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 expansionThen

*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 3Let be a univalent function with Taylor expansionShow 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 5Show 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 thatWhen 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 8Show 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 9The 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 theconformal radiusof 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 thatand 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 these 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

Whenever one has a family of simply connected proper open subsets of containing with for , 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 taht 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 thatfor 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

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 20Let 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 of . 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:* Let be a countable dense subset of . From Lemma 19, the function is Lipschitz continuous, and thus differentiable almost everywhere, for each . Thus there exists a Lebesgue measure zero set such that is differentiable in outside of for each . From the Koebe distortion theorem is also locally Lipschitz (hence locally uniformly equicontinuous) in the variable, so in fact is differentiable in outside of for all . Without loss of generality we may assume contains zero.

Let , and let . Then as approaches from below, we have

uniformly; 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 22The 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 23Let 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 25Show 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

*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, thuswhere 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

Using and telescoping series, it thus suffices to prove the identity

But this follows from observing that

and that

for all .

Exercise 28Show 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 hasfor 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 32If is the Koebe function (1), thenso 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

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

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 33Let .

- (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.

## 13 comments

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2 May, 2018 at 3:17 pm

AnonymousIt 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

AnonymousIt 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 TaoI 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

AulaIn 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 AsimovI always wanted to understand the Bieberbach conjecture. With a 1-page proof this may be possible.

6 May, 2018 at 7:10 am

ConradFor “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 LiangReblogged this on Countable Infinity.

11 May, 2018 at 7:09 am

John MangualI 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 MangualThis 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 SharmaI 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

AnonymousFor 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!