©2025 Ada Academica https://adac.eeEur. J. Math. Anal. 5 (2025) 4doi: 10.28924/ada/ma.5.4
Correspondences Among Inner Functions, Functions With Non-Negative Real Parts and

Conformal Mappings

Ronen Peretz
Department of Mathematics, Ben Gurion University of the Negev, Beer-Sheva, 84105, Israelmensahyaogan2@gmailcom
Abstract. We study an interesting family of dynamical systems on the set of the singular innerfunctions (defined on the unit disk). Starting with an inner function S0(z), we obtain new singularinner functions S1(z), S2(z), . . .. This sequence converges to a holomorphic self-map of the unit diskwhich we call S. The convergence is proved with the aid of a fixed-point theorem, a special caseof the Earle-Hamilton Theorem. The function S itself is not a singular inner function as zS(z) isa conformal map. This conformal map has the surprising property that its inverse (which is a prioridefined on a proper subset of the disk) extends to the entire disk. The motivating question for thisresearch is whether z times a singular inner function can have an omitted value in the unit disk. Thisquestion appears within a book on the Krzyz problem written by the author. This question is stillopen.

1. Correspondences that involve inner functions
Let us recall few correspondences that involve inner functions in H2(U). We denote the unitdisk in C by U . We will denote the (multiplicative) group of inner functions in H∞(U) by Inn.Its subgroup which contains all the singular inner functions will be denoted by SInn. We followthe notations in, [2]. Finally, the (additive) group of holomorphic functions in U which have non-negative real parts and which have finite radial limits almost everywhere on T which are purelyimaginary will be denoted by RP. Later on we will add one condition to this definition of RP butfor now this definition suffices. Here are a few elementary facts that are well known:(1) f ∈ RP ⇔ ∃w ∈ Inn such that f = 1+w

1−w .This is a bijection since
f =
1 + w

1− w ⇔ f · (1− w) = 1 + w ⇔ w · (f + 1) = f − 1⇔ w =
f − 1
f + 1

.

(2) g ∈ SInn⇔ ∃ f ∈ RP such that g = exp(−f ).The correspondence RP→ SInn, f → g is not one-to-one. The kernel is 2πiZ.
Received: 13 Aug 2024.
Key words and phrases. singular inner functions; conformal mapping; Krzyz problem; complex dynamical system;Earle-Hamilton fixed-point theorem; Löwner equation. 1

https://adac.ee
https://doi.org/10.28924/ada/ma.5.4
https://orcid.org/0000-0002-6321-479X


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 2(3) g ∈ SInn⇔ ∃w ∈ Inn such that g = exp (−1+w1−w ).The correspondence Inn→ SInn, w → g is not one-to-one.
1 + w

1− w + 2πik =
1 + v

1− v ⇔ v =

(
1 + w

1− w + 2πik − 1
)/(

1 + w

1− w + 2πik + 1
)
=

=
πik + (1− πik)w
(1 + πik)− πikw .So if we denote by φk(z) the fractional linear function

φk(z) =
πik + (1− πik)z
(1 + πik)− πikzand if we denote M : Inn→ SInn, M(w) = g where

M(w) = exp

(
−
1 + w

1− w

)
,

then M−1(g) = {φk(w) | k ∈ Z}.
2. An example of our construction

It will be convenient to first demonstrate the construction on a particular case where concretecomputations are possible. This construction was motivated by a problem that appeared in thebook, [4]: let S(z) be a singular inner function (S ∈ SInn). Is it true that the inner function
z · S(z) is a surjection U → U?

Theorem 2.1. Let {Sn(z)}∞n=0 be a sequence of singular inner functions defined recursively by:
S0 ∈ SInn (an arbitrary initial point),

Sn+1(z) = exp

(
−
1 + z · Sn(z)
1− z · Sn(z)

)
for n ∈ Z≥0.

Then limn→∞ Sn = S uniformly on compact subsets of U . S(z) is in H∞(U) and it satisfies the
fixed-point equation

S = exp

(
−
1 + z · S
1− z · S

)
.

Also the mapping z · S(z) ∈ H∞(U) is injective U → Im(z · S(z)) ⊂ U but it can not be an inner
function.

Proof.Since S0 ∈ SInn and since an inductive argument shows that if Sn ∈ SInn then
Sn+1 = exp

(
−
1 + z · Sn
1− z · Sn

)
∈ SInn for n ∈ Z≥0,

it follows that the sequence {Sn(z)}∞n=0 is a sequence of singular inner functions. The family offunctions in the sequence is a normal family. Even more, for a fixed-point z ∈ U the function of
t ∈ U given by

exp

(
−
1 + z · t
1− z · t

)
,

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 3is a contraction and so by the fixed-point theorem of S. Banach iterations of this contractionconverge to a unique fixed-point S(z). So limn→∞ Sn = S uniformly on compact subsets of U , and
S(z) satisfies the fixed-point equation

S(z) = exp

(
−
1 + z · S(z)
1− z · S(z)

)
.

