Abstract
In this paper, we initiate the study of the geometric function theory for slice starlike functions over quaternions and its subclasses. This allows us to answer negatively some questions about the Bieberbach conjecture, the growth, distortion, and covering theorems for slice regular functions. Precisely, we find that the Bieberbach conjecture holds true for slice starlike functions in contrast to the fact that the Bieberbach conjecture fails for biholomorphic starlike mappings in higher dimensio
Results & Lemmas (38)
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Theorem 3.2
Theorem 3.2). Moreover, it is proven that the slice regular extension of any holomorphic starlike function belongs to the function class…
Theorem 3.2). Moreover, it is proven that the slice regular extension of any holomorphic starlike function belongs to the function class S∗(see Example 3.4). Hence, the results obtained for S∗in this paper extend corresponding results for holomorphic starlike functions to high dimensions.
Corollary 3.8
Corollary 3.8) and find that the so-called slice starlike in [16, Definition 3.17] and algebraically starlike in [27, Definition 5.20] are…
Corollary 3.8) and find that the so-called slice starlike in [16, Definition 3.17] and algebraically starlike in [27, Definition 5.20] are equivalent. Let us introduce two important subclasses of slice regular functions. Denote by S the unit 2-sphere of purely imaginary quaternions. For every I ∈S, denote by CI the complex plane R ⊕IR, isomorphic to C. Denote by BI the intersection B ∩CI. One subclass of slice regular functions is given by N(B) = {f : B →H : f is slice regular such that f(BI) ⊂CI f
Theorem 1.1.
Theorem 1.1. Let f(q) = q + P+∞ n=2 qnan ∈S∗. Then |an| ≤n, for all n = 2, 3,.... Equality |an| = n for a given n ≥2 holds if and only if…
Theorem 1.1. Let f(q) = q + P+∞ n=2 qnan ∈S∗. Then |an| ≤n, for all n = 2, 3, . . . . Equality |an| = n for a given n ≥2 holds if and only if f(q) = q(1 −qu)−∗2, ∀q ∈B, for some u ∈∂B. Note that, by Theorem 1.1 and Example 3.1, one can find that the Bieberbach conjecture holds for many injective slice regular functions which are not in V(B). Moreover, the Bieberbach conjecture over quaternions still holds for slice close-to-convex functions (see Theorem 4.3). Related to the Bieberbach conjecture,
Theorem 1.2.
Theorem 1.2. Let f(q) = q + P+∞ n=2 qnan ∈S∗. Then, for any λ ∈H, |a3 −λa2 2| ≤max 1, |4λ −3|. This estimate is sharp for each λ. Equality…
Theorem 1.2. Let f(q) = q + P+∞ n=2 qnan ∈S∗. Then, for any λ ∈H, |a3 −λa2 2| ≤max{1, |4λ −3|}. This estimate is sharp for each λ. Equality occurs if f(q) = q(1 −qu)−∗2, q ∈B, for any u ∈∂B. Note that if λ = 0 in Theorem 1.2, then this result is already obtained in Theorem 1.1. As a very important consequence of Theorem A for the second-order coefficient, one may deduce the following well-known growth and distortion theorems. Theorem B. Let F(z) = z + Σ+∞ n=2anzn be an injective holomorphic functi
Theorem 1.3.
Theorem 1.3. Let f(q) = q +P+∞ n=2 qnan be a slice regular function on B such that qf ′(q) ∈S∗. Then |an| ≤1, for all n = 2, 3,....…
Theorem 1.3. Let f(q) = q +P+∞ n=2 qnan be a slice regular function on B such that qf ′(q) ∈S∗. Then |an| ≤1, for all n = 2, 3, . . . . Equality |an| = 1 for a given n ≥2 holds if and only if f(q) = q(1 −qu)−∗, ∀q ∈B, for some u ∈∂B.
