Previous set of notes: Notes 2. Next set of notes: Notes 4.
We now approach conformal maps from yet another perspective. Given an open subset of the complex numbers
, define a univalent function on
to be a holomorphic function
that is also injective. We will primarily be studying this concept in the case when
is the unit disk
.
Clearly, a univalent function on the unit disk is a conformal map from
to the image
; in particular,
is simply connected, and not all of
(since otherwise the inverse map
would violate Liouville’s theorem). In the converse direction, the Riemann mapping theorem tells us that every open simply connected proper subset
of the complex numbers is the image of a univalent function on
. Furthermore, if
contains the origin, then the univalent function
with this image becomes unique once we normalise
and
. Thus the Riemann mapping theorem provides a one-to-one correspondence between open simply connected proper subsets of the complex plane containing the origin, and univalent functions
with
and
. We will focus particular attention on the univalent functions
with the normalisation
and
; such functions will be called schlicht functions.
One basic example of a univalent function on is the Cayley transform
, which is a Möbius transformation from
to the right half-plane
. (The slight variant
is also referred to as the Cayley transform, as is the closely related map
, which maps
to the upper half-plane.) One can square this map to obtain a further univalent function
, which now maps
to the complex numbers with the negative real axis
removed. One can normalise this function to be schlicht to obtain the Koebe function
which now maps to the complex numbers with the half-line
removed. A little more generally, for any
we have the rotated Koebe function
that is a schlicht function that maps to the complex numbers with the half-line
removed.
Every schlicht function has a convergent Taylor expansion
for some complex coefficients with
. For instance, the Koebe function has the expansion
and similarly the rotated Koebe function has the expansion
Intuitively, the Koebe function and its rotations should be the “largest” schlicht functions available. This is formalised by the famous Bieberbach conjecture, which asserts that for any schlicht function, the coefficients should obey the bound
for all
. After a large number of partial results, this conjecture was eventually solved by de Branges; see for instance this survey of Korevaar or this survey of Koepf for a history.
It turns out that to resolve these sorts of questions, it is convenient to restrict attention to schlicht functions that are odd, thus
for all
, and the Taylor expansion now reads
for some complex coefficients with
. One can transform a general schlicht function
to an odd schlicht function
by observing that the function
, after removing the singularity at zero, is a non-zero function that equals
at the origin, and thus (as
is simply connected) has a unique holomorphic square root
that also equals
at the origin. If one then sets
it is not difficult to verify that is an odd schlicht function which additionally obeys the equation
Conversely, given an odd schlicht function , the formula (4) uniquely determines a schlicht function
.
For instance, if is the Koebe function (1),
becomes
which maps to the complex numbers with two slits
removed, and if
is the rotated Koebe function (2),
becomes
De Branges established the Bieberbach conjecture by first proving an analogous conjecture for odd schlicht functions known as Robertson’s conjecture. More precisely, we have
Theorem 1 (de Branges’ theorem) Let
be a natural number.
- (i) (Robertson conjecture) If
is an odd schlicht function, then
- (ii) (Bieberbach conjecture) If
is a schlicht function, then
It is easy to see that the Robertson conjecture for a given value of implies the Bieberbach conjecture for the same value of
. Indeed, if
is schlicht, and
is the odd schlicht function given by (3), then from extracting the
coefficient of (4) we obtain a formula
for the coefficients of in terms of the coefficients of
. Applying the Cauchy-Schwarz inequality, we derive the Bieberbach conjecture for this value of
from the Robertson conjecture for the same value of
. We remark that Littlewood and Paley had conjectured a stronger form
of Robertson’s conjecture, but this was disproved for
by Fekete and Szegö.
To prove the Robertson and Bieberbach conjectures, one first takes a logarithm and deduces both conjectures from a similar conjecture about the Taylor coefficients of , known as the Milin conjecture. Next, one continuously enlarges the image
of the schlicht function to cover all of
; done properly, this places the schlicht function
as the initial function
in a sequence
of univalent maps
known as a Loewner chain. The functions
obey a useful differential equation known as the Loewner equation, that involves an unspecified forcing term
(or
, in the case that the image is a slit domain) coming from the boundary; this in turn gives useful differential equations for the Taylor coefficients of
,
, or
. After some elementary calculus manipulations to “integrate” this equations, the Bieberbach, Robertson, and Milin conjectures are then reduced to establishing the non-negativity of a certain explicit hypergeometric function, which is non-trivial to prove (and will not be done here, except for small values of
) but for which several proofs exist in the literature.
The theory of Loewner chains subsequently became fundamental to a more recent topic in complex analysis, that of the Schramm-Loewner equation (SLE), which is the focus of the next and final set of notes.
— 1. The area theorem and its consequences —
We begin with the area theorem of Grönwall.