Clearly S(z) is a non-vanishing function in H∞(U). Next, let us consider the following holomorphicfunction of w , defined on the once punctured plane as follows:
f : C− {1} → C, f(w) = w exp

(
1+w
1−w

)
.

Then by the fixed-point equation satisfied by S(z) we get f (z · S(z)) = z . So f is a left inverseof z · S(z) and hence z · S(z) : U → Im(z · S(z)) is an injection. More concretely, if we denote
g(z) = z · S(z) then the assumption g(z1) = g(z2) implies that z1 = f (g(z1)) = f (g(z2)) = z2.Since S(z) can not be a constant function (by the fixed-point equation), z · S(z) can not be aninner function (see [3], remarked by Raymond Mortini). �

3. A generalization
Definition 3.1. We will denote by RP, the family of all the F ∈ H(U), that satisfy the followingfour conditions:(i) <F (z) ≥ 0, ∀ z ∈ U .(ii) <F (e iθ) = 0 almost everywhere on T with respect to the Lebesgue measure on T.(iii) The function of t ∈ U given by exp (−F (z · t)) is a contraction (with respect to the Euclideanmetric) where z ∈ U is fixed.(iv) F is a non-constant function.
Remark 3.2. ∀F,G ∈ RP, ∀ a, b ∈ R≥0, such that 0 < a + b ≤ 1, we have a · F + b · G ∈ RP.Also if F (z) = 1+w(z)

1−w(z) , where, as always w(z) ∈ Inn, then F ′(z) = 2w ′(z)
(1−w(z))2 . By

d

dt
exp (−F (z · t)) = −zF ′(z · t) exp (−F (z · t)) ,

it follows by (iii) ∣∣∣∣z 2w ′(z · t)
(1− w(z · t))2 exp (−F (z · t))

∣∣∣∣ ≤ c < 1.In particular we obtain that the generating inner function w(z) of F (z) satisfies:
|z |

|w ′(z · t)|
|1− w(z · t)|2 exp

(
−
1− |w(z · t)|2

|1− w(z · t)|2

)
≤
c

2
<
1

2
.

Thus we conclude that ∀ z ∈ U and ∀w ∈ Inn, such that 1+w1−w ∈ RP we have:
|z |

|w ′(z)|
|1− w(z)|2 exp

(
−
1− |w(z)|2

|1− w(z)|2

)
<
1

2
.

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 4

The construction. Let F ∈ RP. We will use F (z) to define a sequence {Sn(z)}∞n=0 of singular innerfunctions. The definition will use the following recursion: S0 ∈ SInn (an arbitrary initial point).
Sn+1(z) = exp (−F (z · Sn(z))) for n ∈ Z≥0.

Theorem 3.3. The limit limn→∞ Sn(z) = S(z) exists and is uniform on compact subsets of U .
S ∈ H(U) satisfies |S(z)| ≤ 1∀ z ∈ U , and satisfies the following fixed-point equation, S =
exp(−F (z · S)).
The function z · S(z) ∈ H∞(U) is a conformal mapping z · S(z) : U → Im(z · S) ⊆ U but it is not
an inner function.

Proof.Since S0 ∈ SInn and since an inductive argument shows that if Sn ∈ SInn, then Sn+1 = exp(−F (z ·
Sn)) ∈ SInn for n ∈ Z≥0, it follows that all the members of the sequence {Sn}∞n=0 belong to
SInn. The reason for the validity of the inductive argument is that |z · Sn| = |z ||Sn| ≤ |z | forall z ∈ U , using the induction hypothesis Sn ∈ SInn. Thus z · Sn ∈ BH∞ , the unit ball of H∞.Also |e iθ · Sn(e iθ)| = 1 almost everywhere on T with respect to the Lebesgue measure on T.Also this follows by the induction hypothesis on Sn. Hence <F (z · Sn(z)) ≥ 0 ∀ z ∈ U and
< (e iθ · Sn(e iθ)) = 0 almost everywhere on T (recall that F ∈ RP satisfies by the definition
<F (e iθ) = 0 almost everywhere on T). Hence

| exp (−F (z · Sn(z))) | = exp (−<F (z · Sn(z))) ≤ 1 ∀ z ∈ Uand also ∣∣exp (−F (e iθ · Sn(e iθ)))∣∣ = 1 almost everywhere on T.We just proved that Sn+1(z) = exp (−F (z · Sn(z))) ∈ SInn for n ∈ Z≥0. Hence the family offunctions in the sequence {Sn}∞n=0 is a normal family. Moreover, by condition (iii) in Definition3.1, for a fixed z ∈ U iterations of the function of t ∈ U given by exp(−F (z · t)) converge(by Banach fixed-point theorem ) to a unique fixed-point t0 = S(z). So limn→∞ Sn(z) = S(z)uniformly on compact subsets of U , and S(z) satisfies the fixed-point equation S(z) = exp(−F (z ·
S(z))). Clearly, the H∞(U) function is a non-vanishing function that belongs to the unit ball
BH∞(U). Next, let us consider the following holomorphic function of w , defined on U as follows:
f : U → C, f(w) = w exp(F(w)). Then by the fixed-point equation satisfied by S(z) we get:
f (z · S(z)) = z . The reason is that

f (z · S(z)) = z · S(z) exp(F (z · S(z))) = z · S(z) · S(z)−1 = z.

