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Results & Lemmas (60)

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Theorem 3.1. Theorem 3.1. Let Ω⊂C and Π ⊂C be two domains of hyperbolic type. If F ∈A(Ω, Π), then |F ′(z)| ⩽R(F(z), Π) R(z, Ω) at each point z ∈Ω. If z…
Theorem 3.1. Let Ω⊂C and Π ⊂C be two domains of hyperbolic type. If F ∈A(Ω, Π), then |F ′(z)| ⩽R(F(z), Π) R(z, Ω) at each point z ∈Ω. If z ̸= ∞and F(z) ̸= ∞, then |F ′(z)| = R(F(z), Π) R(z, Ω) if and only if F is a locally conformal mapping of Ωonto Π. Several generalizations of the Schwarz–Pick inequality are known. We present only three results, due to Avkhadiev and Wirths, on the extension of this inequality to higher derivatives of order n ⩾2. Following [25], for derivatives of an arbitrary
Theorem 3.2 Theorem 3.2 ([25]). Let Ω⊂C and Π ⊂C be convex domains such that Ω̸= C and Π ̸= C. Then for each n ⩾2, at each point z ∈Ω, for any function…
Theorem 3.2 ([25]). Let Ω⊂C and Π ⊂C be convex domains such that Ω̸= C and Π ̸= C. Then for each n ⩾2, at each point z ∈Ω, for any function F ∈A(Ω, Π) 1 n!|F (n)(z)| ⩽2n−1 R(F(z), Π) Rn(z, Ω) , where 2n−1 is the sharp constant, namely, Cn(Ω, Π) = 2n−1 for any pair of convex domains Ω⊂C and Π ⊂C such that Ω̸= C and Π ̸= C. The reader can find several analogues of this theorem in [25], which are related to various geometric assumptions about Ωand Π. In particular, if Ω⊂C and Π ⊂C are simply connec
Theorem 3.3. Theorem 3.3. The following equality holds for arbitrary domains Ω⊂C and Π ⊂ C of hyperbolic type: C2(Ω, Π) = 1 2  sup z∈Ω |∇R(z, Ω)| + sup…
Theorem 3.3. The following equality holds for arbitrary domains Ω⊂C and Π ⊂ C of hyperbolic type: C2(Ω, Π) = 1 2  sup z∈Ω |∇R(z, Ω)| + sup w∈Π |∇R(w, Π)|  . The condition supz∈Ω|∇R(z, Ω)| < ∞is a criterion for the uniform perfectness of the boundary of the domain Ω⊂C of hyperbolic type (for instance, see [18] or [25]). Hence Theorem 3.3 has the following consequence.
Corollary 3.3.1. Corollary 3.3.1. For two domains of hyperbolic type Ω⊂C and Π ⊂C, C2(Ω, Π) < ∞ if and only if the boundaries of both Ωand Π are uniformly…
Corollary 3.3.1. For two domains of hyperbolic type Ω⊂C and Π ⊂C, C2(Ω, Π) < ∞ if and only if the boundaries of both Ωand Π are uniformly perfect sets. We return to domains Ω⊂C with uniformly perfect boundaries and their geo- metric characterisations in terms of the maximum moduli of domains in § 6, when we describe integral inequalities of Hardy and Rellich type. If one of the domains Ω⊂C and Π ⊂C contains a point at infinity, then C2(Ω, Π) = C2(Π, Ω) = ∞by Theorem 3.3, because in a neighbourho
Theorem 3.4 Theorem 3.4 ([25]). Let Ω⊂C and Π ⊂C be simply connected domains of hyper- bolic type. Let ∞∈Ωand ∞∈Π. Then for each n ⩾2, at each point z…
Theorem 3.4 ([25]). Let Ω⊂C and Π ⊂C be simply connected domains of hyper- bolic type. Let ∞∈Ωand ∞∈Π. Then for each n ⩾2, at each point z ∈Ω, for any F ∈A(Ω, Π) 1 n! |F (n)(z)| ⩽  q + 1 q + p + 1 p n−1 R(F(z), Π) Rn(z, Ω) , (7)
Theorem 3.5. Theorem 3.5. If f(z) = P∞ k=0 akzk is a function in the class B and 0 < p ⩽2, then Mp(r) = max a∈[0,1]  ap + r(1 −a2)p 1 −rap  for 0 ⩽r…
Theorem 3.5. If f(z) = P∞ k=0 akzk is a function in the class B and 0 < p ⩽2, then Mp(r) = max a∈[0,1]  ap + r(1 −a2)p 1 −rap  for 0 ⩽r ⩽2p/2−1. (10) Moreover, Mp(r) <  1
Theorem 3.6. Theorem 3.6. The Bohr radius in the class of odd bounded holomorphic functions is precisely r∗, where r∗= 1 4 r B −2 6 + 1 2 s 3 r 6 B −2…
Theorem 3.6. The Bohr radius in the class of odd bounded holomorphic functions is precisely r∗, where r∗= 1 4 r B −2 6 + 1 2 s 3 r 6 B −2 −B 24 −1
Theorem 3.7. Theorem 3.7. Let f(z) and g(z) be analytic functions in D with Taylor expansions f(z) = P∞ k=0 akzk and g(z) = P∞ k=0 bkzk for z ∈D. If…
Theorem 3.7. Let f(z) and g(z) be analytic functions in D with Taylor expansions f(z) = P∞ k=0 akzk and g(z) = P∞ k=0 bkzk for z ∈D. If f(z) ≺q g(z), then ∞ X k=0 |ak|rk ⩽ ∞ X k=0 |bk|rk for all r ⩽1 3 . Judging by this result, the following conjecture looks plausible.
