You are currently browsing the tag archive for the ‘Jensen’s formula’ tag.

Previous set of notes: 246A Notes 5. Next set of notes: Notes 2.

** — 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; it can be viewed as a more precise assertion of the subharmonicity of (see Exercises 60(ix) and 61 of 246A Notes 5). 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 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 holomorphic function on , 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.)

We will shortly turn to the complex-analytic approach to multiplicative number theory, which relies on the basic properties of complex analytic functions. In this supplement to the main notes, we quickly review the portions of complex analysis that we will be using in this course. We will not attempt a comprehensive review of this subject; for instance, we will completely neglect the conformal geometry or Riemann surface aspect of complex analysis, and we will also avoid using the various boundary convergence theorems for Taylor series or Dirichlet series (the latter type of result is traditionally utilised in multiplicative number theory, but I personally find them a little unintuitive to use, and will instead rely on a slightly different set of complex-analytic tools). We will also focus on the “local” structure of complex analytic functions, in particular adopting the philosophy that such functions behave locally like complex polynomials; the classical “global” theory of entire functions, while traditionally used in the theory of the Riemann zeta function, will be downplayed in these notes. On the other hand, we will play up the relationship between complex analysis and Fourier analysis, as we will incline to using the latter tool over the former in some of the subsequent material. (In the traditional approach to the subject, the Mellin transform is used in place of the Fourier transform, but we will not emphasise the role of the Mellin transform here.)

We begin by recalling the notion of a holomorphic function, which will later be shown to be essentially synonymous with that of a complex analytic function.

Definition 1 (Holomorphic function)Let be an open subset of , and let be a function. If , we say that iscomplex differentiableat if the limitexists, in which case we refer to as the (complex)

derivativeof at . If is differentiable at every point of , and the derivative is continuous, we say that isholomorphicon .

Exercise 2Show that a function is holomorphic if and only if the two-variable function is continuously differentiable on and obeys the Cauchy-Riemann equation

Basic examples of holomorphic functions include complex polynomials

as well as the complex exponential function

which are holomorphic on the entire complex plane (i.e., they are entire functions). The sum or product of two holomorphic functions is again holomorphic; the quotient of two holomorphic functions is holomorphic so long as the denominator is non-zero. Finally, the composition of two holomorphic functions is holomorphic wherever the composition is defined.

- (i) Establish Euler’s formula
for all . (

Hint:it is a bit tricky to do this starting from the trigonometric definitions of sine and cosine; I recommend either using the Taylor series formulations of these functions instead, or alternatively relying on the ordinary differential equations obeyed by sine and cosine.)- (ii) Show that every non-zero complex number has a complex logarithm such that , and that this logarithm is unique up to integer multiples of .
- (iii) Show that there exists a unique principal branch of the complex logarithm in the region , defined by requiring to be a logarithm of with imaginary part between and . Show that this principal branch is holomorphic with derivative .

In real analysis, we have the fundamental theorem of calculus, which asserts that

whenever is a real interval and is a continuously differentiable function. The complex analogue of this fact is that

whenever is a holomorphic function, and is a contour in , by which we mean a piecewise continuously differentiable function, and the contour integral for a continuous function is defined via change of variables as

The complex fundamental theorem of calculus (2) follows easily from the real fundamental theorem and the chain rule.

In real analysis, we have the rather trivial fact that the integral of a continuous function on a closed contour is always zero:

In complex analysis, the analogous fact is significantly more powerful, and is known as Cauchy’s theorem:

Theorem 4 (Cauchy’s theorem)Let be a holomorphic function in a simply connected open set , and let be a closed contour in (thus ). Then .

Exercise 5Use Stokes’ theorem to give a proof of Cauchy’s theorem.

A useful reformulation of Cauchy’s theorem is that of contour shifting: if is a holomorphic function on a open set , and are two contours in an open set with and , such that can be continuously deformed into , then . A basic application of contour shifting is the Cauchy integral formula:

Theorem 6 (Cauchy integral formula)Let be a holomorphic function in a simply connected open set , and let be a closed contour which is simple (thus does not traverse any point more than once, with the exception of the endpoint that is traversed twice), and which encloses a bounded region in the anticlockwise direction. Then for any , one has

*Proof:* Let be a sufficiently small quantity. By contour shifting, one can replace the contour by the sum (concatenation) of three contours: a contour from to , a contour traversing the circle once anticlockwise, and the reversal of the contour that goes from to . The contributions of the contours cancel each other, thus

By a change of variables, the right-hand side can be expanded as

Sending , we obtain the claim.

The Cauchy integral formula has many consequences. Specialising to the case when traverses a circle around , we conclude the mean value property

whenever is holomorphic in a neighbourhood of the disk . In a similar spirit, we have the maximum principle for holomorphic functions:

Lemma 7 (Maximum principle)Let be a simply connected open set, and let be a simple closed contour in enclosing a bounded region anti-clockwise. Let be a holomorphic function. If we have the bound for all on the contour , then we also have the bound for all .

*Proof:* We use an argument of Landau. Fix . From the Cauchy integral formula and the triangle inequality we have the bound

for some constant depending on and . This ostensibly looks like a weaker bound than what we want, but we can miraculously make the constant disappear by the “tensor power trick“. Namely, observe that if is a holomorphic function bounded in magnitude by on , and is a natural number, then is a holomorphic function bounded in magnitude by on . Applying the preceding argument with replaced by we conclude that

and hence

Sending , we obtain the claim.