Thus f is a left is a left inverse of z · S(z) and hence the mapping: z · S(z) : U → Im(z · S(z))is an injection. More concretely, if we denote g(z) = z · S(z) then the assumption g(z1) = g(z2)implies that z1 = f (g(z1)) = f (g(z2)) = z2. The function S(z) can not be a constant function, forif S(z) = e iθ0 , then
e iθ0 = exp

(
−F (e iθ · S(e iθ0))

)
= exp

(
−F (e i(θ+θ0))

)
.

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 5But any function such as F in RP is non-constant by the definition. Since the only injective innerfunctions are Blaschke factors, [3], z · S(z) can not be an inner function (for that would imply that
S(z) is a unimodular constant). �

4. A parametrization of a family of conformal mappings by the functions in RP
We note that we may replace F (z) by any of the functions in the sequence F (z) + 2πiZ. All ofthese are members of RP that will generate S(z) just as F (z) does. However, if G(z)−F (z) 6∈ 2πiZand G(z) (like F (z)) belongs to RP, then exp(−F ) and exp(−G) are different holomorphic functions.Can they share the same fixed-point S(z)? That is, can the following be true?

S(z) = exp (−F (z · S(z))) = exp (−G(z · S(z))) z ∈ U.

By the Permanence Principle for holomorphic functions this holds if and only if G − F ∈ 2πiZ(which is not the case). So the assignment: [F ] := F + 2πiZ → S is an injection of the quotientspace of RP, namely of RP/2πiZ onto the family of functions S[F ](z) in the unit ball of H∞(U),such that z · S[F ](z) : U → Im(z · S[F ](z)) ⊆ U is a conformal mapping, where S[F ] is the usual
F (z) fixed-point: S[F ] = exp(−F (z · S[F ](z))). Here (in the notation of section 3) [F ] stands forany of the members of the equivalence class [F ] in RP/2πiZ. We got our parametrization thatthe title of this section refers to. [F ] determines the conformal mapping z · S[F ] via the fixed-pointequation. So [F ] is the parameter of the conformal mapping z · S[F ](z). If we define the conformalmapping by w = f[F ](z) = z · S[F ](z), then f[F ] : U → Im(f[F ]), is invertible, so that z = f −1

[F ]
(w).Using the fixed-point equation: − log S[F ] = F (z ·S[F ]) we see that S[F ] determines its parameter

[F ] by:
F (w) = − log

(w
z

)
= − log

(
w

f −1(w)

)
.

The family RP is algebraically easy to understand, unlike the family of the conformal mappings:{
S[F ] | [F ] ∈ RP/2πiZ

}
.

For example, see our Remark 3.2: RP is closed for taking linear combinations with coefficients
a, b ∈ R≥0, such that 0 < a+b ≤ 1, i.e. ∀F,G ∈ RP(or RP/2πiZ), a ·F+b ·G ∈ RP(or RP/2πiZ).Geometrically we are dealing with cones. Let (as usual) S[F ] = exp(−F (z · S[F ])), S[G] =
exp(−G(z · S[G])), so that z · S[F ] : U → Im(z · S[F ]), z · S[G] : U → Im(z · S[G]) are con-formal mappings.Then we make the following:
Definition 4.1. ∀ a, b ∈ R≥0, 0 < a+b ≤ 1 we define the conic linear combination by the equation:

a · (z · S[F ])+̂b · (z · S[G]) = z · S[a·F+b·G].

This definition (and a similar one for multiplication by a real non-negative scalar) induces onthe family of our conformal mappings the same conic structure as the one we easily have on RP(or RP/2πiZ).

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 6

Remark 4.2. It is well known that there exist natural and elementary parametrizations between
RP and SInn and also between Inn and SInn. See the explanations in Section 1. Thus, our lesselementary parametrization of the family of conformal mappings z · S[F ](z) by RP/2πiZ can nowbe related to the tree of correspondences among SInn, Inn and RP.
We naturally inquire as to what are the conformal members that form the family of conformalmappings in the tree of correspondences. We will deal with that on the next section. As expecteda main ingredient of these conformal mappings will be the boundary behaviour of the inverseconformal mappings Im(z · S[F ])→ U .

5. The family of conformal mappings CONF
Our family of conformal mappings is clearly given by the following:

Definition 5.1.

