Sets of Univalence in Some Classes of Analytic Functions

This paper attempts to determine the “largest” sets of local univalence for a given class and the “largest” open sets in which all functions belonging to a given class are univalent. We establish general properties of sets of univalence for analytic functions with typical normalization. Moreover, we determine some examples of the sets of univalence for some particular subclasses of analytic functions.


Introduction
Suppose that A is the family of all functions that are analytic in the unit disc D := {z ∈ C : |z| < 1} and normalized by f (0) = 0, f (0) = 1.
By capital letters A, B, . . . we denote compact subclasses of the class A. By capital letters A, B, . . . we denote sets of the complex plane. Let cl A be the closure of the set A and ∂A-the boundary of the set A. Moreover, for the function f ∈ A and for the set A let f θ (z) := e iθ f (e −iθ z) and A θ := {e iθ a : a ∈ A} for θ ∈ R. Let D(p, r) := {z ∈ C : |z − p| < r} (and so D(0, 1) = D).
In complex analysis, many authors investigated the problem of determining the "largest" discs at 0 in which all functions belonging to a given class are locally univalent or univalent. By the "largest" disc we mean the disc with the largest radius. For a compact class A these discs are unique and are called the disc of local univalence of the class A and the disc of univalence of the class A, respectively. The radii of these discs, we call the radius of local univalence of the class A and the radius of univalence of the class A, and are denoted by r LU (A) and r U (A), respectively.
The goal of this paper is to generalize the above problems, which means to determine the "largest" sets in which all functions belonging to a given class are locally univalent or univalent.
1. all functions belonging to A are locally univalent in A, 2. for every set B such that A ⊂ B ⊂ D and A = B there exists a function in A that is not locally univalent in B.
The set of local univalence for the class A is unique and we denote it by LU(A). We have For a compact class A, the set LU(A) is open. Certainly, 0 ∈ LU(A), as a result of normalization (i.e., for all functions f ∈ A we have f (0) = 1 = 0). If a class is not compact, then the set of local univalence can be closed, for example LU(A) = {0}. We consider only compact classes.
More problems appear while discussing the set of univalence for a given class. To avoid considerations about univalence at the boundary of a set, we use the following definition. Let us notice that for the class of functions locally univalent in D with typical normalization, the set of univalence does not exist (is an empty set), although the set of local univalence is D. Let f (z) := (e az − 1)/a, z ∈ D, a ∈ C\{0}. The function f is locally univalent in D, because we have f (z) = e az and f (z) = 0 for all z ∈ D. Furthermore, f is univalent in the disc D(0, π/|a|) and is not univalent in each disc with a larger radius. Since a is any complex number not equal to 0, in any disc at 0 there exist functions which are not univalent in this disc. Analogously, we can prove that this property holds for any disc at a point z 0 , z 0 ∈ D, which is included in D. Thus, the set of univalence for the class of functions locally univalent in D with typical normalization does not exist.

Definition 2.2. An open set
If  (3) it follows that the set of domain univalence for the class A does not have to be the "largest" set of univalence for this class.
The questions arise as to whether sets of univalence are domains and whether sets of univalence are unique. The answer to the first question is negative, i.e., there exist sets of univalence which are not domains, as will be shown in Example 1. The answer to the second question leads to the following theorem.

Proof. Suppose that A is the set of univalence for the class A. We have A ⊂ LU(A) and A = LU(A). But A = LU(A). Contradiction. Thus, LU(A) is the only set of univalence for the class A.
Assume now that A is the set of univalence for the class A not equal to LU(A). We have A ⊂ LU(A) and A = LU(A). Therefore, there exists a point z 0 ∈ LU(A) and z 0 / ∈ A. Since the set LU(A) is open, there exists a neighborhood of the point z 0 , D(z 0 , r 1 ) included in LU(A) and such that D(z 0 , r 1 )∩A = ∅. From the facts that the class A is compact and all functions from the class A are locally univalent in this neighborhood, we conclude that there exists a neighborhood of the point z 0 , D(z 0 , r 2 ), r 2 ≤ r 1 , in which all functions from the class A are univalent. This means that there exists the set of univalence containing D(z 0 , r 2 ), in which all functions from the class A are univalent. Let us denote this set by B. We have proved that there exist the sets A and B, A = B, which are the sets of univalence for the class A, so the proof is complete.