Theorem 3.8. Theorem 3.8. Let f(z) = P∞ k=0 akzk be an analytic function in D such that |f(z)| ⩽1 in D. Then ∞ X k=0 |ak|rk + 16 9 Sr π + λ Sr π 2 ⩽1…
Theorem 3.8. Let f(z) = P∞ k=0 akzk be an analytic function in D such that |f(z)| ⩽1 in D. Then ∞ X k=0 |ak|rk + 16 9 Sr π + λ Sr π 2 ⩽1 for r ⩽1
Theorem 3.9. Theorem 3.9. Let f be a holomorphic function such that |f| ⩽1 in D, and let f(z) = P∞ n=0 anzn. Then Cf(r) ⩽1 r log 1 1 −r for r ⩽R, where…
Theorem 3.9. Let f be a holomorphic function such that |f| ⩽1 in D, and let f(z) = P∞ n=0 anzn. Then Cf(r) ⩽1 r log 1 1 −r for r ⩽R, where R = 0.5335 . . . is the positive root of the equation 2x = 3(1 −x) log 1 1 −x. The value of R is best possible. The question on the behaviour of Cf(r) as r →1 arises in a natural way. If f is a holomorphic function such that |f| ⩽1 in D and f(z) = P∞ n=0 anzn,
Theorem 3.10. Theorem 3.10. If p ∈[0, 2], then the following inequality is best possible: |a0|p + ∞ X k=1 (|ak|p + |bk|p)rk ⩽∥h∥∞max a∈[0,1]  ap + 2r(1…
Theorem 3.10. If p ∈[0, 2], then the following inequality is best possible: |a0|p + ∞ X k=1 (|ak|p + |bk|p)rk ⩽∥h∥∞max a∈[0,1]  ap + 2r(1 −a2)p 1 −rap  for all r ⩽(21/(p−2) + 1)p/2−1. For p > 2 the following inequality holds: |a0|p + ∞ X
Theorem 4.1 Theorem 4.1 ([107], 1985). Let p ∈(1, ∞). Then cp(Ω) is positive if and only if there exists a finite constant Sp = Sp(Ω) such that for…
Theorem 4.1 ([107], 1985). Let p ∈(1, ∞). Then cp(Ω) is positive if and only if there exists a finite constant Sp = Sp(Ω) such that for each compact set K ⊂Ω Z K dx ρp(x, ∂Ω) ⩽Sp(Ω) Capp(K, Ω); (17) then, in addition, (p −1)p−1 ppSp(Ω) ⩽cp(Ω) ⩽ 1 Sp(Ω) , where Sp(Ω) ∈(0, ∞] is the smallest possible constant in (17). For p > n we can write out explicit estimates for the constant in Maz’ya’s
Theorem 4.2 Theorem 4.2 ([9], 2006). Let p ∈[1, ∞), n ⩾2, s ∈[n, ∞), and let Ω⊂Rn be an arbitrary domain satisfying the single condition Ω̸= Rn. Then…
Theorem 4.2 ([9], 2006). Let p ∈[1, ∞), n ⩾2, s ∈[n, ∞), and let Ω⊂Rn be an arbitrary domain satisfying the single condition Ω̸= Rn. Then the following Hardy-type inequality holds: Z Ω |∇u(x)|p dx ρs−p(x, ∂Ω) ⩾(s −n)p pp Z Ω |u(x)|p dx ρs(x, ∂Ω) ∀u ∈C1 0(Ω). There exist domains Ω⊂Rn, Ω̸= Rn, for which (s−n)p/pp is the greatest possible
Corollary 4.2.1. Corollary 4.2.1. Let n ⩾2, and let Ω⊂Rn be a domain such that Ω̸= Rn. If p ∈(n, ∞), then the following Maz’ya isoperimetric inequality…
Corollary 4.2.1. Let n ⩾2, and let Ω⊂Rn be a domain such that Ω̸= Rn. If p ∈(n, ∞), then the following Maz’ya isoperimetric inequality holds for each compact set K ⊂Ω: Z K dx ρp(x, ∂Ω) ⩽ pp (p −n)p Capp(K, Ω). Clearly, the value of Sp(Ω) and Corollary 4.2.1 can be refined for a number of domains for which we know the sharp values of cp(Ω). Here is an example. It is known that for each p ∈(1, ∞) we have cp(Ω) = (p −1)p/pp for all convex domains Ω⊂Rn, Ω̸= Rn (see [26] and [19]). Hence the followin
Corollary 4.2.2. Corollary 4.2.2. For n ⩾2 let Ω⊂Rn be a convex domain such that Ω̸= Rn. Then for each p ∈(1, ∞) and each compact set K ⊂Ω Z K dx ρp(x, ∂Ω)…
Corollary 4.2.2. For n ⩾2 let Ω⊂Rn be a convex domain such that Ω̸= Rn. Then for each p ∈(1, ∞) and each compact set K ⊂Ω Z K dx ρp(x, ∂Ω) ⩽ pp (p −1)p Capp(K, Ω). Note that Miklyukov and Vuorinen [111] (1999) considered generalized Hardy- type inequalities on Riemannian manifolds of dimension n ⩾2 and proved an analogue of Theorem 4.1 for these inequalities. In place of p-capacity, they used certain geometric quantities related to the isoperimetric profile of the Riemannian manifold. In what fo
Theorem 4.3 Theorem 4.3 ([8], 1986). Let Ω⊂C be a domain such that Ω̸= C. Then c2(Ω) > 0 if and only if the boundary of Ωis uniformly ∆-regular. The…
Theorem 4.3 ([8], 1986). Let Ω⊂C be a domain such that Ω̸= C. Then c2(Ω) > 0 if and only if the boundary of Ωis uniformly ∆-regular. The following lemma of Ancona is essential for the proof.