Theorem 1.3
Theorem 1.3 implies directly the following growth theorem: |f(q)| ≤ |q| 1 −|q|, ∀q ∈B, for any slice regular function f on B such that f(0)…
Theorem 1.3 implies directly the following growth theorem: |f(q)| ≤ |q| 1 −|q|, ∀q ∈B, for any slice regular function f on B such that f(0) = 0 and qf ′(q) ∈S∗. For normalized convex functions F on D, there holds a sharper growth theorem |z| 1 + |z| ≤|F(z)| ≤ |z| 1 −|z|, ∀z ∈D, (1.5) which was generalized to Cn (see e.g., [28, 25, 37]). Usually, inequality (1.5) is just a consequence of the following distortion theorem in one and several
Theorem 1.4.
Theorem 1.4. Let f be a slice regular function on B such that qf ′(q) ∈S∗. Then 1 (1 + |q|)2 ≤|f ′(q)| ≤ 1 (1 −|q|)2, ∀q ∈B. Estimates are…
Theorem 1.4. Let f be a slice regular function on B such that qf ′(q) ∈S∗. Then 1 (1 + |q|)2 ≤|f ′(q)| ≤ 1 (1 −|q|)2 , ∀q ∈B. Estimates are sharp. For each q ∈B \ {0}, equality occurs if f(q) = q(1 −qu)−∗, q ∈B, for some u ∈∂B. However, the method adopted in one and several complex variables cannot be used to our setting due to higher dimensions as well as the non-commutativity of quaternions and we do not know if the image of qf ′(q) is convex for any f ∈S∗. Fortunately, we can obtain a Koebe t
Theorem 1.5. · radius
Theorem 1.5. Let f be a slice regular function on B with convex image and f ′(0) = 1. Then f(B) contains an open ball centered at f(0) of…
Theorem 1.5. Let f be a slice regular function on B with convex image and f ′(0) = 1. Then f(B) contains an open ball centered at f(0) of radius 1/2. Moreover, the constant 1/2 is optimal. It is worth mentioning that many methods in complex cases may fail in the quaternionic setting since the regularity does not keep under usual product and composition of two slice regular functions due to the non-commutativity of quaternions. For example, the subordination argument in [29] is not suitable to pr
Theorem 2.3.
Theorem 2.3. Let f: B →H be a slice regular function. Then f(q) = ∞ X n=0 qnan, with an = f (n)(0) n! for all q ∈B. In fact, there are some…
Theorem 2.3. Let f : B →H be a slice regular function. Then f(q) = ∞ X n=0 qnan, with an = f (n)(0) n! for all q ∈B. In fact, there are some limitations in the expansion given in Theorem 2.3. Hence, a new type of expansion occurs in [47]. Definition 2.4. Let f, g : B →H be two slice regular functions of the form f(q) = ∞ X
Proposition 2.7.
Proposition 2.7. Let f and g be slice regular functions on B. Then for all q ∈B Zfs, f −∗∗g(q) = f(Tf(q))−1g(Tf(q)), where Tf: B →B is…
Proposition 2.7. Let f and g be slice regular functions on B. Then for all q ∈B \ Zfs, f −∗∗g(q) = f(Tf(q))−1g(Tf(q)), where Tf : B\Zfs →B\Zfs is defined by Tf(q) = f c(q)−1qf c(q). Furthermore, Tf and Tfc are mutual inverses so that Tf is a diffeomorphism. In one complex variable, the starlike function of order α ∈[0, 1] was first introduced by Robertson [43]. Now we introduce its corresponding version for slice regular functions. Definition 2.8. Let α < 1. We shall say that f is a slice starlike f
Theorem 3.11.
Theorem 3.11. 3. Examples and Lemmas In this section, we shall offer some examples in the function class C and give some useful lemmas to…
Theorem 3.11. 3. Examples and Lemmas In this section, we shall offer some examples in the function class C and give some useful lemmas to prove Theorems 1.1 and 1.2. In addition, we establish a Rogosinski lemma for slice regular functions. Example 3.1. Let f(q) = q + P+∞ n=2 qnan be a slice regular function on B such that P+∞ n=2 n|an| ≤1. Then f is injective on B and is in S∗.