Theorem 2 (Grönwall area theorem) Let
be a univalent function with a convergent Laurent expansion
Then
Proof: By shifting we may normalise
. By hypothesis we have
for any
; by replacing
with
and using a limiting argument, we may assume without loss of generality that the
have some exponential decay as
(in order to justify some of the manipulations below).
Let be a large parameter. If
, then
and
. The area enclosed by the simple curve
is equal to
crucially, the error term here goes to zero as . Meanwhile, by the change of variables formula (using monotone convergence if desired to work in compact subsets of the annulus
initially) and Plancherel’s theorem, the area of the region
is
Comparing these bounds we conclude that
sending to infinity, we obtain the claim.
Exercise 3 Let
be a univalent function with Taylor expansion
Show that the area of
is equal to
. (In particular,
has finite area if and only if
.)
Corollary 4 (Bieberbach inequality)
- (i) If
is an odd schlicht function, then
.
- (ii) If
is a schlicht function, then
.
Proof: For (i), we apply Theorem 2 to the univalent function defined by
, which has a Laurent expansion
, to give the claim. For (ii), apply part (i) to the square root of
with first term
.
Exercise 5 Show that equality occurs in Corollary 4(i) if and only if
takes the form
for some
, and in Corollary 4(ii) if and only if
takes the form of a rotated Koebe function
for some
.
The Bieberbach inequality can be rescaled to bound the second coefficient of univalent functions:
Exercise 6 (Rescaled Bieberbach inequality) If
is a univalent function, show that
When does equality hold?
The Bieberbach inequality gives a useful lower bound for the image of a univalent function, known as the Koebe quarter theorem:
Corollary 7 (Koebe quarter theorem) Let
be a univalent function. Then
contains the disk
.
Proof: By applying a translation and rescaling, we may assume without loss of generality that is a schlicht function, with Taylor expansion
Our task is now to show that for every , the equation
has a solution in
. If this were not the case, then the function
is invertible on
, with inverse being univalent and having the Taylor expansion
Applying Exercise 6 we then have
while from the Bieberbach inequality one also has . Hence by the triangle inequality
, which is incompatible with the hypothesis
.
Exercise 8 Show that the radius
is best possible in Corollary 7 (thus,
does not contain any disk
with
) if and only if
takes the form
for some complex numbers
and real
.
Remark 9 The univalence hypothesis is crucial in the Koebe quarter theorem. Consider for instance the functions
defined by
. These are locally univalent functions (since
is holomorphic with non-zero derivative) and
,
, but
avoids the point
.
Exercise 10 (Koebe distortion theorem) Let
be a schlicht function, and let
have magnitude
.
- (i) Show that
(Hint: compose
on the right with a Möbius automorphism of
that sends
to
and then apply the rescaled Bieberbach inequality.)
- (ii) Show that
(Hint: use (i) to control the radial derivative of
.)
- (iii) Show that
- (iv) Show that
(This cannot be directly derived from (ii) and (iii). Instead, compose
on the right with a Mobius automorphism that sends
to
and
to
, rescale it to be schlicht, and apply (iii) to this function at
.)
- (v) Show that the space of schlicht functions is a normal family. In other words, if
is any sequence of schlicht functions, then there is a subsequence
that converges locally uniformly on compact sets.
- (vi) (Qualitative Bieberbach conjecture) Show that for each natural number
there is a constant
such that
whenever
is a schlicht function with Taylor expansion
Exercise 11 (Conformal radius) If
is a non-empty simply connected open subset of
that is not all of
, and
is a point in
, define the conformal radius of
at
to be the quantity
, where
is any conformal map from
to
that maps
to
(the existence and uniqueness of this radius follows from the Riemann mapping theorem). Thus for instance the disk
has conformal radius
around
.
- (i) Show that the conformal radius is strictly monotone in
: if
are non-empty simply connected open subsets of
, and
, then the conformal radius of
around
is strictly greater than that of
.
- (ii) Show that the conformal radius of a disk
around an element
of the disk is given by the formula
.
- (iii) Show that the conformal radius of
around
lies between
and
, where
is the radius of the maximal disk
that is contained in
.
- (iv) If the conformal radius of
around
is equal to
, show that for all sufficiently small
, the ring domain
has modulus
, where
denotes a quantity that goes to zero as
, and the modulus of a ring domain was defined in Notes 2.
We can use the distortion theorem to obtain a nice criterion for when univalent maps converge to a given limit, known as the Carathéodory kernel theorem.
Theorem 12 (Carathéodory kernel theorem) Let
be a sequence of simply connected open proper subsets of
containing the origin, and let
be a further simply connected open proper subset of
containing
. Let
and
be the conformal maps with
and
(the existence and uniqueness of these maps are given by the Riemann mapping theorem). Then the following are equivalent:
- (i)
converges locally uniformly on compact sets to
.