Lemma 4.1 Lemma 4.1 ([8], 1986). Let Ω⊂C be a domain such that Ω̸= C. Then c2(Ω) > 0 if and only if there exist a positive superharmonic function…
Lemma 4.1 ([8], 1986). Let Ω⊂C be a domain such that Ω̸= C. Then c2(Ω) > 0 if and only if there exist a positive superharmonic function v(z) in Ωand a positive constant ε = ε(Ω) such that ∆v(z) + εv(z) ρ2(z, ∂Ω) ⩾0 in Ωin the distributional sense, that is, for any non-negative function ψ ∈C∞ 0 (Ω) ZZ Ω  ∆v(z) + εv(z) ρ2(z, ∂Ω) 
Theorem 4.4 Theorem 4.4 ([31], 1978). For a domain Ω⊂C of hyperbolic type the quan- tity α(Ω) is positive if and only if each doubly connected domain G…
Theorem 4.4 ([31], 1978). For a domain Ω⊂C of hyperbolic type the quan- tity α(Ω) is positive if and only if each doubly connected domain G ⊂Ωseparating the boundary of Ωhas conformal modulus at most L = L(Ω). In what follows M(Ω) is the smallest of the constants L = L(Ω). We call M(Ω) the conformal maximum modulus. For each simply connected domain Ω⊂C of hyperbolic type we have α(Ω) > 0 and set M(Ω) = 0 (also see Definition 5.1 below). With this convention there is no need to distinguish the ca
Theorem 4.5 Theorem 4.5 ([69], 1989). Let Ω⊂C be a domain of hyperbolic type. Then c2(Ω) is positive if and only if M(Ω) < ∞. In particular, Ancona [8]…
Theorem 4.5 ([69], 1989). Let Ω⊂C be a domain of hyperbolic type. Then c2(Ω) is positive if and only if M(Ω) < ∞. In particular, Ancona [8] proved that for each simply connected domain Ω⊂C of hyperbolic type we have c2(Ω) ⩾1/16. In the next section we present the definition of the Euclidean maximum modulus M0(Ω), which is easier to evaluate or estimate than the quantity M(Ω). 5. Conformal and Euclidean maximum moduli Recall that, for a doubly connected domain Ω2 ⊂C which is conformally equiv- al
Theorem 5.1 Theorem 5.1 ([17], 2019). Let Ω⊂C be a domain of hyperbolic type, and let f: D →Ωbe a locally conformal covering from the disc D = z ∈C:…
Theorem 5.1 ([17], 2019). Let Ω⊂C be a domain of hyperbolic type, and let f : D →Ωbe a locally conformal covering from the disc D = {z ∈C: |z| < 1} onto Ω(in particular, if Ωis simply connected, then f : D →Ωis a univalent conformal mapping of D onto Ω). Then R3(z, Ω) 4 ∆2R(z, Ω) ≡(1 −|ζ|2)4|Sf(ζ)|2, where z = f(ζ) ∈Ωand Sf(ζ) is the Schwarzian derivative of f at ζ ∈D. Calculating the precise value of Mod(Ω) is usually a complicated task. We present an example based on well-known results of Teic
Proposition 6.1. Proposition 6.1. Let Ω⊂C, Ω̸= C, be a domain. If the Euclidean maximum modulus satisfies M0(Ω) < ∞, then ZZ Ω |∇u(z)|2 dx dy ⩾ 1 16(πM0(Ω)…
Proposition 6.1. Let Ω⊂C, Ω̸= C, be a domain. If the Euclidean maximum modulus satisfies M0(Ω) < ∞, then ZZ Ω |∇u(z)|2 dx dy ⩾ 1 16(πM0(Ω) + γ0)4 ZZ Ω |u(z)|2 dx dy ρ2(z, ∂Ω) ∀u ∈C1 0(Ω).
Theorem 6.1 Theorem 6.1 ([9], 2006). Let p ∈[1, ∞), and let Ω⊂C be a domain such that Ω̸= C. Then M0(Ω) < ∞ ⇐⇒ cp(2, Ω) > 0, so that cp(2, Ω) is…
Theorem 6.1 ([9], 2006). Let p ∈[1, ∞), and let Ω⊂C be a domain such that Ω̸= C. Then M0(Ω) < ∞ ⇐⇒ cp(2, Ω) > 0, so that cp(2, Ω) is positive if and only if the boundary of Ω⊂C is a uniformly perfect set. Then the following inequality holds: 1 2p pp(πM0(Ω) + γ0)2p ⩽cp(2, Ω) ⩽ 1 min{2p, pp}M p 0 (Ω) . Now consider the following Rellich-type inequality: ZZ Ω
Theorem 6.2 Theorem 6.2 ([12], 2016). Let Ω⊂C be a domain such that Ω̸= C. Then C∗ 2(2, Ω) > 0 ⇐⇒ M0(Ω) < ∞ ⇐⇒ C∗ 2(4, Ω) > 0 and the following…
Theorem 6.2 ([12], 2016). Let Ω⊂C be a domain such that Ω̸= C. Then C∗ 2(2, Ω) > 0 ⇐⇒ M0(Ω) < ∞ ⇐⇒ C∗ 2(4, Ω) > 0 and the following estimates hold: q C∗ 2(2, Ω) ⩾c2(2, Ω) = c2(Ω) and C∗ 2(4, Ω) ⩾c2(2, Ω) = c2(Ω).