CONF =
{
z · S[F ] ∈ H(U)

∣∣ [F ] ∈ RP/2πiZ, S[F](z) = exp (−F(z · S[F](z))) , ∀z ∈ U} .
We recall the following facts:(1) z · S[F ] : U → Im(z · S[F ]) is an injection.(2) z · S[F ] ∈ BH∞(U) − Inn. ( [3]).(3) If w = f[F ](z) = z · S[F ](z), (z ∈ U), then

F (w) = − log
(w
z

)
= − log

(
w

f −1
[F ]
(w)

)
, w ∈ Im(z · S[F ](z)).

We would like to characterize the conformal family CONF without any reference to the family ofthe parameters RP/2πiZ.
Theorem 5.2. The family CONF consists of all the holomorphic functions f (z) ∈ H(U) that satisfy
the following:
(a) f : U → Im(f ) ⊆ U is a conformal mapping.
(b) f (0) = 0.
(c) The function:

F (w) = − log
(

w

f −1(w)

)
, w ∈ Im(f ),

can be analytically be defined on all of U (not just on Im(f )), and it satisfies <F (w) ≥ 0 for all
w ∈ Im(f ) and <F (f −1(w)) = 0 for all w ∈ ∂ Im(f ). (d) For a fixed z ∈ U , the function z ·t

f −1(z ·t)
is a contraction in t ∈ U for which z · t ∈ Im(f ). It is a contraction with respect to the Euclidean
metric.

Proof.Let us denote the family of all the mappings f ∈ H(U) that satisfy (a), (b), (c) and (d) by A. Weneed to prove that CONF = A where the definition of CONF is given in Definition 5.1.

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 7(I) CONF ⊆ A:Let f[F ](z) = z · S[F ](z) ∈ CONF. By fact (1) after Definition 5.1, f[F ] : U → Im(f[F ]) ⊆ U is aninjection. Clearly f[F ](0) = 0. By fact (3) after Definition 5.1,
F (w) = − log

(
w

f −1
[F ]
(w)

)
, ∀w ∈ Im(f ).

Finally, by Definition 3.1 (iii) we have: (iii) The function of t ∈ U given by exp (−F (z · t)) is acontraction (with respect to the Euclidean metric) where z ∈ U is fixed. But exp (−F (z · t)) =
z ·t

f −1(z ·t) which proved (d) and completes the proof of CONF ⊆ A.(II) A ⊆ CONF:Let f : U → Im(f ) ⊆ U be an elelment of A. Then f (0) = 0 and |f (z)| ≤ 1 for all z ∈ U . By theSchwarz Lemma we get |f (z)| ≤ |z | for all z ∈ U . Hence for all w ∈ U where w = f (z) we have∣∣∣w
z

∣∣∣ ≤ 1 so − log ∣∣∣w
z

∣∣∣ ≥ 0.
If we define

Ff (w) = − log
(w
z

)
= − log

(
w

f −1(w)

)
,

then <Ff (w) ≥ 0. Now (ii) and (iii) in Definition 3.1 follow. Hence f ∈ CONF. �

Here is an interesting consequence on the conformal mappings of the family CONF:
Corollary 5.3. Let f , g ∈ CONF and let a, b ∈ R≥0, 0 < a+ b ≤ 1. Then ∃ ha,b ∈ CONF such that
we have the following multiplicative relation among these three conformal mappings:(

h−1a,b(w)

w

)
=

(
f −1(w)

w

)a (
g−1(w)

w

)b
.

Proof.By the proof of Theorem 5.2 it follows that,
f , g ∈ CONF⇔ − log

(
w

f −1(w)

)
,− log

(
w

g−1(w)

)
∈ RP.

We note that in f −1(w) we have w ∈ Im(f ) and in g−1(w) we have w ∈ Im(g). By property (5)after Definition 5.1 we know that the two-dimensional Lebesgue measures of the sets U − Im(f )and U − Im(g) are zero. Hence Im(f ) ∩ Im(g) is an open subset of U and the two-dimensionalLebesgue measure of the set U− (Im(f ) ∩ Im(g)) is zero. So Im(f )∩ Im(g) is a large open subsetof U on which both holomorphic functions
− log

(
w

f −1(w)

)
, and − log

(
w

g−1(w)

)
,

are defined and belong to RP. I.e. these are the restrictions of RP functions to the intersection ofthe images of the conformal mappings f and g. Since:
a

{
− log

(
w

f −1(w)

)}
+ b

{
− log

(
w

g−1(w)

)}
= − log

(
w

f −1(w)

)a ( w

g−1(w)

)b
∈ RP,

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 8it follows by the proof of Theorem 5.2 that ∃ ha,b ∈ CONF such that:(
w

f −1(w)

)a ( w

g−1(w)

)b
=

(
w

h−1a,b(w)

)
.