- (ii) For every subsequence
of the
,
is the set of all
such that there is an open connected set containing
and
that is contained in
for all sufficiently large
.
If conclusion (ii) holds, is known as the kernel of the domains
.
Proof: Suppose first that converges locally uniformly on compact sets to
. If
, then
for some
. If
, then the holomorphic functions
converge uniformly on
to the function
, which is not identically zero but has a zero in
. By Hurwitz’s theorem we conclude that
also has a zero in
for all sufficiently large
; indeed the same argument shows that one can replace
by any element of a small neighbourhood of
to obtain the same conclusion, uniformly in
. From compactness we conclude that for sufficiently large
,
has a zero in
for all
, thus
for sufficiently large
. Since
is open connected and contains
and
, we see that
is contained in the set described in (ii).
Conversely, suppose that is a subsequence of the
and
is such that there is an open connected set
containing
and
that is contained in
for sufficiently large
. The inverse maps
are holomorphic and bounded, hence form a normal family by Montel’s theorem. By refining the subsequence we may thus assume that the
converge locally uniformly to a holomorphic limit
. The function
takes values in
, but by the open mapping theorem it must in fact map to
. In particular,
. Since
converges to
, and
converges locally uniformly to
, we conclude that
converges to
, thus
and hence
. This establishes the derivation of (ii) from (i).
Now suppose that (ii) holds. It suffices to show that every subsequence of has a further subsequence that converges locally uniformly on compact sets to
(this is an instance of the Urysohn subsequence principle). Then (as
contains
) in particular there is a disk
that is contained in the
for all sufficiently large
; on the other hand, as
is not all of
, there is also a disk
which is not contained in the
for all sufficiently large
. By Exercise 11, this implies that the conformal radii of the
around zero is bounded above and below, thus
is bounded above and below.
By Exercise 10(v), and rescaling, the functions then form a normal family, thus there is a subsequence
of the
that converges locally uniformly on compact sets to some limit
. Since
is positive and bounded away from zero,
is also positive, so
is non-constant. By Hurwitz’s theorem,
is therefore also univalent, and thus maps
to some region
. By the implication of (ii) from (i) (with
replaced by
) we conclude that
is the set of all
such that there is an open connected set containing
and
that is contained in
for all sufficiently large
; but by hypothesis, this set is also
. Thus
, and then by the uniqueness part of the Riemann mapping theorem,
as desired.
The condition in Theorem 12(ii) indicates that “converges” to
in a rather complicated sense, in which large parts of
are allowed to be “pinched off” from
and disappear in the limit. This is illustrated in the following explicit example:
Exercise 13 (Explicit example of kernel convergence) Let
be the function from (5), thus
is a univalent function from
to
with the two vertical rays from
to
, and from
to
, removed. For any natural number
, let
and let
, and define the transformed functions
.
- (i) Show that
is a univalent function from
to
with the two vertical rays from
to
, and from
to
, removed, and that
and
.
- (ii) Show that
converges locally uniformly to the function
, and that this latter map is a univalent map from
to the half-plane
. (Hint: one does not need to compute everything exactly; for instance, any terms of the form
can be written using the
notation instead of expanded explicitly.)
- (iii) Explain why these facts are consistent with the Carathéodory kernel theorem.
As another illustration of the theorem, let be two distinct convex open proper subsets of
containing the origin, and let
be the associated conformal maps from
to
respectively with
and
. Then the alternating sequence
does not converge locally uniformly to any limit. The set
is the set of all points that lie in a connected open set containing the origin that eventually is contained in the sequence
; but if one passes to the subsequence
, this set of points enlarges to
, and so the sequence
does not in fact have a kernel.
However, the kernel theorem simplifies significantly when the are monotone increasing, which is already an important special case:
Corollary 14 (Monotone increasing case of kernel theorem) Let the notation and assumptions be as in Theorem 12. Assume furthermore that
and that
. Then
converges locally uniformly on compact sets to
.
Loewner observed that the kernel theorem can be used to approximate univalent functions by functions mapping into slit domains. More precisely, define a slit domain to be an open simply connected subset of formed by deleting a half-infinite Jordan curve
connecting some finite point
to infinity; for instance, the image
of the Koebe function is a slit domain.
Theorem 15 (Loewner approximation theorem) Let
be a univalent function. Then there exists a sequence
of univalent functions whose images
are slit domains, and which converge locally uniformly on compact subsets to
.