Theorem 6.3 Theorem 6.3 ([14], 2018). Let m ∈N, m ⩾2. For a domain Ω⊂C, Ω̸= C, the constant Am(Ω) is positive if and only if ∂Ωis a uniformly perfect…
Theorem 6.3 ([14], 2018). Let m ∈N, m ⩾2. For a domain Ω⊂C, Ω̸= C, the constant Am(Ω) is positive if and only if ∂Ωis a uniformly perfect set. In addition, Am(Ω) ⩾ (m −1)! 2c2(Ω). Using the estimates from Theorems 6.1 and 6.3 we can prove the following.
Proposition 6.2. Proposition 6.2. Let m ∈N, m ⩾2, and let Ω⊂C be a domain such that Ω̸= C. If M0(Ω) < ∞, then ZZ Ω |∆m/2u(z)|2 dx dy ⩾ ((m −1)!)2 16(πM0(Ω)…
Proposition 6.2. Let m ∈N, m ⩾2, and let Ω⊂C be a domain such that Ω̸= C. If M0(Ω) < ∞, then ZZ Ω |∆m/2u(z)|2 dx dy ⩾ ((m −1)!)2 16(πM0(Ω) + γ0)4 ZZ Ω |u(z)|2 dx dy ρ2m(z, ∂Ω) ∀u ∈C∞ 0 (Ω), where γ0 = Γ4(1/4)/(4π2). It looks like a promising idea to consider Lp-versions of (23) and (24) for p ∈
Theorem 6.4 Theorem 6.4 ([20], 2022). Let p ∈[2, ∞) and let Ω⊂C be a domain such that Ω̸= C. Then C∗ p(2, Ω) > 0 ⇐⇒ M0(Ω) < ∞ ⇐⇒ C∗∗ p (2, Ω) > 0,
Theorem 6.4 ([20], 2022). Let p ∈[2, ∞) and let Ω⊂C be a domain such that Ω̸= C. Then C∗ p(2, Ω) > 0 ⇐⇒ M0(Ω) < ∞ ⇐⇒ C∗∗ p (2, Ω) > 0,
Proposition 6.3. Proposition 6.3. Let Ω⊂C be a domain such that Ω̸= C. If p ∈[2, ∞) and the Euclidean maximum modulus of Ωsatisfies M0(Ω) < ∞, then ZZ Ω…
Proposition 6.3. Let Ω⊂C be a domain such that Ω̸= C. If p ∈[2, ∞) and the Euclidean maximum modulus of Ωsatisfies M0(Ω) < ∞, then ZZ Ω |∆u|p dx dy ρ2−2p(z, ∂Ω) ⩾ (p −1)p 4pp2p(πM0(Ω) + γ0)4p ZZ Ω |u|p dx dy ρ2(z, ∂Ω) for each function u ∈C∞ 0 (Ω).
Proposition 6.4. Proposition 6.4. Let Ω⊂C, Ω̸= C, be a domain. If p ∈[2, ∞) and the Euclidean maximum modulus of Ωsatisfies M0(Ω) < ∞, then ZZ Ω |∆u|p dx dy…
Proposition 6.4. Let Ω⊂C, Ω̸= C, be a domain. If p ∈[2, ∞) and the Euclidean maximum modulus of Ωsatisfies M0(Ω) < ∞, then ZZ Ω |∆u|p dx dy ρ2−2p(z, ∂Ω) ⩾ (p −1)p 4p−1p2p−2(πM0(Ω) + γ0)4p−4 ZZ Ω |u|p−2|∇u|2 dx dy for each function u ∈C∞ 0 (Ω). Next we sketch the proof of a number of refined estimates for the constants in Hardy and Rellich-type inequalities for simply and doubly connected domains. To
Theorem 6.5 Theorem 6.5 (see [11]). If p ∈[1, ∞) and Ωis a simply or doubly connected domain of hyperbolic type, then for each real-valued function u…
Theorem 6.5 (see [11]). If p ∈[1, ∞) and Ωis a simply or doubly connected domain of hyperbolic type, then for each real-valued function u ∈C∞ 0 (Ω) ZZ Ω |∇u(z)|p dx dy R2−p(z, Ω) ⩾2p pp ZZ Ω |u(z)|p dx dy R2(z, Ω) , z = x + iy,
Proposition 6.5. Proposition 6.5. Let Ω⊂C be a simply connected domain of hyperbolic type. Then the following hold: (i) if p ∈[1, 2], then 2p−4/pp ⩽cp(2, Ω)…
Proposition 6.5. Let Ω⊂C be a simply connected domain of hyperbolic type. Then the following hold: (i) if p ∈[1, 2], then 2p−4/pp ⩽cp(2, Ω) ⩽24−p/pp; (ii) if p ∈[2, ∞), then 1/(2p)p ⩽cp(2, Ω) ⩽2p/pp. In the proofs of these results we can use Theorem 6.5 and the following result: for a simply connected domain Ω⊂C of hyperbolic type the inequalities R(z, Ω)/4 ⩽ ρ(z, ∂Ω) ⩽R(z, Ω), z ∈Ω, hold by Koebe’s 1/4-theorem and the principle of the hyperbolic metric. We deduce these estimates with details on
Proposition 6.6. Proposition 6.6. Let Ω⊂C be a doubly connected domain of hyperbolic type, and let M0(Ω) < ∞. Then the following hold: (i) if p ∈[1, 2],…