Theorem 3.2.
Theorem 3.2. Let F: D →C be a holomorphic function with F(0) = 0 and F ′(0) = 1. Then F is starlike with respect to 0 if and only if Re zF…
Theorem 3.2. Let F : D →C be a holomorphic function with F(0) = 0 and F ′(0) = 1. Then F is starlike with respect to 0 if and only if Re zF ′(z) F (z) > 0 on D. We recall also a so-called convex combination identity for slice regular functions which was used to establish sharp growth and distortion theorems for a subclass of slice regular functions (see [41] or [42]).
Lemma 3.3.
Lemma 3.3. Let f be a slice regular function on B such that f(BI) ⊆CI for some I ∈S. Then for every reJθ ∈B with J ∈S, (3.2) f(reJθ) 2 = 1…
Lemma 3.3. Let f be a slice regular function on B such that f(BI) ⊆CI for some I ∈S. Then for every reJθ ∈B with J ∈S, (3.2) f(reJθ) 2 = 1 + ⟨I, J⟩ 2 f(reIθ) 2 + 1 −⟨I, J⟩ 2 f(re−Iθ) 2, where ⟨I, J⟩= −Re (IJ) ∈[−1, 1]. From Theorem 3.2, Lemmas 3.3 and 3.7 below, one can show that Example 3.4. Fix I ∈S. Let F(z) = z + P+∞ n=2 znan, an ∈CI for n = 2, 3, 4 . . . be a starlike function
Proposition 3.6.
Proposition 3.6. Let f, g be two slice regular functions on B such that Zf = ∅. Then we have Re f(q)−1g(q) > 0 on B ⇔Re f(q)−∗∗g(q) >…
Proposition 3.6. Let f, g be two slice regular functions on B such that Zf = ∅. Then we have Re f(q)−1g(q) > 0 on B ⇔Re f(q)−∗∗g(q) > 0 on B, (3.4) and |g(q)| < |f(q)| on B ⇔|f(q)−∗∗g(q)| < 1 on B. (3.5)
Lemma 3.7.
Lemma 3.7. Let f be a slice regular function on B such that Zf = 0 and f ′(0) = 1. Then the following statements are equivalent: (1) Re…
Lemma 3.7. Let f be a slice regular function on B such that Zf = {0} and f ′(0) = 1. Then the following statements are equivalent: (1) Re f(q)−1qf ′(q) > α on B \ {0}; (2) Re f(q)−∗∗(qf ′(q)) > α on B \ {0}; (3) For any u ∈∂B, the function M(r) = |f(ru)|/rα is strictly increasing on (0, 1).
Corollary 3.8.
Corollary 3.8. Let F be holomorphic on D such that F(0) = 0 and F ′(0) = 1. Then the following statements are equivalent. (1) F is starlike…
Corollary 3.8. Let F be holomorphic on D such that F(0) = 0 and F ′(0) = 1. Then the following statements are equivalent. (1) F is starlike with respect to 0; (2) Re F(z)−1zF ′(z)) > 0 on D; (3) for any θ ∈[0, 2π), the function M(r) = |F(reiθ)| is strictly increasing on [0, 1).
Theorem 4.1
Theorem 4.1] which is useful in the sequel.
Theorem 4.1] which is useful in the sequel.
Theorem 3.10.
Theorem 3.10. If f: B →B be slice regular such that f(0) = f ′(0) =... = f (m−1)(0) = 0 (m ≥1). Then |f(q)| ≤|q|m, ∀q ∈B, and |f (m)(0)|…
Theorem 3.10. If f : B →B be slice regular such that f(0) = f ′(0) = . . . = f (m−1)(0) = 0 (m ≥1). Then |f(q)| ≤|q|m, ∀q ∈B, and |f (m)(0)| ≤m!. Both inequalities are strict (except q = 0) unless f(q) = qmu for some u ∈∂B. In fact, we can establish a sharpened form of the Schwarz lemma for slice regular functions. Its complex version is due to Rogosinski (see e.g., [13, p. 200]). For q0 ∈H, denote B(q0, r) = {q ∈H : |q −q0| < r}.