Proof: First suppose that extends to a univalent function on a slightly larger disk
for some
. Then the image
of the unit circle is a Jordan curve enclosing the region
in the interior. Applying the Jordan curve theorem (and the Möbius inversion
), one can find a half-infinite Jordan curve
from
to infinity that stays outside of
. For any
, one can concatenate this curve with the arc
to obtain another half-infinite Jordan curve
, whose complement
is a slit domain which has
as kernel (why?). If we let
be the conformal maps from
to
with
and
, we conclude from the Carathéodory kernel theorem that
converges locally uniformly on compact sets to
.
If is just univalent on
, then it is the locally uniform limit of the dilations
, which are univalent on the slightly larger disks
. By the previous arguments, each
is in turn the locally uniform limit of univalent functions whose images are slit domains, and the claim now follows from a diagonalisation argument.
— 2. Loewner chains —
The material in this section is based on lecture notes of Contreras.
An important tool in analysing univalent functions is to study one-parameter families of univalent functions, parameterised by a time parameter
, in which the images
are increasing in
; roughly speaking, these families allow one to study an arbitrary univalent function
by “integrating” along such a family from
back to
. Traditionally, we normalise these families into (radial) Loewner chains, which we now define:
Definition 16 (Loewner chain) A (radial) Loewner chain is a family
of univalent maps
with
and
(so in particular
is schlicht), such that
for all
. (In these notes we use the prime notation exclusively for differentiation in the
variable; we will use
later for differentiation in the
variable.)
A key example of a Loewner chain is the family
of dilated Koebe functions; note that the image of each
is the slit domain
, which is clearly monotone increasing in
. More generally, we have the rotated Koebe chains
for any real .
Whenever one has a family of simply connected proper open subsets of
containing
with
for
, we can use the Riemann mapping theorem to uniquely define univalent functions
with
,
, and
. By definition,
is then the conformal radius of
around
, which is a strictly increasing function of
by Exercise 11. If this conformal radius
is equal to
at
and increases continuously to infinity as
, then one can reparameterise the
variable so that
, at which point one obtains a Loewner chain.
From the Koebe quarter theorem we see that each image in a Loewner chain contains the disk
. In particular the
increase to fill out all of
:
.
Let be a Loewner chain, Let
. The relation
is sometimes expressed as the assertion that
is subordinate to
. It has the consequence that one has a composition law of the form
for a univalent function , uniquely defined as
, noting that
is well-defined on
. By construction, we have
and
as well as the composition laws
for . We will refer to the
as transition functions.
From the Schwarz lemma, we have
for , with strict inequality when
. In particular, if we introduce the function
for and
, then (after removing the singularity at infinity and using (10)) we see that
is a holomorphic map to the right half-plane
, normalised so that
Define a Herglotz function to be a holomorphic function , thus
is a Herglotz function for all
. A key family of examples of a Herglotz function are the Möbius transforms
for
. In fact, all other Herglotz functions are basically just averages of this one:
Exercise 17 (Herglotz representation theorem) Let
be a Herglotz function, normalised so that
.
- (i) For any
, show that
for
. (Hint: The real part of
is harmonic, and so has a Poisson kernel representation. Alternatively, one can use a Taylor expansion of
.)
- (ii) Show that there exists a (Radon) probability measure
on
such that
for all
. (One will need a measure-theoretic tool such as Prokhorov’s theorem, the Riesz representation theorem, or the Helly selection principle.) Conversely, show that every probability measure
on
generates a Herglotz function
with
by the above formula.
- (iii) Show that the measure
constructed on (ii) is unique.
This has a useful corollary, namely a version of the Harnack inequality:
Exercise 18 (Harnack inequality) Let
be a Herglotz function, normalised so that
. Show that
for all
.
This gives some useful Lipschitz regularity properties of the transition functions and univalent functions
in the
variable:
Lemma 19 (Lipschitz regularity) Let
be a compact subset of
, and let
. Use
to denote a quantity bounded in magnitude by
, where
depends only on
.
- (i) For any
and
, one has
- (ii) For any
and
, one has
One can make the bounds much more explicit if desired (see e.g. Lemma 2.3 of these notes of Contreras), but for our purposes any Lipschitz bound will suffice.
Proof: To prove (i), it suffices from (11) and the Schwarz-Pick lemma (Exercise 13 from Notes 2) to establish this claim when . We can also assume that
since the claim is trivial when
. From the Harnack inequality one has
for , which by (12) and some computation gives
giving the claim.
Now we prove (ii). We may assume without loss of generality that is convex. From Exercise 10 (normalising
to be schlicht) we see that
for
, and hence
has a Lipschitz constant of
on
. Since
, the claim now follows from (13).
As a first application of this we show that every schlicht function starts a Loewner chain.
Lemma 20 Let
be schlicht. Then there exists a Loewner chain
with
.
