∈ H
(Ω )
such that
n∑ (gi)
fi ϕ = 1
i=1
and so ϕ = ∑_{i=1}^{n}f_{i}g_{i}. Therefore,
[ϕ]
⊆
[g ,⋅⋅⋅,g ]
1 n−1
. On the other hand, if
∑_{k=1}^{n−1}h_{k}g_{k}∈
[g,⋅⋅⋅,g ]
1 n− 1
you could define h ≡∑_{k=1}^{n−1}h_{k}
(g ∕ϕ )
k
, an analytic
function with the property that hϕ = ∑_{k=1}^{n−1}h_{k}g_{k} which shows
[ϕ ]
=
[g,⋅⋅⋅,g ]
1 n− 1
.
Therefore,
[g1,⋅⋅⋅,gn] = [ϕ,gn]
Now ϕ has no zeros in common with g_{n} because the zeros of ϕ are contained in the set of
zeros for g_{1},
⋅⋅⋅
,g_{n−1}. Now consider a zero, α of ϕ. It is not a zero of g_{n} and so near α,
these functions have the form
∑∞ k ∑∞ k
ϕ (z) = ak (z − α) ,gn(z) = bk(z − α) ,b0 ⁄= 0.
k=m k=0
I want to determine coefficients for an analytic function, h such that
m (1− hgn,α) ≥ m (ϕ,α ). (56.3.20)
(56.3.20)
Let
∞
h(z) = ∑ c (z − α )k
k=0 k
and the c_{k} must be determined. Using Merten’s theorem, the power series for 1 −hg_{n} is
of the form
∞ ( j )
1− b0c0 −∑ ∑ bj−rcr (z − α)j.
j=1 r=0
First determine c_{0} such that 1 − c_{0}b_{0} = 0. This is no problem because b_{0}≠0. Next you
need to get the coefficients of
(z − α)
to equal zero. This requires
bc + b c = 0.
1 0 0 1
Again, there is no problem because b_{0}≠0. In fact, c_{1} =
(− b1c0∕b0)
. Next consider the
second order terms if m ≥ 2.
b2c0 + b1c1 +b0c2 = 0
Again there is no problem in solving, this time for c_{2} because b_{0}≠0. Continuing this way,
you see that in every step, the c_{k} which needs to be solved for is multiplied by b_{0}≠0.
Therefore, by Corollary 56.1.7 there exists an analytic function, h satisfying 56.3.20.
Therefore,
(1 − hgn)
∕ϕ has a removable singularity at every zero of ϕ and so may be
considered an analytic function. Therefore,
1− hgn
1 = --ϕ---ϕ + hgn ∈ [ϕ,gn] = [g1⋅⋅⋅gn]
which shows
[g ⋅⋅⋅g ]
1 n
= H
(Ω )
=
[1]
. It follows the claim is established.
Now suppose
{g1⋅⋅⋅gn}
are just elements of H
(Ω)
. As explained above, it can be
assumed they all have zeros of finite order and the zeros have no limit point in Ω since if
these occur, you can delete the function from the list. By Corollary 56.1.7 there exists
ϕ ∈ H
(Ω)
such that m
(ϕ,z)
≤ min
{m (gi,z) : i = 1,⋅⋅⋅,n}
. Then g_{k}∕ϕ has a removable
singularity at each zero of g_{k} and so can be regarded as an analytic function. Also, as
before, there is no point which is a zero of each g_{k}∕ϕ and so by the first part of this
argument,
[g1∕ϕ⋅⋅⋅gn∕ϕ]
= H
(Ω)
. As in the first part of the argument, this implies
[g1⋅⋅⋅gn]
=
[ϕ]
which proves the theorem.
[g1⋅⋅⋅gn]
is a principal ideal as
claimed.
The following corollary follows from the above theorem. You don’t need to assume Ω
is connected.
Corollary 56.3.5Every finitely generated ideal in H
(Ω )
for Ω an open set is aprincipal ideal.
Proof:Let
[g1,⋅⋅⋅,gn]
be a finitely generated ideal in H
(Ω )
. Let
{Uk}
be the
components of Ω. Then applying the above to each component, there exists h_{k}∈ H
(Uk)
such that restricting each g_{i} to U_{k},
[g1,⋅⋅⋅,gn]
=
[hk]
. Then let h
(z)
= h_{k}
(z)
for
z ∈ U_{k}. This is an analytic function which works.