Proposition 6.6. Let Ω⊂C be a doubly connected domain of hyperbolic type, and let M0(Ω) < ∞. Then the following hold: (i) if p ∈[1, 2], then 2p pp(4M0(Ω) + 2 + 2 √ 2 )2 ⩽cp(2, Ω) ⩽min 2p pp , 1 2−pM0(Ω)p  ; (ii) if p ∈[2, ∞), then 1
Theorem 6.5 Theorem 6.5 and the following result: for a doubly connected domain Ω⊂C of hyperbolic type R(z, Ω) 4M0(Ω) + 2 + 2 √ 2 ⩽ρ(z, ∂Ω) ⩽R(z, Ω), z…
Theorem 6.5 and the following result: for a doubly connected domain Ω⊂C of hyperbolic type R(z, Ω) 4M0(Ω) + 2 + 2 √ 2 ⩽ρ(z, ∂Ω) ⩽R(z, Ω), z ∈Ω. The left-hand bound was proved in [17]. It is asymptotically sharp as M(Ω) →∞ in the following sense: for any sequence of doubly connected domains Ωn ⊂C with finite moduli M(Ωn) such that M(Ωn) →∞as n →∞, the following equality holds (see [17]): lim n→∞ 1 4M0(Ωn) sup
Proposition 6.7. Proposition 6.7. Let m ∈N, m ⩾2, and let Ω⊂C be a simply connected domain if hyperbolic type. Then for each real function u ∈C∞ 0 (Ω) ZZ Ω…
Proposition 6.7. Let m ∈N, m ⩾2, and let Ω⊂C be a simply connected domain if hyperbolic type. Then for each real function u ∈C∞ 0 (Ω) ZZ Ω |∆m/2u(z)|2 dx dy ⩾((m −1)!)2 16 ZZ Ω |u(z)|2 dx dy ρ2m(z, ∂Ω) and ZZ Ω ρ2(z, ∂Ω)|∆u(z)|2 dx dy ⩾
Proposition 6.8. Proposition 6.8. Let m ∈N, m ⩾2, and let Ω⊂C be a doubly connected domain of hyperbolic type. If the Euclidean maximum modulus of…
Proposition 6.8. Let m ∈N, m ⩾2, and let Ω⊂C be a doubly connected domain of hyperbolic type. If the Euclidean maximum modulus of Ωsatisfies M0(Ω) < ∞, then for each real function u ∈C∞ 0 (Ω) ZZ Ω |∆m/2u(z)|2 dx dy ⩾ ((m −1)!)2 (4M0(Ω) + 2 + 2 √ 2 )2 ZZ Ω |u(z)|2 dx dy ρ2m(z, ∂Ω)
Theorem 6.6 Theorem 6.6 ([21], 2022). Let p ∈[2, ∞), and let Ω⊂C be a non-convex domain such that ρ(Ω) < ∞. If Ωis λ-close to convex, where λ ⩾ρ(Ω)/Λ2,…
Theorem 6.6 ([21], 2022). Let p ∈[2, ∞), and let Ω⊂C be a non-convex domain such that ρ(Ω) < ∞. If Ωis λ-close to convex, where λ ⩾ρ(Ω)/Λ2, then ZZ Ω |∇u(z)|p dx dy ρ2−p(z, ∂Ω) ⩾1 pp ZZ Ω |u(z)|p dx dy ρ2(x, ∂Ω) ∀u ∈C1 0(Ω)
Theorem 6.6. · radius Theorem 6.6. Example 6.1 ([21]). Let Ω2 ⊂C be a ‘cross-like’ domain bounded by hyperbolae and defined by Ω2 =  z = x + iy ∈C: −∞< x < ∞,…
Theorem 6.6. Example 6.1 ([21]). Let Ω2 ⊂C be a ‘cross-like’ domain bounded by hyperbolae and defined by Ω2 =  z = x + iy ∈C: −∞< x < ∞, |y| < 1 |x|  . Clearly, the unit disc with centre at the origin lies in Ω2, while any other disc in this domain has radius less than one. Hence the radius of the maximal open disc in Ω2 is ρ(Ω2) = 1. It is easy to see that Ω2 is λ-close to convex for λ = min R(x), where R(x) is the curvature radius of the hyperbola y = 1/x, 0 < x < ∞, at the point (x, 1/x). B
Theorem 6.7. Theorem 6.7. Let Ω⊂C be a domain λ-close to convex for λ = λ0(Ω) ∈(0, ∞). Let p ∈[1, ∞) and assume that ρ(Ω) ∈(0, ∞). Then for each…
Theorem 6.7. Let Ω⊂C be a domain λ-close to convex for λ = λ0(Ω) ∈(0, ∞). Let p ∈[1, ∞) and assume that ρ(Ω) ∈(0, ∞). Then for each function u ∈C∞ 0 (Ω) ZZ Ω |∇u(z)|p ρ2−p(z, ∂Ω) dx dy ⩾ 1 pp(1 + log γ)p ZZ Ω |u(z)|p ρ2(z, ∂Ω) dx dy, where γ = 1 + ρ(Ω)/λ0(Ω). Note the following useful estimate established in [22]:
Theorem 7.1. Theorem 7.1. Let p > 1. If condition (33) holds, then the series g∞(t) = ∞ X k=1 1 t −zk is convergent in Lp(R). Conversely, if this series…
