Abstract
Marx and Strohhäcker showed around in 1933 that $f(z)/z$ is subordinate to $1/(1-z)$ for a normalized convex function $f$ on the unit disk $|z|<1.$ Brickman, Hallenbeck, MacGregor and Wilken proved in 1973 further that $f(z)/z$ is subordinate to $k_α(z)/z$ if $f$ is convex of order $α$ for $1/2\leα<1$ and conjectured that this is true also for $0<α<1/2.$ Here, $k_α$ is the standard extremal function in the class of normalized convex functions of order $α$ and $k_0(z)=z/(1-z).$ We prove the conje
Results & Lemmas (14)
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Theorem 1.1.
Theorem 1.1. The function hα(z) = kα(z)/z maps D univalently onto a convex domain for each 0 ≤α < 1. We remark that, in the context of the…
Theorem 1.1. The function hα(z) = kα(z)/z maps D univalently onto a convex domain for each 0 ≤α < 1. We remark that, in the context of the hypergeometric function, this follows also from results of K¨ustner in [5] (see the remark at the end of Section 2 for more details). Anyway, the conjecture has been confirmed:
Corollary 1.2.
Corollary 1.2. Let 0 ≤α < 1. Then, for f ∈K(α), the following subordination holds: f(z) z ≺kα(z) z on D. In view of the form, it is easy to…
Corollary 1.2. Let 0 ≤α < 1. Then, for f ∈K(α), the following subordination holds: f(z) z ≺kα(z) z on D. In view of the form, it is easy to see that kα is bounded on D if and only if α > 1/2. By analyzing the shape of the image of D under the mapping hα(z) = kα(z)/z, we obtain the following more refined result.
Theorem 1.3.
Theorem 1.3. Let 0 ≤α < 1 and f ∈K(α). Then the following hold: (i) kα(−r) −r ≤Re f(z) z ≤kα(r) r for |z| = r < 1. (ii) When 0 < α < 1/2,…
Theorem 1.3. Let 0 ≤α < 1 and f ∈K(α). Then the following hold: (i) kα(−r) −r ≤Re f(z) z ≤kα(r) r for |z| = r < 1. (ii) When 0 < α < 1/2, the asymptotic lines of the boundary curve of hα(D) are given by v = ± cot(πα)(u − 1 2α−1). In particular, the values of f(z)/z for z ∈D are contained in the sector S = {u + iv : |v| < cot(πα)(u − 1 2α−1)}.
Theorem 10
Theorem 10] and the right-hand one follows also from Robertson’s theorem (see Lemma 3.1 below). A much simpler proof of (i) is now…
Theorem 10] and the right-hand one follows also from Robertson’s theorem (see Lemma 3.1 below). A much simpler proof of (i) is now available thanks to Corollary 1.2. The proof of this theorem and more information about the constant M(α) will be given in Section 3. We also provide an application of our results to an extremal problem for K(α) in Section 3. Styer and Wright [10] studied (non-)univalence of a convex combination of two convex functions. Among other things, the following result is mos
Theorem 1.4.
Theorem 1.4. (f + g)/2 ∈S∗for f, g ∈K(0.6). The proof will be given in Section 4. Note that the constant 0.6 = 3/5 is not best possible. We…
Theorem 1.4. (f + g)/2 ∈S∗for f, g ∈K(0.6). The proof will be given in Section 4. Note that the constant 0.6 = 3/5 is not best possible. We remark that the claim (1.1) for an odd convex function f is not necessarily true. An example will be given in Section 5. 2. Proof of Theorem 1.1 We now show that the function hα(z) = kα(z)/z is convex (univalent) on D for each 0 ≤α < 1. To this end, we only need to see that 1 + zh′′ α(z)/h′ α(z) has positive real
Lemma 2.1
Lemma 2.1 (K¨ustner). For non-zero real numbers a, b, c with −1 < a ≤b < c, let F(z) = 2F1(a, b; c; z). Then inf z∈D 1 + zF ′′(z) F ′(z)…
Lemma 2.1 (K¨ustner). For non-zero real numbers a, b, c with −1 < a ≤b < c, let F(z) = 2F1(a, b; c; z). Then inf z∈D 1 + zF ′′(z) F ′(z) = 1 + −F ′′(−1) F ′(−1) ≥1 −(a + 1)(b + 1) b + c + 2 Since 2F1(a, b; c; z) = 2F1(b, a; c; z), we can apply the above lemma to our function hα(z) = 2F1(β, 1; 2; z) for 0 < α < 1; equivalently, for 0 < β < 2. Hence, by (2.1), we obtain
Lemma 3.1
Lemma 3.1 (Robertson). Let 0 ≤α < 1 and f ∈K(α). Then, −kα(−r) ≤|f(z)| ≤kα(r) for |z| = r < 1. In particular, the image domain f(D)…
Lemma 3.1 (Robertson). Let 0 ≤α < 1 and f ∈K(α). Then, −kα(−r) ≤|f(z)| ≤kα(r) for |z| = r < 1. In particular, the image domain f(D) contains the disk |w| < −kα(−1). We will use also the following simple fact.
