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** — 1. Jensen’s formula — **

Suppose is a non-zero rational function , then by the fundamental theorem of algebra one can write

for some non-zero constant , where ranges over the zeroes of (counting multiplicity) and ranges over the zeroes of (counting multiplicity), and assuming avoids the zeroes of . Taking absolute values and then logarithms, we arrive at the formula as long as avoids the zeroes of both and . (In this set of notes we use for the natural logarithm when applied to a positive real number, and for the standard branch of the complex logarithm (which extends ); the multi-valued complex logarithm will only be used in passing.) Alternatively, taking logarithmic derivatives, we arrive at the closely related formula again for avoiding the zeroes of both and . Thus we see that the zeroes and poles of a rational function describe the behaviour of that rational function, as well as close relatives of that function such as the log-magnitude and log-derivative . We have already seen these sorts of formulae arise in our treatment of the argument principle in 246A Notes 4.

Exercise 1Let be a complex polynomial of degree .

- (i) (Gauss-Lucas theorem) Show that the complex roots of are contained in the closed convex hull of the complex roots of .
- (ii) (Laguerre separation theorem) If all the complex roots of are contained in a disk , and , then all the complex roots of are also contained in . (
Hint:apply a suitable Möbius transformation to move to infinity, and then apply part (i) to a polynomial that emerges after applying this transformation.)

There are a number of useful ways to extend these formulae to more general meromorphic functions than rational functions. Firstly there is a very handy “local” variant of (1) known as Jensen’s formula:

Theorem 2 (Jensen’s formula)Let be a meromorphic function on an open neighbourhood of a disk , with all removable singularities removed. Then, if is neither a zero nor a pole of , we have where and range over the zeroes and poles of respectively (counting multiplicity) in the disk .

One can view (3) as a truncated (or localised) variant of (1). Note also that the summands are always non-positive.

*Proof:* By perturbing slightly if necessary, we may assume that none of the zeroes or poles of (which form a discrete set) lie on the boundary circle . By translating and rescaling, we may then normalise and , thus our task is now to show that

An important special case of Jensen’s formula arises when is holomorphic in a neighborhood of , in which case there are no contributions from poles and one simply has

This is quite a useful formula, mainly because the summands are non-negative. Here are some quick applications of this formula:

Exercise 3Use (6) to give another proof of Liouville’s theorem: a bounded holomorphic function on the entire complex plane is necessarily constant.

Exercise 4Use Jensen’s formula to prove the fundamental theorem of algebra: a complex polynomial of degree has exactly complex zeroes (counting multiplicity), and can thus be factored as for some complex numbers with . (Note that the fundamental theorem was invoked previously in this section, but only for motivational purposes, so the proof here is non-circular.)

Exercise 5 (Shifted Jensen’s formula)Let be a meromorphic function on an open neighbourhood of a disk , with all removable singularities removed. Show that for all in the open disk that are not zeroes or poles of , where and . (The function appearing in the integrand is sometimes known as the Poisson kernel, particularly if one normalises so that and .)

Exercise 6 (Bounded type)

- (i) If is a bounded holomorphic function on that is not identically zero, show that .
- (ii) If is a meromorphic function on that is the ratio of two bounded holomorphic functions that are not identically zero, show that . (Functions of this form are said to be of bounded type and lie in the
Nevanlinna classfor the unit disk .)

Exercise 7 (Smoothed out Jensen formula)Let be a meromorphic function on an open set , and let be a smooth compactly supported function. Show that where range over the zeroes and poles of (respectively) in the support of . Informally argue why this identity is consistent with Jensen’s formula.

When applied to entire functions , Jensen’s formula relates the order of growth of near infinity with the density of zeroes of . Here is a typical result:

Proposition 8Let be an entire function, not identically zero, that obeys a growth bound for some and all . Then there exists a constant such that has at most zeroes (counting multiplicity) for any .

Entire functions that obey a growth bound of the form for every and (where depends on ) are said to be of order at most . The above theorem shows that for such functions that are not identically zero, the number of zeroes in a disk of radius does not grow much faster than . This is often a useful preliminary upper bound on the zeroes of entire functions, as the order of an entire function tends to be relatively easy to compute in practice.