Our proof is now completed. �

One can conclude more surprising properties on the conformal mappings of the family CONF.
6. The geometry of the image of conformal mappings in CONF

Let F ∈ RP. We chose an arbitrary singular inner function S0(z) ∈ SInn and we gener-ated an infinite sequence of singular inner functions using the following recursion: Sn+1(z) =
exp (−F (z · Sn(z))) for n ∈ Z≥0. The limit S(z) = limn→∞ Sn(z) exists for all z ∈ U . Theconvergence is uniform on compact subsets of U . S(z) satisfies the fixed-point equation S(z) =
exp (−F (z · S(z))), z ∈ U . The function g(z) = z · S(z) is a conformal mapping U → Im(g) ⊆ U .To see that we defined the following holomorphic function defined on U .

f : U → C, f(w) = w exp (F(w)) .

It easily follows by the fixed-point equation that f (g(z)) = z . Thus g has a left inverse f and hence
g is injective. That was the first surprise. The sequence {Sn(z)}∞n=1 of singular inner functions,each of which covers U infinitely many times, produced the limit S(z) so that g(z) = z · S(z) wasinjective. The complete opposite behavior. g covers each point of U at most once.In this section we will describe the image Im(g) by trying to identify its boundary. At first onemight expect a very wild boundary because of the limit above. We will encounter here our secondsurprise. The boundary ∂Im(g) will turn out to be completely tame. It will be composed of curveswhich are the zero sets of certain planar harmonic functions plus very few corners in between thedifferent zero sets. Thus a piecewise smooth closed Jordan curve.
Theorem 6.1. The image Im(g) = Im(z · S(z)) of a conformal mapping in CONF is a piecewise
smooth closed Jordan curve. It is composed of arcs on T and of arcs which are subsets of the zero
set of the planar harmonic function <F (w) + log |w |, plus a small number of corners.

Proof.Since f (w) = g−1(w) and g : U → Im(g) ⊆ U is conformal, in order to to identify Im(g), we needto identify those arcs in the closure of the unit disk U that are mapped by f into the unit circle
T = ∂U. So we want to solve for all w ∈ U that satisfy |f (w)| = 1. I.e.

|w exp (F (w))| = 1.

We recall that F ∈ RP and hence, by Definition 3.1, condition (ii) we know that <F (e iθ) = 0 almosteverywhere on T with respect to the Lebesgue measure on T. Since |exp (F (w))| = exp (<F (w))it follows that: ∣∣e iθ exp (F (e iθ))∣∣ = 1

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 9almost everywhere on T. So we want to solve for all w ∈ U that satisfy |f (w)| = 1. This means,to find all w ∈ U for which |w exp(F (w))| = 1, that means |w | exp(<F (w)) = 1, i.e. those w ∈ Uthat satisfy:
<F (w) = − log |w |.This equation has harmonic functions on both sides. Alternatively we look for the zero set in U ofthe planar harmonic function <F (w) + log |w |. This proves our theorem. �

In the case of our first example:
F (w) =

1 + w

1− w .In this case:
1 + w

1− w =
1− |w |2

|1− w |2 +
w − w
|1− w |2 .Our equation <F (w) + log |w | = 0 becomes:

1− |w |2

|1− w |2 + log |w | = 0.We note that any w ∈ T−{1} solves this equation. In particular both ±i are solutions. We mentionthose two in particular because we will see soon that they are the two zeros of the derivative of ourholomorphic function in interest and hence this function is not injective exactly at those two points.We can obtain the equation of this curve either in Cartesian coordinates: x = <w , y = =w ,
|w |2 = x2 + y2.

1− x2 − y2

1 + x2 + y2 − 2x +
1

2
log(x2 + y2) = 0,or better in polar coordinates: x = r cos θ, y = r sin θ.

1− r2

1 + r2 − 2r cos θ + log r = 0.Solving for cos θ this is:
cos θ =

1

2r

{
1 + r2 +

1− r2

log r

}
.

Remark 6.2. We note that:
lim
r→1−

1

2r

{
1 + r2 +

1− r2

log r

}
= 0,

so the equation above has exactly two solutions in [−π, π], and these are ±π2 . These correspondto ±i .
The zero set within U is the given by

θ(r) = cos−1
{
1

2r

(
1 + r2 +

1− r2

log r

)}
.

This intersects the x-axis to the right of 0, for θ = 0 so cos θ = 1:
1 =

1

2r

{
1 + r2 +

1− r2

log r

}
⇒ (1− r)

{
1− r +

1 + r

log r

}
= 0.

https://doi.org/10.28924/ada/ma.5.4


Eur. J. Math. Anal. 10.28924/ada/ma.5.4 10One solution is r = x = 1 and others we obtain by:
1− x +

1 + x

log x
= 0 or 1 + x + (1− x) log x = 0.