Lemma 3.2.
Lemma 3.2. Let Ωbe an unbounded convex domain in C whose boundary is parametrized positively by a Jordan curve w(t) = u(t) + iv(t), 0 < t <…
Lemma 3.2. Let Ωbe an unbounded convex domain in C whose boundary is parametrized positively by a Jordan curve w(t) = u(t) + iv(t), 0 < t < 1, with w(0+) = w(1−) = ∞. Suppose that u(0+) = +∞and that v(t) has a finite limit as t →0+. Then v(t) ≤v(0+) for 0 < t < 1.
Lemma 3.2.
Lemma 3.2. We have thus proved assertion (iii). □ We indicate how to compute the value of M(α) for 1/2 < α < 1. Set c = γ/2 = α −1/2 ∈(0,…
Lemma 3.2. We have thus proved assertion (iii). □ We indicate how to compute the value of M(α) for 1/2 < α < 1. Set c = γ/2 = α −1/2 ∈(0, 1/2). Since hα(D) is a bounded convex domain symmetric in R, it is easy to see that vγ(θ) has a unique critical point, say, θα at which vγ attains its maximum so that M(α) = vγ(θα). Here, θ = θα is a unique solution of the equation (3.1) c cot θ 2 + (1 −c) cot(cπ + (1 −c)θ) 2 sin θ 2 2c sin(cπ + (1 −c)θ) −cos θ = 0
Theorem 3.3.
Theorem 3.3. For 0 < α < 1, the function ϕα(θ) = θ + arg h′ α(eiθ) maps the interval (0, π] onto (π(1 −α), π] homeomorphically.…
Theorem 3.3. For 0 < α < 1, the function ϕα(θ) = θ + arg h′ α(eiθ) maps the interval (0, π] onto (π(1 −α), π] homeomorphically. Furthermore, the following hold. (i) Suppose α = 0. Then, Q0(0) = 1/2 and Q0(t) = −∞for 0 < t ≤π.
Lemma 4.1.
Lemma 4.1. Let f ∈S. Suppose that f(D) contains the disk Dρ for some ρ > 0. Then Dρ ⊂fa(D) for a ∈D.
Lemma 4.1. Let f ∈S. Suppose that f(D) contains the disk Dρ for some ρ > 0. Then Dρ ⊂fa(D) for a ∈D.
Lemma 4.2.
Lemma 4.2. Let ρ be a positive constant. Suppose that two functions f, g ∈K satisfy the following two conditions: (1) f(D) and g(D) both…
Lemma 4.2. Let ρ be a positive constant. Suppose that two functions f, g ∈K satisfy the following two conditions: (1) f(D) and g(D) both contain the disk Dρ, and (2) | Im [f(z)/z]| < ρ and | Im [g(z)/z]| < ρ on D. Then (f + g)/2 ∈S∗.
Lemma 5.1
Lemma 5.1 (Alexander). The function f(z) = z + a2z2 + a3z3 +... is convex univalent on D if ∞ X n=1 n2|an| ≤1. We also need the following…
Lemma 5.1 (Alexander). The function f(z) = z + a2z2 + a3z3 + . . . is convex univalent on D if ∞ X n=1 n2|an| ≤1. We also need the following auxiliary result which is a special case of Theorem 5 of Ruscheweyh [9] with n = 1.
Lemma 5.2
Lemma 5.2 (Ruscheweyh). The function qγ(z) = ∞ X j=1 γ + 1 γ + j zj belongs to K for Re γ ≥0. In particular, the function H1 given in (1.2)…
Lemma 5.2 (Ruscheweyh). The function qγ(z) = ∞ X j=1 γ + 1 γ + j zj belongs to K for Re γ ≥0. In particular, the function H1 given in (1.2) is univalent because H1 = 1 + q1/2/3. We now consider the function f(z) = z + z3 100 + z5 50. Then, by Alexander’s lemma, f is an odd convex function. Secondly, we observe that f has a non-zero fixed point z0 in D. Indeed, by solving the algebraic equation f(z) = z, we
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