4.4.4 Classifying Growth of Entire Functions
Lemma 4.4.4.
Let (where ) be a nowhere-vanishing holomorphic function. It follows that
Proof.
Without loss of generality, assume . Since is non-vanishing and is simply connected, we may define the holomorphic logarithm as
for any fixed and all , where is any piecewise smooth curve from to .
Hence, and is therefore harmonic. The assertion then follows from the mean-value property.
Theorem 4.4.11 (Jensen's Formula).
Let be meromorphic such that . If , are the zeros and poles of in , counted with multiplicities and orders, respectively, then
Proof.
For simplicity, assume are the zeros in and are the zeros on . Similarly, let be the poles in and be the poles on . Let
where is holomorphic and non-vanishing on . Since
by Lemma 4.4.4 on ,
where , , and the leftover integrals (up until and ) for interior points vanish by Cauchy–Goursat (Theorem 3.1.7), since has a removable singularity at .
We now are left to prove that the remaining integral vanishes as well, which is not as immediate since the integrand does not extend continuously to the boundary. Let , , then (by )
Splitting at and using with a substitution for the second integral then yields
Changing back to , we have
As an immediate consequence, we have:
Corollary 4.4.2 (Jensen's Inequality).
Let be holomorphic on such that and . It follows that
Theorem 4.4.12 (Poisson–Jensen Formula).
Suppose is a meromorphic function on such that on and is non-vanishing and non-infinity on . Let and be the zeros and poles of in , counted with multiplicity and order, respectively (multiplicities and orders count as multiple zeros or poles). Then it follows that
where , , and is the Poisson kernel in (4.4.6).
Proof.
For fixed not at zeros or poles, let
where is the unit disk automorphism sending 0 to . Then maps 0 to , and has zeros at or and poles at or . By Jensen’s formula (Theorem 4.4.11),
Lemma 4.4.5.
Let be a non-constant bounded holomorphic function whose zeros are , counted according to their multiplicities, ordered such that for all . Then,
is convergent.
Proof.
First assume and choose such that on . Let count the number of zeros of , according to multiplicities, inside . By Jensen’s Formula (Theorem 4.4.11), we have
For any fixed positive integer , choose such that . Then and
since each for . Therefore,
Rearranging,
Now let with . Since , it follows that
This holds for every . Since for all , the partial sums are decreasing and bounded below by , hence converge to some finite limit, and
or equivalently,
For any , we have . Hence,
If has a zero of multiplicity at , then the argument applies to .
Theorem 4.4.13 (Blaschke Product).
Let be a sequence such that the series is convergent (known as the Blaschke condition). Then the Blaschke product, defined by
(where is a Möbius transformation in the form of (4.4.1)), locally uniformly converges to an analytic function on such that on , and its only zeros are precisely at each of , counted according to multiplicities.
Proof.
If it can be shown that
locally uniformly converges, we can use Lemma 4.4.3 to show that the infinite product converges uniformly on compact subsets of . Let be a compact subset. The summand can be bounded with
Since
is convergent (Blaschke condition), by the Weierstrass -Test (Theorem 2.3.2), converges uniformly on . By Lemma 4.4.3, the infinite product in (4.4.13) converges uniformly on compact subsets of . The properties of its zeros follow from the lemma.
Lastly, since and each partial product is bounded by 1, it follows that on .
Remark.
A more general Blaschke product has an additional factor of to account for a zero at the origin, similar to the case of the Weierstrass product.
Corollary 4.4.3.
Let be bounded and holomorphic whose multiplicity of the zero at 0 is (if does not vanish at 0, then ). If are its zeros in , counting multiplicities, then
where is bounded, holomorphic, and non-vanishing on . Moreover,
Proof.
Let
By construction, extends to its removable singularities to a holomorphic function that does not vanish. Because
it follows that
The partial products
give for fixed , , the existence of such that implies
Then by the Maximum Modulus Principle (Theorem 3.4.1),
as . Letting , gives
which in conjunction with (4.4.14), completes the final assertion.
From the results above, a recurring theme in complex analysis is hinted at; the rate of growth of functions provides insight towards the distribution of its zeros.
The subjects to be discussed here are relevant and preliminary to Nevanlinna theory, or the study of holomorphic value distribution.
For an entire function , let (by the Maximum Modulus Principle in Theorem 3.4.1).
Definition 4.4.4 (Growth Order of Entire Functions).
An entire function is said to be of finite order if there exists such that
or in loose terms, is of finite order if it grows at most exponentially for large . The order of , or is defined to be the infimum of all satisfying the previous condition.
Proposition 4.4.2.
Let be entire; then if there exist such that
then .
Proof.
For , since as , for any , there exists such that
for . There exists such that
For simplicity, let the value be denoted by . Then
By assumption, we have
Letting , the assertion follows.
Theorem 4.4.14.