We easily check that (1+ x +(1− x) log x)′ > 0 for 0 < x < 1 and so there is exactly one solution
x = x0 of

1− x +
1 + x

log x
= 0,

in 0 < x < 1. A similar computation shows that the curve:
1− x2 − y2

1 + x2 + y2 − 2x +
1

2
log(x2 + y2) = 0,

determines x as a function of y in [−1, 1]. It connects −i = (0,−1) to i = (0, 1). If goes through
(x0, 0) and is symmetric with respect to the x-axis. It is strictly monotonic decreasing from (x0, 0)to (0, 1) and by symmetry with respect to the x-axis it is strictly monotonic increasing from (0,−1)to (x0, 0). Thus U is divided into two parts by that zero set. The part in U to the left of the curveand the part to the right of that zero set. Since the left part contains the origin (by x0 > 0) it isthat left part that is the image of our conformal mapping in this example, that corresponds to thefunction in RP given by:

F (z) =
1 + z

1− z .

7. Combining two dynamical systems
Next we will make use of two dynamical systems. The first is the discrete dynamical systemwe used above. It is controlled by a simple recursion which is generated by a function in RP.The second is the continuous dynamical system of Löwner type that is controlled by the partialdifferential equation for B, the class of bounded non-vanishing functions. In fact B = SInn the classof the singular inner functions. The notation B as well as its differential equation were describedin Section 2 of the basic paper [1]. The notation SInn was used in [2]. We recall facts from Section2 of [1]. Suppose f ∈ B has the Herglotz representation

f (z) = exp

(
−
∫ 2π
0

e iθ + z

e iθ − z h(θ)dθ
)
,

where h(θ) ≥ 0. The collection of such functions is dense in the subfamily of B consisting offunctions for which f (0) > 0. Changing variable by the substitution τ = τ(θ) = ∫ θ0 h(φ)dφ, andputting k(τ) = e iθ leads to the formula,
f (z) = exp

(
−
∫ t0

0

1 + k(τ)z

1− k(τ)z dτ
)
, (7.1)

where t0 = τ(2π) = − log f (0). Conversely, if k(τ) is a measurable function of τ which satisfies
|k(τ)| = 1, τ ∈ R, then equation (7.1) defines a function of class B. Given f (z) as in equation(7.1), we set

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Eur. J. Math. Anal. 10.28924/ada/ma.5.4 11

f (z, t) = exp

(
−
∫ t

0

1 + k(τ)z

1− k(τ)z dτ
)
, 0 ≤ t ≤ t0. (7.2)

Then f (z, t) ∈ B for all t ∈ [0, t0], f (z, t0) = f (z), and f (z, 0) = 1. It follows from equation (7.2)that for almost all t ,
∂f (z, t)

∂t
= −f (z, t) ·

1 + k(t) · z
1− k(t) · z . (7.3)

This is the differential equation for B.We recall our discrete dynamical system:Let G ∈ RP. We will use the function G(z) to generate a sequence {Sn}∞n=0 of singular innerfunctions. It is controlled by the following recursion, S0(z) ∈ SInn (an arbitrary initial point).
Sn+1(z) = exp (−G(z · Sn(z))) for n ∈ Z≥0. (7.4)

We proved in Theorem 3.3, the following:The limit S(z) = limn→∞ Sn(z) exists and is uniform on compact subsets of U . S ∈ H(U) satisfies
|S(z)| ≤ 1 ∀ z ∈ U , and satisfies the following fixed-point equation, S(z) = exp (−G(z · S(z))).The function z ·S(z) ∈ BH∞(U), the unit ball of H∞(U). z ·S(z) is a conformal mapping (it belongsto CONF). Thus z ·S(z) : U → Im(z ·S(z)) ⊆ U , but it is not an inner function, see for example [3].Let us denote the following correspondence by F :

F : SInn→ CONF, F (S0) = z · S(z).

One result that we will demonstrate below is that the correspondence F is, in fact, a constant. Wewill give two different proofs for that result. This result might seem to be surprising at first. But itis not really surprising.
Remark 7.1. We clearly have ∀ n ∈ Z≥0, F (Sn) = z · S(z). So the correspondence F is certainlyconstant on the sequence {Sn(z)}∞n=0 which is the output of our recursion, that generates thediscrete dynamical system. So we can view F as a correspondence SInn/{{Sn(z)}∞n=0} → CONF.However, since we will prove that F is a constant correspondence (given a G ∈ RP) we willconclude that the truly interesting correspondence is not SInn→ CONF, but is T : RP→ CONF,
T (G(z)) = z · S(z).
We combine the continuous dynamical system that was described in equation (7.3), with our G-discrete dynamical system (G ∈ RP) that was described in equation (7.4), as follows:

F : {f (z, t) | 0 ≤ t ≤ t0} → CONF, F (f (z, t)) = z · S(z, t).

Here the starting point of the recursion is S0(z, t) = f (z, t) and Sn+1(z, t) = exp (−G(z · Sn(z, t)))for n ∈ Z≥0. S(z, t) = limn→∞ Sn(z, t) for z ∈ U (as was mentioned above), also S(z, t) =
exp (−G(z · S(z, t))) for z ∈ U , and z · S(z, t) ∈ CONF for each 0 ≤ t ≤ t0.