Theorem 7.1. Let p > 1. If condition (33) holds, then the series g∞(t) = ∞ X k=1 1 t −zk is convergent in Lp(R). Conversely, if this series is convergent in Lp(R), the sequence |yn| is arranged in increasing order, and |zk| ⩽C|yk|, then (33) holds. It should be noted that, although condition (33) for convergence is close to (31), it is not a consequence of it. A suitable example is easy to give. 8. Moduli of quadrilaterals and doubly connected domains Conformal moduli play an important role in g
Theorem 8.1 Theorem 8.1 (Ahlfors–Warshawski). Let f and g be continuously differentiable functions. Then the conformal modulus of the quadrilateral Q =…
Theorem 8.1 (Ahlfors–Warshawski). Let f and g be continuously differentiable functions. Then the conformal modulus of the quadrilateral Q = (Q; z1, z2, z3, z4) satisfies the inequalities Z b a dx θ(x) ⩽Mod(Q) ⩽ Z b a dx θ(x) + Z b a φ′(x)2 + θ′(x)2/12 θ(x)
Theorem 8.2. Theorem 8.2. The conformal modulus QH has the following asymptotic behav- iour: Mod(QH) ∼cH, where c = Z b a dx θ(x). Moreover, Mod(QH) ⩾cH…
Theorem 8.2. The conformal modulus QH has the following asymptotic behav- iour: Mod(QH) ∼cH, where c = Z b a dx θ(x) . Moreover, Mod(QH) ⩾cH and Mod(QH) −cH = O(H−1), H →∞. Note that in Theorem 8.2 the functions f and g are not assumed to be smooth. For the first time it was in fact established in [46]; a shorter proof using Theorem 8.1 was given in [120].
Theorem 8.3. Theorem 8.3. As H →∞, the asymptotic relation Mod(GH) ∼ 1 (c1 + c2)H holds, where c1 = Z d c dx f1(x) −g1(x) and c2 = Z d c dx
Theorem 8.3. As H →∞, the asymptotic relation Mod(GH) ∼ 1 (c1 + c2)H holds, where c1 = Z d c dx f1(x) −g1(x) and c2 = Z d c dx
Theorem 8.4. Theorem 8.4. If D = z ∈C: r < |z| < R, then Mod(DH) ∼ 1 αH, H →∞, where α = 1 R2 −r2  πr2 + 2R2 arcsin r R + 2r p R2 −r2 
Theorem 8.4. If D = {z ∈C: r < |z| < R}, then Mod(DH) ∼ 1 αH , H →∞, where α = 1 R2 −r2  πr2 + 2R2 arcsin r R + 2r p R2 −r2 
Theorem 8.3 Theorem 8.3 can be extended to domains in wider classes. For example, it holds when the outer boundary component is unbounded (the function…
Theorem 8.3 can be extended to domains in wider classes. For example, it holds when the outer boundary component is unbounded (the function g is defined on the whole axis and can take the infinite value) and also in the case when the boundary components are not the graphs of single-valued functions and can even not be curves, but rather continua of a fairly arbitrary shape. In [120] and [122] the authors considered the Vuorinen problem for doubly con- nected domains Ωcontaining the point at infi
Theorem 8.5. Theorem 8.5. Let ΩH = fH(Ω). Then Mod(ΩH) ∼ 1 γH, H →∞, where γ = Z d c dx f1(x) −f2(x).
Theorem 8.5. Let ΩH = fH(Ω). Then Mod(ΩH) ∼ 1 γH , H →∞, where γ = Z d c dx f1(x) −f2(x) .
Theorem 9.1. Theorem 9.1. The Robin capacities σ(γ+) and σ(γ−) with respect to C γ are equal to σ(γ+) = d sin2 φ 4 and σ(γ−) = d cos2 φ 4, respectively,…
Theorem 9.1. The Robin capacities σ(γ+) and σ(γ−) with respect to C \ γ are equal to σ(γ+) = d sin2 φ 4 and σ(γ−) = d cos2 φ 4 , respectively, where d is the transfinite diameter of γ. Furthermore, σ(γ+) + σ(γ−) = d and σ(γ+) −σ(γ−) = P, where P is the dimensionless aerodynamical lift of the arc γ. It was noted in [113] that Theorem 9.1 also holds for arbitrary not necessarily infinitesimally thin airfoils. Variations of Robin capacities.