Proof: This will be similar to the proof of Theorem 15. First suppose that extends to be univalent on
for some
, then
is a Jordan curve. Then by Carathéodory’s theorem (Theorem 20 of Notes 2) (and the Möbius inversion
) one can find a conformal map
from the exterior of
to the exterior of
that sends infinity to infinity. If we define
for
to be the region enclosed by the Jordan curve
, then the
are increasing in
with conformal radius going to infinity as
. If one sets
to be the conformal maps with
and
, then
(by the uniqueness of Riemann mapping) and by the Carathéodory kernel theorem,
converges locally uniformly to
as
. In particular, the conformal radii
are continuous in
. Reparameterising in
one can then obtain the required Loewner chain.
Now suppose is only univalent on
. As in the proof of Theorem 15, one can express
as the locally uniform limit of schlicht functions
, each of which extends univalently to some larger disk
. By the preceding discussion, each of the
extends to a Loewner chain
. From the Lipschitz bounds (and the Koebe distortion theorem) one sees that these chains are locally uniformly equicontinuous in
and
, uniformly in
, and hence by Arzela-Ascoli we can pass to a subsequence that converges locally uniformly in
to a limit
; one can also assume that the transition functions
converge locally uniformly to limits
. It is then not difficult by Hurwitz theorem to verify the limiting relations (9), (11), and that
is a Loewner chain with
as desired.
Suppose that are close to each other:
. Then one heuristically has the approximations
and hence by (12) and some rearranging
and hence on applying , (9), and the Newton approximation
This suggests that the should obey the Loewner equation
for some Herglotz function . This is essentially the case:
Theorem 21 (Loewner equation) Let
be a Loewner chain. Then, for
outside of an exceptional set
of Lebesgue measure zero, the functions
are differentiable in time for each
, and obey the equation (14) for all
and
, and some Herglotz function
for each
with
. Furthermore, the maps
are measurable for every
.
Proof: From Lemma 19, the function is Lipschitz continuous, and thus differentiable almost everywhere, for each
. Here we run into a technical issue in that the exceptional set of times
in which differentiability fails can depend on
, and there are uncountably many such
‘s. However one can avoid this issue using the Cauchy integral formula as follows. For any fixed
and
in a compact range
, the functions
are uniformly bounded on
by the Koebe distortion theorem, and by Fubini’s theorem we have for almost all
in this range that the
are differentiable at
for almost all
. By the Cauchy integral formula and dominated convergence we then conclude that for all such times
, we in fact have
differentiable at
for every
. Sending
to
, we can thus find a universal measure zero set
of times
, outside of which
is differentiable for all
. The argument in fact shows that this differentiability is locally uniform in
.
Let , and let
. Then as
approaches
from below, we have
locally uniformly in ; from (9) and Newton approximation we thus have
which implies that
Also we have
and hence by (12)
Taking limits, we see that the function is Herglotz with
, giving the claim. It is also easy to verify the measurability (because derivatives of Lipschitz functions are measurable).
Example 22 The Loewner chain (7) solves the Loewner equation with the Herglotz function
. With the rotated Koebe chains (8), we instead have
.
Although we will not need it in this set of notes, there is also a converse implication that for every family of Herglotz functions depending measurably on
, one can associate a Loewner chain.
Let us now Taylor expand a Loewner chain at each time
as
as , we have
. As
is differentiable in almost every
for each
, and is locally uniformly continuous in
, we see from the Cauchy integral formulae that the
are also differentiable almost everywhere in
. If we similarly write
for all outside of
, then
, and we obtain the equations
and so forth. For instance, for the Loewner chain (7) one can verify that and
for
solve these equations. For (8) one instead has
and
.
We have the following bounds on the first few coefficients of :
Exercise 23 Let
be a Herglotz function with
. Let
be the measure coming from the Herglotz representation theorem.
- (i) Show that
for all
. In particular,
for all
. Use this to give an alternate proof of the upper bound in the Harnack inequality.
- (ii) Show that
.
We can use this to establish the first two cases of the Bieberbach conjecture:
Theorem 24 (
cases of Bieberbach) If
is schlicht, then
and
.
The bound is not new, and indeed was implicitly used many times in the above arguments, but we include it to illustrate the use of the equations (15), (16).
Proof: By Lemma 20, we can write (and
) for some Loewner chain
.
We can write (15) as . On the other hand, from the Koebe distortion theorem applied to the schlicht functions
, we have
, so in particular
goes to zero at infinity. We can integrate from
to infinity to obtain
From Harnack’s inequality we have , giving the required bound
.
In a similar vein, writing (16) as
we obtain
As , we may integrate from
to infinity to obtain the identity
Taking real parts using Exercise 23(ii) and (17), we have
Since , we thus have
where . By Cauchy-Schwarz, we have
, and from the bound
, we thus have
Replacing by the schlicht function
(which rotates
by
) and optimising in
, we obtain the claim
.