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Eur. J. Math. Anal. 10.28924/ada/ma.5.4 128. More results
Those results will be summarized in three theorems and one corollary. We begin with thecorresponding computations. By the differential equation for B, in equation (7.3) and by therecursion, in equation (7.4) we have,

∂S1(z, t)

∂t
=

∂

∂S0
{exp (−G(z · S0))} ·

∂S0(z, t)

∂t
=

= S1(z, t) ·
{
−z ·

∂G(w)

∂w
|w=z ·S0(z,t)

}
·
∂S0(z, t)

∂t
=

= S1(z, t) ·
{
−z ·

∂G(w)

∂w
|w=z ·S0(z,t)

}
·
{
−S0(z, t) ·

1 + k(t)z

1− k(t)z

}
=

= S0(z, t) · S1(z, t) · z ·
∂G(w)

∂w
|w=z ·S0(z,t) ·

{
1 + k(t)z

1− k(t)z

}
.

Next,
∂S2(z, t)

∂t
=

∂

∂S1
{exp (−G(z · S1))} ·

∂S1(z, t)

∂t
=

= S2(z, t) ·
{
−z ·

∂G(w)

∂w
|w=z ·S1(z,t)

}
·
∂S1(z, t)

∂t
=

= S2(z, t) ·
{
−z ·

∂G(w)

∂w
|w=z ·S1(z,t)

}
·S0(z, t) ·S1(z, t) · z ·

∂G(w)

∂w
|w=z ·S0(z,t) ·

{
1 + k(t)z

1− k(t)z

}
=

= −S0(z, t) · S1(z, t) · S2(z, t) · z2 ·
∂G(w)

∂w
|w=z ·S0(z,t) ·

∂G(w)

∂w
|w=z ·S1(z,t) ·

1 + k(t)z

1− k(t)z .Inductive arguments prove:
Theorem 8.1.

∂Sn(z, t)

∂t
= (−1)n+1 ·

 n∏
j=0

Sj(z, t)

 · zn ·
n−1∏
j=0

(
∂G(w)

∂w
|w=z ·Sj (z,t)

) ·
{
1 + k(t)z

1− k(t)z

}
.

Theorem 8.2. There exists a unique S(z) ∈ H(U) such that it is the only fixed-point of the
function exp (−G(z · w)), i.e. S(z) = exp (−G(z · S(z))). Moreover, ∀S0(z) ∈ SInn, the recur-
sion Sn+1(z) = exp (−G(z · Sn(z))), n ∈ Z≥0, defines a sequence of singular inner functions
{Sn(z)}∞n=0. This sequence converges uniformly on compact subsets of U to the fixed-point S(z),
i.e. S(z) = limn→∞ Sn(z) uniformly on compact subsets of U .
So S(z) is determined by the recursion but independently of the initial singular inner function
S0(z).
Thus ∀S0(z), T0(z) ∈ SInn, Sn+1(z) = exp (−G(z · Sn(z))), Tn+1(z) = exp (−G(z · Tn(z))) and
we have: limn→∞ Sn(z) = limn→∞ Tn(z) = S(z), uniformly on compact subsets of U .

Proof.We will outline two proofs. The first proof is using the Banach fixed-point theorem. Namely,
∀ z ∈ U , the function of w ∈ U given by: exp (−G(z · w)) is a contraction and so by the theoremof Banach it has a unique fixed-point w = S(z). Moreover, this fixed-point is the limit of the

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Eur. J. Math. Anal. 10.28924/ada/ma.5.4 13sequence, defined by the recursion wn+1 = exp (−G(z · wn)), independently of the initial point w0.From this we get our conclusions.
A second proof uses the differential equation of B, namely we start at the beginning of chain
S0(z, t) = f (z, t) and generate the sequence of singular inner functions by our recursion: Sn+1(z) =
exp (−G(z · Sn(z))). We obtain the limit uniformly on compact subsets of U , S(z, t) = limn→∞ Sn(z, t).
S(z, t) is a fixed-point

S(z, t) = exp (−G(z · S(z, t))) .We apply the operator ∂
∂t to both sides of the fixed-point equation (justified by our assumptionson S0(z, t)). We obtain:

∂S(z, t)

∂t
= −z · S(z, t) ·

{
∂G(w)

∂w
|w=z ·S(z,t)

}
·
∂S(z, t)

∂t
.

We claim that ∂S(z,t)
∂t = 0 for all t . For if there were an open non-empty interval of t , over which

∂S(z,t)
∂t 6= 0, then by the equation above:{

w ·
∂G(w)

∂w
|w=z ·S(z,t)

}
= −1.

So z ·S(z, t) can be one of a discrete set which are the zeros of the non-zero holomorphic function
w ·

∂G(w)

∂w
+ 1.