Theorem 9.2. Theorem 9.2. The following inequalities hold for reduced moduli and the corre- sponding Robin capacities: µ(At; Dt) −µ(A; D) ⩽−1 4π2 Σt and…
Theorem 9.2. The following inequalities hold for reduced moduli and the corre- sponding Robin capacities: µ(At; Dt) −µ(A; D) ⩽−1 4π2 Σt and σ(At; Dt) −σ(A; D) ⩾1 2π σ(A; D)Σt, where Σt is the area of Ωt in the metric ρ(z) = |g′(z)/g(z)|, that is, Σt = ZZ Ωt ρ2(z) dx dy. Assuming additionally that η lies in an open smooth subarc A and the family ηt, 0 < t ⩽t0, consists of smooth arcs satisfying condition ( eR), the following formulae hold:
Theorem 9.3. Theorem 9.3. Among all smooth arcs of length l with curvature satisfying K(s) ⩽ψ(s) a.e., 0 ⩽s ⩽l, where s is the natural parameter on the…
Theorem 9.3. Among all smooth arcs of length l with curvature satisfying K(s) ⩽ψ(s) a.e., 0 ⩽s ⩽l, where s is the natural parameter on the arc and ψ is a non-negative continuous function such that K0l ⩽c0, where K0 = max 0⩽s⩽l ψ(s) and c0 = 0.30284265 . . . , an arc with curvature K(s) ≡ψ(s), s ∈[0, l], has the greatest aerodynamical lift. Note that the constant c0 in Theorem 9.3 gives in fact an estimate for the varia- tion of the slope angle of the tangent to the arc in the motion along this a
Theorem 10.1 Theorem 10.1 ([115] and [117]). The family of functions R(z, t) satisfies the dif- ferential equation ˙R(z, t) R′(z, t) = M X l=2…
Theorem 10.1 ([115] and [117]). The family of functions R(z, t) satisfies the dif- ferential equation ˙R(z, t) R′(z, t) = M X l=2 Pl,ml−2(z) (z −al)ml−1 ˙Al, where Pl,j is the Taylor polynomial of degree j of the function Hl(z) = QN j=1(z −bj)nj+1 Q 1⩽k⩽M,k̸=l(z −ak)mk−1
Theorem 10.1 Theorem 10.1 yields the following.
Theorem 10.1 yields the following.
Theorem 10.2 Theorem 10.2 ([117]). The critical points al, 1 ⩽l ⩽M, and poles bj, 1 ⩽j ⩽M, of the functions R(z, t) satisfy the system of differential…
Theorem 10.2 ([117]). The critical points al, 1 ⩽l ⩽M , and poles bj , 1 ⩽j ⩽M , of the functions R(z, t) satisfy the system of differential equations ˙al = H(ml−1) l (al) (ml −1)! ˙Al + X 2⩽k⩽M,k̸=l G(mk−2) kl (ak) (mk −2)! ˙Ak, ˙bj =
Theorem 10.3 Theorem 10.3 ([119]). The one-parameter family of functions f(z, t) defined by (40) satisfies the equation ˙f(z, t) f ′(z, t) = 1 c N X k=1…
Theorem 10.3 ([119]). The one-parameter family of functions f(z, t) defined by (40) satisfies the equation ˙f(z, t) f ′(z, t) = 1 c N X k=1 ˙Ak (mk −1)! ∂mk−1 ∂ξmk−1 Z(ξ, z)Gk(ξ)  ξ=ak
Theorem 10.3 Theorem 10.3 yields the following result.
Theorem 10.3 yields the following result.
Theorem 10.4 Theorem 10.4 ([119]). The parameters in the integral representation (40) satisfy the system of ordinary differential equations c ˙ak = ˙Ak…
Theorem 10.4 ([119]). The parameters in the integral representation (40) satisfy the system of ordinary differential equations c ˙ak = ˙Ak G(mk) k (ak) mk! −∂mk−1 ∂ξmk−1 Gk(ξ)˜Lk(ξ) (mk −1)!
Theorem 10.5. Theorem 10.5. The one-parameter family of functions f(z, t) satisfies the partial differential equation ˙f(z, t) f ′(z, t) = h(z, t), where…
Theorem 10.5. The one-parameter family of functions f(z, t) satisfies the partial differential equation ˙f(z, t) f ′(z, t) = h(z, t), where h(z, t) = 4 X j=1 γj(t)[ζ(z −zj(t)) −ζ(z0(t) −zj(t)) −η1(t)(z −z0(t))] −˙z0(t), γk(t) = ˙Ak(t) Dk(t) and Dk(t) = c(t) exp{γ(t)zk(t)}
Theorem 10.6 Theorem 10.6 ([48]). If ex ⩽l/2, then for a varying from −∞to ex −l/2 the conformal modulus of the domain G(A1, A2, A3, A4) decreases from…
Theorem 10.6 ([48]). If ex ⩽l/2, then for a varying from −∞to ex −l/2 the conformal modulus of the domain G(A1, A2, A3, A4) decreases from +∞to 0. On the other hand, if ex > l/2, then the conformal modulus decreases from +∞to a positive quantity as a varies from −∞to 0. 11. Critical values of polynomials In 1981 Smale [140] stated the following famous conjecture on critical values of polynomials. Consider polynomials f of degree n ⩾2 such that f(0) = 0 and f ′(0) = 1. Set S(f) = min  f(ζ) ζ : f
Theorem 11.1. Theorem 11.1. Let f be a polynomial of degree n ⩾3 normalized so that f(0) = 0 and f ′(0) = 1. Assume that all of its critical points are…
Theorem 11.1. Let f be a polynomial of degree n ⩾3 normalized so that f(0) = 0 and f ′(0) = 1. Assume that all of its critical points are real. Then there is a critical point ζ such that
Theorem 11.2. Theorem 11.2. Let f be a polynomial of degree n ⩾2 such that f(0) = 0 and f ′(0) = 1. Assume that the critical points of f lie in the…
Theorem 11.2. Let f be a polynomial of degree n ⩾2 such that f(0) = 0 and f ′(0) = 1. Assume that the critical points of f lie in the sector {reiθ : r > 0, |θ| ⩽π/6}. Then S(f) ⩽1/2 and equality holds if and only if n = 2.