Theorem 3.11.
Theorem 3.11. Let q0, b ∈B. For the set of all slice regular functions f: B →B with f(0) = 0 and f ′(0) = b, the range of values of f(q0)…
Theorem 3.11. Let q0, b ∈B. For the set of all slice regular functions f : B →B with f(0) = 0 and f ′(0) = b, the range of values of f(q0) is the closed ball B(c, r), where c = q0b(1 −|q0|2) 1 −|q0b|2 , r = |q0|2(1 −|b|2) 1 −|q0b|2 .
Corollary 3.12.
Corollary 3.12. Let f be a slice regular function on B with f(0) = 0 and |f(q)| < 1 for all q ∈B. Then |q||f ′(0)| −|q| 1 −|qf ′(0)|…
Corollary 3.12. Let f be a slice regular function on B with f(0) = 0 and |f(q)| < 1 for all q ∈B. Then |q||f ′(0)| −|q| 1 −|qf ′(0)| ≤|f(q)| ≤|q||q| + |f ′(0)| 1 + |qf ′(0)| , ∀q ∈B. Equality holds at some point q0 ̸= 0 if and only if f(q) = qϕa(q)u for some a ∈B, u ∈∂B. As an important generalization of the Schwarz lemma, the classical Schwarz-Pick lemma states that if F is a holomorphic self-mapping on D, then (3.10) |F ′(z)| ≤1 −|F(z)|2 1 −|z|2 , ∀z ∈D. However, this classical version fails e
Lemma 3.13.
Lemma 3.13. Let f be a slice regular function on B such that |f(q)| ≤1 for all q ∈B. Then |f ′(0)| ≤1 −|f(0)|2. (3.11) Equality holds in…
Lemma 3.13. Let f be a slice regular function on B such that |f(q)| ≤1 for all q ∈B. Then |f ′(0)| ≤1 −|f(0)|2. (3.11) Equality holds in (3.11) if and only if f is of the form f(q) = (1 −qa)−∗∗(a −q)u, ∀q ∈B, for some a ∈B, u ∈∂B.
Lemma 3.13.
Lemma 3.13.
Lemma 3.13.
Theorem 4.1.
Theorem 4.1. Let f(q) = 1 + ∞ P n=1 qnan be a function in P. Then a2 −a2 1 2 ≤2 −|a1|2 2. (4.1) Equality holds in (4.1) if and only if f(q)…
Theorem 4.1. Let f(q) = 1 + ∞ P n=1 qnan be a function in P. Then a2 −a2 1 2 ≤2 −|a1|2 2 . (4.1) Equality holds in (4.1) if and only if f(q) = (qϕa(q)u + 1) ∗(1 −qϕa(q)u)−∗, for some a ∈B and u ∈∂B.
Theorem 4.2.
Theorem 4.2. Let f(q) = 1 + ∞ P n=1 qnan be a slice regular function in P. Then 1 −|q| 1 + |q| ≤Ref(q) ≤|f(q)| ≤1 + |q| 1 −|q|, ∀q ∈B,…
Theorem 4.2. Let f(q) = 1 + ∞ P n=1 qnan be a slice regular function in P. Then 1 −|q| 1 + |q| ≤Ref(q) ≤|f(q)| ≤1 + |q| 1 −|q|, ∀q ∈B, (4.2) and |an| ≤2, n = 1, 2, . . . . (4.3) Moreover, |a1| = 2 or equality holds for the first or third inequality in (4.2) at some q0 ̸= 0 if and only
Theorem 4.3.