Exercise 25 Show that equality in the above bound
is only attained when
is a rotated Koebe function.
The Loewner equation (14) takes a special form in the case of slit domains. Indeed, let be a slit domain not containing the origin, with conformal radius
around
, and let
be the Loewner chain with
. We can parameterise
so that the sets
have conformal radius
around
for every
, in which case we see that
must be the unique conformal map from
to
with
and
. For instance, for the chain (7) we would have
.
Theorem 26 (Loewner equation for slit domains) In the above situation, we have the Loewner equation holding with
for almost all
and some measurable
.
Proof: Let be a time where the Loewner equation holds. For
, the function
extends continuously to the boundary, and is two-to-one on the split
, except at the tip
where there is a single preimage
on the unit circle; this can be seen by taking a holomorphic square root of
, using a Möbius transformation to map the resulting image to a set bounded by a Jordan curve, and applying Carathéodory’s theorem (Theorem 20 from Notes 2) to the resulting conformal map. The image
is then
with a Jordan arc
removed, where
is a point on the boundary of the sphere. Applying Carathéodory’s theorem to a holomorphic square root of
, we see that
extends continuously to be a map from
to
, with an arc
on the boundary mapping (in two-to-one fashion) to the arc
, and the endpoints of this arc mapping to
. From this and (12), we see that
converges to zero outside of the arc
, which by the Herglotz representation theorem implies that the measure
associated to
is supported on the arc
. An inspection of the proof of Carathéodory’s theorem also reveals that the
are equicontinuous on
as
, and thus converge uniformly to
(which is the identity function) as
. This implies that
must converge to the point
as
approaches
, and so
converges vaguely to the Dirac mass at
. Since
converges locally uniformly to
, we conclude the formula (18). As
depends measurably in
, we conclude that
does also.
In fact one can show that extends to a continuous function
, and that the Loewner equation holds for all
, but this is a bit trickier to show (it requires some further distortion estimates on conformal maps, related to the arguments used to prove Carathéodory’s theorem in the previous notes) and will not be done here. One can think of the function
as “driving force” that incrementally enlarges the slit via the Loewner equation; this perspective is often used when studying the Schramm-Loewner evolution, which is the topic of the next (and final) set of notes.
— 3. The Bieberbach conjecture —
We now turn to the resolution of the Bieberbach (and Robertson) conjectures. We follow the simplified treatment of de Branges’ original proof, due to FitzGerald and Pommerenke, though we omit the proof of one key ingredient, namely the non-negativity of a certain hypergeometric function.
The first step is to work not with the Taylor coefficients of a schlicht function or with an odd schlicht function
, but rather with the (normalised) logarithm
of a schlicht function
, as the coefficients end up obeying more tractable equations. To transfer to this setting we need the following elementary inequalities relating the coefficients of a power series with the coefficients of its exponential.
Lemma 27 (Second Lebedev-Milin inequality) Let
be a formal power series with complex coefficients and no constant term, and let
be its formal exponential, thus
where
is the formal series
. Then for any
, one has
Proof: If we formally differentiate (19) in , we obtain the identity
extracting the coefficient for any
, we obtain the formula
By Cauchy-Schwarz, we thus have
which we can rearrange as
Using and telescoping series, it thus suffices to prove the identity
But this follows from observing that
and that
for all .
Exercise 28 Show that equality holds in (20) for a given
if and only if there is
such that
for all
.
Exercise 29 (First Lebedev-Milin inequality) With the notation as in the above lemma, and under the additional assumption
, prove that
(Hint: using the Cauchy-Schwarz inequality as above, first show that the power series
is bounded term-by-term by the power series of
.) When does equality occur?
Exercise 30 (Third Lebedev-Milin inequality) With the notation as in the above lemma, show that
(Hint: use the second Lebedev-Milin inequality and (21), together with the calculus inequality
for all
.) When does equality occur?
Using these inequalities, one can reduce the Robertson and Bieberbach conjectures to the following conjecture of Milin, also proven by de Branges:
Theorem 31 (Milin conjecture) Let
be a schlicht function. Let
be the branch of the logarithm of
that equals
at the origin, thus one has
for some complex coefficients
. Then one has
for all
.
Indeed, if
is an odd schlicht function, let be the schlicht function given by (4), then
Applying Lemma 27 with , we obtain the Robertson conjecture, and the Bieberbach conjecture follows.
Example 32 If
is the Koebe function (1), then
so in this case
and
. Similarly, for the rotated Koebe function (2) one has
and again
. If one works instead with the dilated Koebe function
, we have
, thus the time parameter only affects the constant term in
. This is already a hint that the coefficients of
could be worth studying further in this problem.