Another basic application of the integral formula is

Corollary 8Every holomorphic function is complex analytic, thus it has a convergent Taylor series around every point in the domain. In particular, holomorphic functions are smooth, and the derivative of a holomorphic function is again holomorphic.

Conversely, it is easy to see that complex analytic functions are holomorphic. Thus, the terms “complex analytic” and “holomorphic” are synonymous, at least when working on open domains. (On a non-open set , saying that is analytic on is equivalent to asserting that extends to a holomorphic function of an open neighbourhood of .) This is in marked contrast to real analysis, in which a function can be continuously differentiable, or even smooth, without being real analytic.

*Proof:* By translation, we may suppose that . Let be a a contour traversing the circle that is contained in the domain , then by the Cauchy integral formula one has

for all in the disk . As is continuously differentiable (and hence continuous) on , it is bounded. From the geometric series formula

and dominated convergence, we conclude that

with the right-hand side an absolutely convergent series for , and the claim follows.

Exercise 9Establish the generalised Cauchy integral formulaefor any non-negative integer , where is the -fold complex derivative of .

This in turn leads to a converse to Cauchy’s theorem, known as Morera’s theorem:

Corollary 10 (Morera’s theorem)Let be a continuous function on an open set with the property that for all closed contours . Then is holomorphic.

*Proof:* We can of course assume to be non-empty and connected (hence path-connected). Fix a point , and define a “primitive” of by defining , with being any contour from to (this is well defined by hypothesis). By mimicking the proof of the real fundamental theorem of calculus, we see that is holomorphic with , and the claim now follows from Corollary 8.

An important consequence of Morera’s theorem for us is

Corollary 11 (Locally uniform limit of holomorphic functions is holomorphic)Let be holomorphic functions on an open set which converge locally uniformly to a function . Then is also holomorphic on .

*Proof:* By working locally we may assume that is a ball, and in particular simply connected. By Cauchy’s theorem, for all closed contours in . By local uniform convergence, this implies that for all such contours, and the claim then follows from Morera’s theorem.

Now we study the zeroes of complex analytic functions. If a complex analytic function vanishes at a point , but is not identically zero in a neighbourhood of that point, then by Taylor expansion we see that factors in a sufficiently small neighbourhood of as

for some natural number (which we call the *order* or *multiplicity* of the zero at ) and some function that is complex analytic and non-zero near ; this generalises the factor theorem for polynomials. In particular, the zero is isolated if does not vanish identically near . We conclude that if is connected and vanishes on a neighbourhood of some point in , then it must vanish on all of (since the maximal connected neighbourhood of in on which vanishes cannot have any boundary point in ). This implies unique continuation of analytic functions: if two complex analytic functions on agree on a non-empty open set, then they agree everywhere. In particular, if a complex analytic function does not vanish everywhere, then all of its zeroes are isolated, so in particular it has only finitely many zeroes on any given compact set.

Recall that a rational function is a function which is a quotient of two polynomials (at least outside of the set where vanishes). Analogously, let us define a meromorphic function on an open set to be a function defined outside of a discrete subset of (the *singularities* of ), which is locally the quotient of holomorphic functions, in the sense that for every , one has in a neighbourhood of excluding , with holomorphic near and with non-vanishing outside of . If and has a zero of equal or higher order than at , then the singularity is removable and one can extend the meromorphic function holomorphically across (by the holomorphic factor theorem (4)); otherwise, the singularity is non-removable and is known as a *pole*, whose order is equal to the difference between the order of and the order of at . (If one wished, one could extend meromorphic functions to the poles by embedding in the Riemann sphere and mapping each pole to , but we will not do so here. One could also consider non-meromorphic functions with essential singularities at various points, but we will have no need to analyse such singularities in this course.) If the order of a pole or zero is one, we say that it is *simple*; if it is two, we say it is *double*; and so forth.

Exercise 12Show that the space of meromorphic functions on a non-empty open set , quotiented by almost everywhere equivalence, forms a field.

By quotienting two Taylor series, we see that if a meromorphic function has a pole of order at some point , then it has a Laurent expansion

absolutely convergent in a neighbourhood of excluding itself, and with non-zero. The Laurent coefficient has a special significance, and is called the residue of the meromorphic function at , which we will denote as . The importance of this coefficient comes from the following significant generalisation of the Cauchy integral formula, known as the residue theorem:

Exercise 13 (Residue theorem)Let be a meromorphic function on a simply connected domain , and let be a closed contour in enclosing a bounded region anticlockwise, and avoiding all the singularities of . Show thatwhere is summed over all the poles of that lie in .

The residue theorem is particularly useful when applied to logarithmic derivatives of meromorphic functions , because the residue is of a specific form:

Exercise 14Let be a meromorphic function on an open set that does not vanish identically. Show that the only poles of are simple poles (poles of order ), occurring at the poles and zeroes of (after all removable singularities have been removed). Furthermore, the residue of at a pole is an integer, equal to the order of zero of if has a zero at , or equal to negative the order of pole at if has a pole at .

Remark 15The fact that residues of logarithmic derivatives of meromorphic functions are automatically integers is a remarkable feature of the complex analytic approach to multiplicative number theory, which is difficult (though not entirely impossible) to duplicate in other approaches to the subject. Here is a sample application of this integrality, which is challenging to reproduce by non-complex-analytic means: if is meromorphic near , and one has the bound as , then must in fact stay bounded near , because the only integer of magnitude less than is zero.

## Recent Comments