Hence z · S(z, t) 6∈ CONF, a contradiction. Hence indeed S(z, t) = S(z) is independent of t .Since the beginning of the chain {f (z, t)} equals the first element of the sequence of the singularinner functions, S0(z, t) = f (z, t), this again, implies the conclusions of Theorem 3.3. �

Theorem 8.3.

lim
n→∞

 n∏
j=0

Sj(z, t)

 · zn ·
n−1∏
j=0

(
∂G(w)

∂w
|w=z ·Sj (z,t)

) = 0.
Proof.By Theorem 6.1 and Theorem 8.1 where we use limn→∞ ∂Sn(z,t)

∂t = 0. �

In particular, if we start our recursion from its fixed-point S0(z, t) = S(z), then our sequenceis stationary, Sj(z, t) = S(z) for all j ∈ Z≥0, and the formula of Theorem 5.2 gives us,
Corollary 8.4.

lim
n→∞

(z · S(z))n ·
{
∂G(w)

∂w
|w=z ·S(z)

}n
= 0 ∀ z ∈ U,

equivalently

lim
n→∞

{(
w ·

∂G(w)

∂w

)
|w=z ·S(z)

}n
= 0 ∀ z ∈ U,

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Eur. J. Math. Anal. 10.28924/ada/ma.5.4 14

equivalently ∣∣∣∣{(w · ∂G(w)∂w

)
|w=z ·S(z)

}∣∣∣∣ < 1 ∀ z ∈ U.
The last inequality can be written as follows:

z · S(z) · G′(z · S(z)) ∈ BH∞(U) where
∣∣z · S(z) · G′(z · S(z))∣∣ < 1 ∀ z ∈ U.

We end our paper with the example G(w) = 1+w
1−w . On the next we will present the formulas weproved, for this particular case.

9. An example
Let us consider

G(w) =
1 + w

1− w ∈ RP.Let S0(z, t) = f (z, t) and
Sn+1(z, t) = exp

(
−
1 + z · Sn(z, t)
1− z · Sn(z, t)

)
for n ∈ Z≥0

S(z, t) = limn→∞ Sn(z, t) uniformly on compact subsets of U , so that
S(z, t) = exp

(
−
1 + z · S(z, t)
1− z · S(z, t)

)
∀ z ∈ U

and z · S(z, t) ∈ CONF, ∀ 0 ≤ t ≤ t0. We have the following results:
(9.5)

∂Sn(z, t)

∂t
= (−1)n+1 ·

 n∏
j=0

Sj(z, t)

 ·
{

(2z)n∏n−1
j=0 (1− z · Sj(z, t))2

}
·
{
1 + k(t)z

1− k(t)z

}
.

This follows by Theorem 2.1.
There exists a unique S(z) ∈ H(U) such that it is the only fixed-point of the function exp (−1+z ·w1−z ·w

),i.e.
S(z) = exp

(
−
1 + z · S(z)
1− z · S(z)

)
.

Moreover ∀S0(z) ∈ SInn, the recursion
Sn+1(z, t) = exp

(
−
1 + z · Sn(z, t)
1− z · Sn(z, t)

)
for n ∈ Z≥0

defines a sequence of singular inner functions {Sn(z)}∞n=0. This sequence converges uniformlyon compact subsets of U to the fixed-point S(z),i.e. S(z) = limn→∞ Sn(z) uniformly on compactsubsets of U .So S(z) is determined by the recursion independently of the initial singular inner function S0(z).This follows by Theorem 3.3.

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Eur. J. Math. Anal. 10.28924/ada/ma.5.4 15

lim
n→∞

(2z)n ·
∏n
j=0 Sj(z, t)∏n−1

j=0 (1− z · Sj(z, t))2
= 0. (9.6)

This follows by Theorem 8.2. ∣∣∣∣ 2z · S(z)
(1− z · S(z))2

∣∣∣∣ < 1 ∀ z ∈ U, (9.7)
equivalently (1− |z | · |S(z)|)2 > 2<{z · S(z)} ∀ z ∈ U . This follows by Corollary 5.3.

References
[1] J.A. Hummel, S. Scheinberg, L. Zalcman, A coefficient problem for bounded nonvanishing functions, J. Anal. Math.31 (1977) 169-190.[2] O. Ivrii, Critical structures of inner functions, J. Funct. Anal. 281 (2021) 109-198.[3] R. McLaughlin, Exceptional Sets for Inner Functions, J. London Math. Soc. s2-4 (4) (1972) 696-700.[4] R. Peretz, The Krzyż Conjecture Theory and Methods, World Scientific, Singapore, 2021.

https://doi.org/10.28924/ada/ma.5.4

	1. Correspondences that involve inner functions
	2. An example of our construction
	3. A generalization
	4. A parametrization of a family of conformal mappings by the functions in RP
	5. The family of conformal mappings CONF
	6. The geometry of the image of conformal mappings in CONF
	7. Combining two dynamical systems
	8. More results
	9. An example
	References