To prove the Milin conjecture, we use the Loewner chain method. It suffices by Theorem 15 and a limiting argument to do so in the case that is a slit domain. Then, by Theorem 26,
is the initial function
of a Loewner chain
that solves the Loewner equation
for all and almost every
, and some function
.
We can transform this into an equation for . Indeed, for non-zero
we may divide by
to obtain
(for any local branch of the logarithm) and hence
Since ,
is equal to
at the origin (for an appropriate branch of the logarithm). Thus we can write
The are locally Lipschitz in
(basically thanks to Lemma 19) and for almost every
we have the Taylor expansions
and
Comparing coefficients, we arrive at the system of ordinary differential equations
for every .
Fix (we will not need to use any induction on
here). We would like to use the system (22) to show that
The most naive attempt to do this would be to show that one has a monotonicity formula
for all , and that the expression
goes to zero as
, as the claim would then follow from the fundamental theorem of calculus. This turns out to not quite work; however it turns out that a slight modification of this idea does work. Namely, we introduce the quantities
where for each ,
is a continuously differentiable function to be chosen later. If we have the initial condition
for all , then the Milin conjecture is equivalent to asking that
. On the other hand, if we impose a boundary condition
for , then we also have
as
, since
is schlicht and hence
is a normal family, implying that the
are bounded in
for each
. Thus, to solve the Milin, Robertson, and Bieberbach conjectures, it suffices to find a choice of weights
obeying the initial and boundary conditions (23), (24), and such that
for almost every (note that
will be Lipschitz, so the fundamental theorem of calculus applies).
Let us now try to establish (25) using (22). We first write , and drop the explicit dependence on
, thus
for . To simplify this equation, we make a further transformation, introducing the functions
(with the convention ); then we can write the above equation as
We can recover the from the
by the formula
It may be worth recalling at this point that in the example of the rotated Koebe Loewner chain (2) one has ,
, and
, for some real constant
. Observe that
has a simpler form than
in this example, suggesting again that the decision to transform the problem to one about the
rather than the
is on the right track.
We now calculate
Conveniently, the unknown function no longer appears explicitly! Some simple algebra shows that
and hence by summation by parts
with the convention .
In the example of the rotated Koebe function, with , the factors
and
both vanish, which is consistent with the fact that
vanishes in this case regardless of the choice of weights
. So these two factors look to be related to each other. On the other hand, for more general choices of
, these two expressions do not have any definite sign. For comparison, the quantity
also vanishes when
, and has a definite sign. So it is natural to see of these three factors are related to each other. After a little bit of experimentation, one eventually discovers the following elementary identity giving such a connection:
Inserting this identity into the above equation, we obtain
which can be rearranged as
We can kill the first summation by fiat, by imposing the requirement that the obey the system of differential equations
for ; then we just have
Hence if we also have the non-negativity condition
for all and
, we will have obtained the desired monotonicity (25).
To summarise, in order to prove the Milin conjecture for a fixed value of , we need to find functions
obeying the initial condition (23), the boundary condition (24), the differential equation (26), and the nonnegativity condition (27), with the convention
. This is a significant reduction to the problem, as one just has to write down an explicit formula for such functions and verify all the properties.
Let us work out some simple cases. First consider the case . Now our task is to solve the system
for all . This is easy: we just take
(indeed this is the unique choice). This gives the
case of the Milin conjecture (which corresponds to the
case of Bieberbach).
Next consider the case . The system is now
Again, a routine computation shows that there is a unique solution here, namely and
. This gives the
case of the Milin conjecture (which corresponds to the
case of Bieberbach). One should compare this argument to that in Theorem 24, in particular one should see very similar weight functions emerging.
Let us now move on to . The system is now
A slightly lengthier calculation gives the unique explicit solution
to the above conditions.
These simple cases already indicate that there is basically only one candidate for the weights that will work. A calculation can give the explicit formula:
Exercise 33 Let
.
- (i) Show there is a unique choice of continuously differentiable functions
that solve the differential equations (26) with initial condition (23), with the convention
. (Use the Picard existence theorem.)
- (ii) For any
, show that the expression
is equal to
when
is even and
when
is odd.
- (iii) Show that the functions
for
obey the properties (23), (26), (24). (Hint: for (23), first use (ii) to show that
is equal to
when
is even and
when
is odd, then use (26).)
The Bieberbach conjecture is then reduced to the claim that
for any and
. This inequality can be directly verified for any fixed
; for general
it follows from general inequalities on Jacobi polynomials by Askey and Gasper, with an alternate proof given subsequently by Gasper. A further proof of (28), based on a variant of the above argument due to Weinstein that avoids explicit use of (28), appears in this article of Koepf. We will not detail these arguments here.