Conjecture 2.6. If the set LU(A)
is not the set of univalence for the class A, then there are infinitely many sets of univalence for this class.
One can notice that the union of all sets of univalence for the class A is equal to LU(A). If LU(A) is the union of infinite number of different sets of univalence for the class A, then Conjecture 2.6 can be proved analogously as we have proved Theorem 2.5 (that sets of univalence are more than one). Therefore, the question arises as to whether LU(A) can be the union of finite number of different sets of univalence for the class A. Example 1 shows that LU(A) can be the union of three sets of univalence for the class A.
Another question arises as to whether the intersection of all sets of univalence for the class A can be an empty set. Example 6 shows that it can be an empty set.
Let us notice that even though all functions are univalent in the disc of univalence of a given class A, but it does not mean that the intersection of all sets of univalence for the class A contains the disc of univalence of this class. Inclusion occurs when the disc of univalence is equal to LU(A).

Definition 2.7.
A class A is called the class invariant under rotation if for each function f ∈ A each function f θ belongs to A, θ ∈ R. Theorem 2.8. If a set A is the set of univalence for the class A invariant under rotation, then A θ , θ ∈ R, are also the sets of univalence for this class.
Proof. Let A be the set of univalence for the class A. It follows that for any function f ∈ A all functions f α , α ∈ R, are univalent in A. We will prove that all functions belonging to the class A are univalent in Since functions f θ+α are univalent in A (because A is the set of univalence for the class A and f θ+α ∈ A) and for z ∈ A θ we have e −iθ z ∈ A, this implies that all functions belonging to the class A are univalent in A θ .

Some Examples
Example 1. Let us consider a class consisting of one function f (z) := z + z 2 , z ∈ D, i.e., the class The properties of the class A: 1. LU(A) = D\{−1/2}, r LU (A) = 1/2. 2. r U (A) = 1/2. Since we have Re{f (z)} > 0 for |z| < 1/2, so according to the Noshiro-Warschawski Theorem (see for example [1]), the function f is univalent in the disc D(0, 1/2). 3. The function f maps the disc D onto the set shown in Fig. 1, which can be presented as a union of two disjoint sets X and Y such that all points of the set X are taken by f only once, and all points of the set Y are taken by f twice. 4. Let E := D ∩ D(−1, 1) (see Fig. 2). The function f maps the set E onto the set shown in Fig. 4; the boundary of this set is the curve Γ : x(θ) = cos θ + cos 2θ y(θ) = sin θ + sin 2θ , θ ∈ [2π/3, 4π/3] .
All points of the set f (E) are taken twice (at points symmetric with respect to the point −1/2). 5. Let F := D\ cl E (see Fig. 3). The function f maps the set F onto the set shown in Fig. 5.   7. The function f is univalent in the set F. Since we have F ∩ h(F) = ∅, so from the property (6) we obtain that f is univalent in F. 8. For γ ∈ [0, 2π) let where C + = {z ∈ C : Im z > 0}, C − = {z ∈ C : Im z < 0} (see Fig. 6). From the property (6) it follows that f is univalent in G(γ) and h(G(γ)). Since h(G(γ)) = G(π + γ) for γ ∈ [0, π), we can restrict ourselves to G(γ). Among the sets G(γ), only the set G(π/2) (see Fig. 8) is the set of domain univalence for the class A, but it is not the set of univalence for the class A (Remark 2.4(2)). While for γ = π/2, the sets G(γ) are not the sets of domain univalence for the class A (so they are not the sets of univalence for the class A, either).  1)). Let us notice that H(π/2) is the set of univalence for the class A containing G(π/2)-the set of domain univalence for this class (Remark  (3)). Furthermore, H(π/2) is the set of univalence for the class A, which is not a domain (see Fig. 9). 10. The union of three sets of univalence for the class A is equal to LU(A).
To show that G(π/2) is not the set of univalence for the class B, it is enough to find some open set B such that G(π/2) ⊂ B ⊂ D and G(π/2) = B, in which each function f θ , θ ∈ [0, 2π) is univalent. Let B = G(π/2) ∪ X, where X = D + (0, 1/2)\(cl D(i, 1) ∪ cl D(−i, 1)) and D + (0, 1/2) is the right half-disc of the disc The set G(π/2) is included in some set of univalence for the class B, which is not the disc of univalence of this class. r U (C) = 1/2 and the set D(0, 1/2) is the set of univalence for the class C. Since for |z| < 1/2 we have Re {f (z)} > 0 for f ∈ C, according to the Noshiro-Warschawski Theorem (see for instance [1]), the function f is univalent in the disc D(0, 1/2). 3. The set D(0, 1/2) is the only one set of univalence for the class C, because it is equal to the set of local univalence for this class (see Theorem 2.5).