Theorem 4.3. Let f(q) = q + P+∞ n=2 qnan ∈C. Then |an| ≤n, for all n = 2, 3,.... Equality |an| = n for a given n ≥2 holds if and only if…
Theorem 4.3. Let f(q) = q + P+∞ n=2 qnan ∈C. Then |an| ≤n, for all n = 2, 3, . . . . Equality |an| = n for a given n ≥2 holds if and only if f(q) = q(1 −qu)−∗2, ∀q ∈B, for some u ∈∂B.
Theorem 5.1.
Theorem 5.1. Let p(q) = 1 + ∞ P n=1 qnan be a slice regular function in P. Then 1 −|q|2 1 + |a1q| + |q|2 ≤Rep(q) ≤|p(q)| ≤1 + |a1q| + |q|2…
Theorem 5.1. Let p(q) = 1 + ∞ P n=1 qnan be a slice regular function in P. Then 1 −|q|2 1 + |a1q| + |q|2 ≤Rep(q) ≤|p(q)| ≤1 + |a1q| + |q|2 1 −|q|2 , ∀q ∈B. (5.1) Moreover, equality holds for the first or third inequality in (5.1) at some q0 ̸= 0 if and only if p(q) = (qϕa(q)u + 1) ∗(1 −qϕa(q)u)−∗, ∀q ∈B, for some u ∈∂B.
Theorem 5.2.
Theorem 5.2. Let f(q) = q + ∞ P n=2 qnan ∈S∗. Then the following inequalities hold for all q ∈B |q| 1 + |a2q| + |q|2 ≤|f(q)| ≤ |q| 1 −|q|2…
Theorem 5.2. Let f(q) = q + ∞ P n=2 qnan ∈S∗. Then the following inequalities hold for all q ∈B |q| 1 + |a2q| + |q|2 ≤|f(q)| ≤ |q| 1 −|q|2 1 + |q| 1 −|q| |a2| 2 ; (5.3) 1 −|q|2
Theorem 5.3.
Theorem 5.3. Let f ∈S∗. Then the following inequalities hold for all q ∈B |q| (1 + |q|)2 ≤|f(q)| ≤ |q| (1 −|q|)2; (5.8) 1 −|q| (1 + |q|)3…
Theorem 5.3. Let f ∈S∗. Then the following inequalities hold for all q ∈B |q| (1 + |q|)2 ≤|f(q)| ≤ |q| (1 −|q|)2 ; (5.8) 1 −|q| (1 + |q|)3 ≤|f ′(q)| ≤ 1 + |q| (1 −|q|)3 ; (5.9) (5.10) 1 −|q| 1 + |q| ≤|qf ′(q)| |f(q)|
Theorem 5.4.
Theorem 5.4. Let m be a positive integer and f(q) = q + P+∞ n=m+1 qnan be a slice regular function on B such that qf ′(q) ∈S∗. Then 1 (1 +…
Theorem 5.4. Let m be a positive integer and f(q) = q + P+∞ n=m+1 qnan be a slice regular function on B such that qf ′(q) ∈S∗. Then 1 (1 + |q|m)2/m ≤|f ′(q)| ≤ 1 (1 −|q|m)2/m , ∀q ∈B.
Theorem 5.5.
Theorem 5.5. Let f(q) = q + P+∞ n=m+1 qnan ∈S∗for some positive integer m. Then |q| (1 + |q|m)2/m ≤|f(q)| ≤ |q| (1 −|q|m)2/m, ∀q ∈B.
Theorem 5.5. Let f(q) = q + P+∞ n=m+1 qnan ∈S∗for some positive integer m. Then |q| (1 + |q|m)2/m ≤|f(q)| ≤ |q| (1 −|q|m)2/m , ∀q ∈B.
Theorem 5.6.
Theorem 5.6. Let f, g be two slice regular functions on B. If f ∈C and g ≺N f, then the following inequalities hold for all q ∈B |g(q)| ≤…
Theorem 5.6. Let f, g be two slice regular functions on B. If f ∈C and g ≺N f, then the following inequalities hold for all q ∈B |g(q)| ≤ |q| (1 −|q|)2 and |g′(q)| ≤ 1 + |q| (1 −|q|)3 . In fact, we can obtain more information related to (5.8). Note that the author in [31] established the Hayman’s regularity theorem making use of complex version of Theorem 5.7 for the p-univalent holomorphic function.
