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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
Exercise 1 Letbe 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) Letbe 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
Exercise 3 Use (6) to give another proof of Liouville’s theorem: a bounded holomorphic functionon the entire complex plane is necessarily constant.
Exercise 4 Use Jensen’s formula to prove the fundamental theorem of algebra: a complex polynomialof 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) Letbe 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 class for the unit disk
.)
Exercise 7 (Smoothed out Jensen formula) Letbe 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 8 Letbe 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) Letbe 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
Suppose has a zero
with
. If we factor
, where
is the Blaschke product (5), then
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
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 thatfor 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) Letdenote 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 13 The factorisation (iii) can be refined further, withbeing 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) Letbe 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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