16 comments
Comments feed for this article
2 May, 2018 at 3:17 pm
Anonymous
It seems clearer to add that the simply connected proper subset
(in line 11) should be also open and nonempty. Also “simply connected” here (with respect to complex plane topology) can be extended to be with respect to the Riemann sphere topology.
[Openness hypothesis added. For this set of notes, only univalent functions taking values in the complex numbers (rather than the Riemann sphere) are considered. – T.]
4 May, 2018 at 11:02 am
Anonymous
It is interesting to observe that the method of proof was used to prove the stronger Milin’s conjecture and apparently is not adapted for a direct proof of the weaker Bieberbach conjecture. Is it possible to modify the method for a direct proof of the Bieberbach conjecture?
4 May, 2018 at 1:59 pm
Terence Tao
I am not aware of any confirmed proof of the Bieberbach conjecture that does not go through the Milin conjecture.
5 May, 2018 at 4:17 am
Aula
In the proof of Lemma 27, “it thus suffices the identity” should be “…suffices to prove the identity” or something similar.
5 May, 2018 at 9:05 am
Dan Asimov
I always wanted to understand the Bieberbach conjecture. With a 1-page proof this may be possible.
6 May, 2018 at 7:10 am
Conrad
For “simple” univalent functions (eg those with real coefficients and more generally the typically real functions on the unit disk which may not be univalent – since they have a simple integral representation with respect to a positive unit measure on the circle, or for that matter close to convex and related univalent functions which also have nice integral representations) Bieberbach’s conjecture is fairly easy to deduce; also Littlewood proved fairly easily in the 20’s the inequality a(n)<en so the right order of the magnitude for the coefficients is not that deep a problem either.
However there are some functionals (related to the Milin one(s) that is used in de Branges' proof) on the S class that have different asymptotic behaviors on odd versus even coefficients (see Grinshpan' survey article in Kuhnau's handbook of Geometric Function Theory I, p 313) that hint to the reasons no one so far managed to prove Bieberbach without going through the 2-symmetrization of the S-class that leads to an odd univalent function and the Robertson conjecture which then follows from the negativity of the Milin functional
8 May, 2018 at 12:01 am
Jasper Liang
Reblogged this on Countable Infinity.
11 May, 2018 at 7:09 am
John Mangual
I was never sure why Bierbach conjecture was so important but I always liked Gronwall area formula since it linked complex analysis and fourier series.
Maybe the floor plan of a room — all floor plans of houses — are conformally equivalent. We can partition a rectangle into various rooms connected by doors. And then we randomly walk around. And these are all related.
15 May, 2018 at 1:48 am
John Mangual
This the first time I read that the function you obtain from the Riemann Mapping Theorem is univalent. Can we tell just looking at the series expansion that a holomorphic function is injective? Let alone if it maps to a specific shape…
Here is something nice. Let
. Then we have
it's like we pinch off a tiny bit of the value at the origin with the derivative. And the shape is a Cardioid. This taken from a paper "Topology of Quadrature Domains" https://arxiv.org/abs/1307.0487
19 May, 2018 at 10:33 pm
Vivek Sharma
I know that You are very busy and occupied Sir, but only One question: When are Notes 4 coming?
29 May, 2018 at 9:30 am
246C notes 4: Brownian motion, conformal invariance, and SLE | What's new
[…] can help gain an intuitive understanding of the Carathéodory kernel theorem (Theorem 12 from Notes 3). Consider for instance the example in Exercise 13 from those notes. It is intuitively clear that a […]
31 May, 2018 at 7:31 pm
Anonymous
For Exercise 18, is the presence of an ‘s’ a typo? If not, what does it refer to?
[This is a typo, now removed. -T]
27 June, 2018 at 2:54 am
U.V. S.
Does anybody else remember the discussion of Herglotz vs. Airy by the wonderful Prof. Bender? I confess I cannot read a publication mentioning Herglotz functions seriously since that day!
27 February, 2021 at 9:48 am
246C notes 2: Circle packings, conformal maps, and quasiconformal maps | What's new
[…] set of notes: Notes 1. Next set of notes: Notes 3. We now leave the topic of Riemann surfaces, and turn now to the (loosely related) topic of […]
23 May, 2022 at 12:38 pm
Anonymous
Link to “lecture note of Contreras” doesn’t work.
[Link revised – T.]
25 May, 2022 at 9:44 pm
Anonymous
Dr. Tao:
are only functions of time, shouldn’t (15) (16) using regular derivatives
instead of
?
Since the coefficients
[I am reserving primes in this post for the complex derivative in the
variable, and am avoiding it for the
variable to reduce confusion – T.]