Theorem 5.7.
Theorem 5.7. Let f ∈S∗and M∞(r, f) = max|q|=r |f(q)| for r ∈(0, 1). Then the function φ(r) = 1 r (1 −r)2M∞(r, f) is decreasing on (0, 1)…
Theorem 5.7. Let f ∈S∗and M∞(r, f) = max|q|=r |f(q)| for r ∈(0, 1). Then the function φ(r) = 1 r (1 −r)2M∞(r, f) is decreasing on (0, 1) and hence tends to a limit α ∈[0, 1]. For any α ∈[0, 1], there exists u ∈∂B such that lim rր1−(1 −r)2|f(ru)| = α.
Theorem 6.1.
Theorem 6.1. Let f(q) = P+∞ n=0 qnan be a slice regular function on B such that f(q) ∈Π:= q ∈H: Re q ≤1 for all q ∈B. Then +∞ X n=1 |qnan|…
Theorem 6.1. Let f(q) = P+∞ n=0 qnan be a slice regular function on B such that f(q) ∈Π := {q ∈H : Re q ≤1} for all q ∈B. Then +∞ X n=1 |qnan| ≤dist (f(0), ∂Π), |q| ≤1 3. (6.2)
Lemma 6.2.
Lemma 6.2. Let gm,r(z) = Pm n=0 rnzn, z ∈D, r ∈R. Then Re gm,r(z) ≥1 2 for all z ∈D and r ∈(0, rm), where r1 = 1/2, r2 = p 3/8, and rm for…
Lemma 6.2. Let gm,r(z) = Pm n=0 rnzn, z ∈D, r ∈R. Then Re gm,r(z) ≥1 2 for all z ∈D and r ∈(0, rm), where r1 = 1/2, r2 = p 3/8, and rm for m ≥3 is the unique positive solution of the equation 1 −r −2rm+1 = 0.
Theorem 6.3.
Theorem 6.3. Let f = g + h: B →H be such that the image f(B) is convex, where g(q) = P+∞ n=0 qnan and h(q) = P+∞ n=0 qnbn are slice regular…
Theorem 6.3. Let f = g + h : B →H be such that the image f(B) is convex, where g(q) = P+∞ n=0 qnan and h(q) = P+∞ n=0 qnbn are slice regular functions on B. Then each partial sum Sm(q) = m X n=0 qnan + m X n=0 qnbn maps B(0, rm) into f(B), where rm are given by Lemma 6.2.
Theorem 6.4.
Theorem 6.4. Let f ∈S∗, then it holds that B(0, 1 4) ⊂f(B). The estimate is precise.
Theorem 6.4. Let f ∈S∗, then it holds that B(0, 1 4) ⊂f(B). The estimate is precise.
Theorem 6.5. · radius
Theorem 6.5. Let f(q) = q + P+∞ n=3 qnan ∈S∗, then it holds that B(0, 1 2) ⊂f(B). 6.4. Bloch-Landau Theorem. In complex analysis, a…
Theorem 6.5. Let f(q) = q + P+∞ n=3 qnan ∈S∗, then it holds that B(0, 1 2) ⊂f(B). 6.4. Bloch-Landau Theorem. In complex analysis, a well-known result of Bloch-Landau theorem says that if F is a holomorphic function in D with the only restriction F ′(0) = 1, then the image of F contains a disc of radius r ≥b, where b is an absolute constant. Now we introduce its corresponding version of slice regular functions. For a ∈B and the slice regular function f on B with f ′(0) = 1, denote by r(a, f) the
Theorem 6.6.
Theorem 6.6. The constant C given in (6.4) has the following estimate 1 2 ≤C ≤π 4.
Theorem 6.6. The constant C given in (6.4) has the following estimate 1 2 ≤C ≤π 4 .
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