The order of an entire may be explicitly given by
Proof.
By assumption, we have , (or simply just by the nature of the exponential) such that
for some and any . Taking logarithms twice we have
as . Moreover, for any , , such that
as . Therefore,
Example 4.4.2.
The function is of order 1, while is not of finite order.
Proof.
We consider the two examples separately:
-
Observe that
For , we have , and hence for , we have
Therefore,
-
Let , then
The utility of is that it gives implications on the rate of which the zeros of an entire function tend to . The order for meromorphic functions is more general and is pertinent in Nevanlinna Theory (@ sec:nevanlinnatheory). This is quantified technically by the convergence range of the sum given by
where each is a zero. Specifically, the infimum of all such under which the prescribed sum converges correlates to this right. For example, let for each . Then for any , the integral test gives the convergence of the series, while if (corresponding to a slower approach to ), the series converges for .
For the following discussions, let count the zeros of in according to multiplicity.
Lemma 4.4.6.
If is entire with , then
Proof.
By Jensen’s formula (Theorem 4.4.11), for , we have
where are the zeros of in , ordered such that each . Then
Theorem 4.4.15.
For a nonzero complex sequence counting multiplicities (such that , etc.), the sum
converges for any
where counts in the closed disk of radius .
Proof.
Choose such that
For sufficiently large ,
For sufficiently large , by the ordering of zeros, it follows that
for sufficiently small . As , we have
By the comparison test, we then have the convergence of
Theorem 4.4.16.
For an entire function () of finite order whose zeros are at counting multiplicities (such that , etc.), the sum
converges for any .
Proof.
By trivial definition, we have
for all and some . Lemma 4.4.6 gives that for any ,
Hence,
as . Then for sufficiently large , we have
For sufficiently large , by the ordering of zeros, it follows that
for sufficiently small . As , we have
The left-hand side as a summation is convergent for or lower, and hence we have the convergence of
Therefore, for any , the series
converges. Then by the Weierstrass Factorization Theorem (Theorem 4.4.7),
locally uniformly converges on , where is entire.
Definition 4.4.5.
The rank of an entire function is the smallest for which the associated sum
converges, where are its zeros in .
The conclusion of Theorem 4.4.16 is that the rank of an entire function with finite order is finite. Moreover, the rank .
Definition 4.4.6.
Let be entire of finite rank . By the Weierstrass Factorization theorem (Theorem 4.4.7),
If is a polynomial of degree , then is said to be of finite genus .
This particular Weierstrass factorization is the Weierstrass canonical factorization of (the portion corresponding to the product of elementary factors itself is the Weierstrass canonical product). Now that we have indulged in the implications of to its zero distribution, we now turn to the function in the exponential.
Lemma 4.4.7.
Let be entire with finite order such that . Let be the zeros of , listed with multiplicities, such that . Suppose ; then for any ,
Proof.
For fixed , let such that lie in . For each we obtain
since by the logarithm formula. Now by definition of , Lemma 4.4.6 gives the estimate for sufficiently large and arbitrarily small :
Thus,
Letting (positive by theorem assumption), we obtain
Theorem 4.4.17.
Let be entire with . Then for ( integer) and ,
Proof.
For fixed , , we have
by the Residue Theorem (Theorem 4.5.1). Since where is an integer, we must have and thus
Therefore,
where the last expression uses the inequality derived from Jensen’s formula (Corollary 4.4.2) on the remaining integral.
Now by assumption, we have
for any and sufficiently large . Hence,
at . Then since , the expression vanishes as .
Proposition 4.4.3.
Let be entire, non-constant, and of finite order such that . Let be the zeros of counted according to multiplicities such that . If is an integer, then
for all .
Proof.
Let . By the Poisson–Jensen Formula (Theorem 4.4.12), at each non-singular point, we have (the kernel representation derived in (4.4.6))
For any holomorphic , we have
Therefore, by differentiation under the integral sign,
Differentiating times from here gives
The first two terms vanish as by Theorem 4.4.17 and Lemma 4.4.7.
Lemma 4.4.8 (Logarithmic Factorization).
Let be entire, non-constant, and of finite order such that . Let
be the associated product. If is an integer, then
for all .
Proof.
Let be the -th partial product of . Then
implying that
Since the polynomial in the rightmost term has degree at most , and because , after derivatives each term of the expression vanishes. For an arbitrarily chosen compact avoiding , some such that for all , we have , . Then the convergence of
from Theorem 4.4.16 implies the absolute convergence of
in . Moreover, it can be shown that the uniform convergence of (from the Weierstrass factorization) and (by the Weierstrass Convergence Theorem, Theorem 4.1.1) in implies that of . Hence, the Weierstrass Convergence Theorem implies that
The two preceding results are similar in conclusion, but Lemma 4.4.8 is not a special case of Proposition 4.4.3 since we have not asserted that the canonical product is of finite order.