Theorem 11.3. Theorem 11.3. Let f be a polynomial of degree n ⩾2 such that f(0) = 0 and f ′(0) = 1. Assume that the critical points of f lie on the ray 1…
Theorem 11.3. Let f be a polynomial of degree n ⩾2 such that f(0) = 0 and f ′(0) = 1. Assume that the critical points of f lie on the ray {1 + reiθ : r ⩾0}, where 0 ⩽θ ⩽π/2. Then S(f) ⩽1/2 and equality holds if and only if n = 2.
Theorem 11.4. Theorem 11.4. Let f be a polynomial of degree n ⩾2 such that f(0) = 0 and f ′(0) = 1. Assume that the critical points of f lie on the union…
Theorem 11.4. Let f be a polynomial of degree n ⩾2 such that f(0) = 0 and f ′(0) = 1. Assume that the critical points of f lie on the union of the rays {1 + re±iθ : r ⩾0}, where 0 < θ ⩽π/2. Then S(f) < 2/3. Of course, if 0 ⩽θ ⩽π/6 in the last theorem, then we are in the case covered by Theorem 11.2. In this case we have the upper bound 1/2 rather than 2/3. Now we describe results related to the so-called dual Smale conjecture. In fact, there is a certain analytic relation for critical values of
Theorem 11.5. Theorem 11.5. Let f be a polynomial of degree six such that f(0) = 0 and f ′(0) = 1. Then there exists a point ζ such that f ′(ζ) = 0 and…
Theorem 11.5. Let f be a polynomial of degree six such that f(0) = 0 and f ′(0) = 1. Then there exists a point ζ such that f ′(ζ) = 0 and |f(ζ)/ζ| ⩾1/6. Moreover, a critical point ζ such that |f(ζ)/ζ| > 1/6 exists, provided f has a form distinct from f(z) = 1 6a 1 −(1 −az)6 , where a ∈C \ {0}.

Definitions (4)

Def 4.1. Definition 4.1. Let Ω⊂C be a domain such that Ω̸= C. For each point z ∈ (∂Ω) ∞ and each radius r > 0, consider the harmonic measure of…
Definition 4.1. Let Ω⊂C be a domain such that Ω̸= C. For each point z ∈ (∂Ω) \ {∞} and each radius r > 0, consider the harmonic measure of Ω∩∂Bz(r) in the open set Ω∩Bz(r). We say that the boundary of Ωis uniformly ∆-regular if there exists a con- stant ε1(Ω) ∈(0, 1) such that for all z ∈(∂Ω) \ {∞} and r > 0 this harmonic measure ωzr(w), w ∈Ω∩Bz(r), satisfies the inequality ωzr(w) ⩽1 −ε1(Ω) for all w ∈Ω∩∂Bz(r/2). Recall that harmonic measures are the subject of the monograph [73] by Garnett and
Def 5.1. Definition 5.1. Let Ω⊂C be a domain with boundary containing at least two points. Then the conformal maximum modulus M(Ω) is defined as…
Definition 5.1. Let Ω⊂C be a domain with boundary containing at least two points. Then the conformal maximum modulus M(Ω) is defined as follows. 1) If Ωis a simply connected domain, then set M(Ω) = 0. 2) If Ωis a doubly connected domain, then M(Ω) is its conformal modulus, that is, M(Ω) := Mod(Ω) = 1 2π log b a ∈(0, ∞], provided that Ωis equivalent to the annulus {z ∈C: a < |z| < b}, 0 ⩽a < b ⩽∞. 3) If Ωis multiply connected, then
Def 5.2. Definition 5.2. Let Ω⊂C be a domain with at least two boundary points. Let Ann(Ω) be the set of annuli introduced above. Then the Euclidean…
Definition 5.2. Let Ω⊂C be a domain with at least two boundary points. Let Ann(Ω) be the set of annuli introduced above. Then the Euclidean maximum modulus ь M0(Ω) is defined as follows. 1) If Ann(Ω) = ∅, then M0(Ω) = 0. 2) If Ann(Ω) is not empty, then M0(Ω) := sup A∈Ann(Ω) 1
Def 6.1. Definition 6.1. Given a positive number λ, we say that a domain Ω⊂C is λ-close to convex if Ω̸= C and for each boundary point ζ ∈(∂Ω) ∞…
Definition 6.1. Given a positive number λ, we say that a domain Ω⊂C is λ-close to convex if Ω̸= C and for each boundary point ζ ∈(∂Ω)\{∞} there exists a point aζ ∈C \ Ωsuch that |ζ −aζ| = λ and the disc Dζ = {z ∈C: |z −aζ| < λ} lies in C \ Ω. If in this definition the domain Ω⊂C is bounded, then Ω̸= C holds automati- cally and (∂Ω) \ {∞} = ∂Ω. It is obvious that if the domain Ω⊂C is convex and Ω̸= C, then this domain is λ-close to a convex one for each λ ∈(0, ∞). The geometric assumptions about
Function classes studied:

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