*Proof:* First suppose that is non-zero. From (6) applied with and one has

Just as (3) and (7) give truncated variants of (1), we can create truncated versions of (2). The following crude truncation is adequate for many applications:

Theorem 9 (Truncated formula for log-derivative)Let be a holomorphic function on an open neighbourhood of a disk that is not identically zero on this disk. Suppose that one has a bound of the form for some and all on the circle . Let be constants. Then one has the approximate formula for all in the disk other than zeroes of . Furthermore, the number of zeroes in the above sum is .

*Proof:* To abbreviate notation, we allow all implied constants in this proof to depend on .

We mimic the proof of Jensen’s formula. Firstly, we may translate and rescale so that and , so we have when , and our main task is to show that

for . Note that if then vanishes on the unit circle and hence (by the maximum principle) vanishes identically on the disk, a contradiction, so we may assume . From hypothesis we then have on the unit circle, and so from Jensen’s formula (3) we see that In particular we see that the number of zeroes with is , as claimed.Suppose has a zero with . If we factor , where is the Blaschke product (5), then

Observe from Taylor expansion that the distance between and is , and hence for . Thus we see from (9) that we may use Blaschke products to remove all the zeroes in the annulus while only affecting the left-hand side of (8) by ; also, removing the Blaschke products does not affect on the unit circle, and only affects by thanks to (9). Thus we may assume without loss of generality that there are no zeroes in this annulus.Similarly, given a zero with , we have , so using Blaschke products to remove all of these zeroes also only affects the left-hand side of (8) by (since the number of zeroes here is ), with also modified by at most . Thus we may assume in fact that has no zeroes whatsoever within the unit disk. We may then also normalise , then for all . By Jensen’s formula again, we have

and thus (by using the identity for any real ) On the other hand, from (7) we have which implies from (10) that and its first derivatives are on the disk . But recall from the proof of Jensen’s formula that is the derivative of a logarithm of , whose real part is . By the Cauchy-Riemann equations for , we conclude that on the disk , as required.

Exercise 10

- (i) (Borel-Carathéodory theorem) If is analytic on an open neighborhood of a disk , show that (
Hint:one can normalise , , , and . Now maps the unit disk to the half-plane . Use a Möbius transformation to map the half-plane to the unit disk and then use the Schwarz lemma.)- (ii) Use (i) to give an alternate way to conclude the proof of Theorem 9.

A variant of the above argument allows one to make precise the heuristic that holomorphic functions locally look like polynomials:

Exercise 11 (Local Weierstrass factorisation)Let the notation and hypotheses be as in Theorem 9. Then show that for all in the disk , where is a polynomial whose zeroes are precisely the zeroes of in (counting multiplicity) and is a holomorphic function on of magnitude and first derivative on this disk. Furthermore, show that the degree of is .

Exercise 12 (Preliminary Beurling factorisation)Let denote the space of bounded analytic functions on the unit disk; this is a normed vector space with norm

- (i) If is not identically zero, and denote the zeroes of in counting multiplicity, show that and
- (ii) Let the notation be as in (i). If we define the Blaschke product where is the order of vanishing of at zero, show that this product converges absolutely to a meromorphic function on outside of the , and that for all . (It may be easier to work with finite Blaschke products first to obtain this bound.)
- (iii) Continuing the notation from (i), establish a factorisation for some holomorphic function with for all .
- (iv) (Theorem of F. and M. Riesz, special case) If extends continuously to the boundary , show that the set has zero measure.

Remark 13The factorisation (iii) can be refined further, with being the Poisson integral of some finite measure on the unit circle. Using the Lebesgue decomposition of this finite measure into absolutely continuous parts one ends up factorising functions into “outer functions” and “inner functions”, giving the Beurling factorisation of . There are also extensions to larger spaces than (which are to as is to ), known as Hardy spaces. We will not discuss this topic further here, but see for instance this text of Garnett for a treatment.

Exercise 14 (Littlewood’s lemma)Let be holomorphic on an open neighbourhood of a rectangle for some and , with non-vanishing on the boundary of the rectangle. Show that where ranges over the zeroes of inside (counting multiplicity) and one uses a branch of which is continuous on the upper, lower, and right edges of . (This lemma is a popular tool to explore the zeroes of Dirichlet series such as the Riemann zeta function.)

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