The Schwarz Type Inequality for Harmonic Mappings of the Unit Disc with Boundary Normalization

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satisfying the Laplace equation In 1959 E. Heinz proved in [5] that for every F ∈ Har(D), if F(D) ⊂ D and F(0) = 0, then |F(z)| ≤ 4 π arctan |z|, z ∈ D; (1. 1) cf. also [4, p. 77] and [8]. Moreover, the above estimate is sharp. The equality in (1.1) holds if and if z ∈ D ∩ R. This result can be treated as a counterpart of the well known Schwarz lemma where holomorphic mappings are replaced by the harmonic ones. The estimation (1.1) can be essentially improved under the additional assumption that F is a quasiconformal mapping of D onto itself; cf. [8,Thm. 3.3] where the bound of |F(z)| depends on the maximal dilatation K of F. In this paper we intend to find a sort of the estimation (1.1) for F ∈ Har(D) where the normalization condition F(0) = 0 is replaced by a certain boundary one described in Corollary 2.2. Its proof appeals to Theorem 2.1 which gives a Schwarz type estimation for F determined by the Poisson integral of a certain function f : T → cl(D). Here and subsequently, the symbol cl(A) stands for the closure of a set A ⊂ C in the Euclidian topology. From Corollary 2.2 we infer Corollary 2.4, which deals with harmonic injective mappings of D onto itself. In particular, Corollary 2.4 leads to the following result: If F is a continuous function in the closed disk cl(D) onto itself, harmonic and injective in D, and normalized by F(e k ) = e k := e 2π ik/3 , k ∈ {0, 1, 2}, (

The Schwarz Type Inequalities
Given an integrable function f : T → C we denote by P[ f ](z) the Poisson integral of f at z ∈ D, i.e., we conclude from (2.1) that then we conclude from the formula (2.2) that Since each sector D k , k ∈ {0, 1, 2}, is closed and convex in C we conclude from the assumption (2.4) and the integral mean value theorem for complex-valued functions that there exist α k ∈ I k and r k ∈ [0; 1] for k ∈ {0, 1, 2} such that (2.14) Therefore, we can assume that all c 0 , c 1 , c 2 are positive, and consequently because r k ∈ [0; 1] for k ∈ {0, 1, 2}. Since α k ∈ I k for k ∈ {0, 1, 2}, we see that If 2π/3 ≤ α ≤ 4π/3, then by (2.15) and (2.16), and so If 4π/3 ≤ α ≤ 2π , then by (2.15) and (2.16), and so Thus in all the cases the inequality (2.14) holds. In order to estimate the last sum in (2.12) we consider the function where χ A is the characteristic function of a set A ⊂ T, i.e., χ A (t) := 1 for t ∈ A and χ A (t) := 0 for t ∈ T\A. Throughout the paper we understand the function log as the inverse function to exp | , where := {z ∈ C : |Imz| < π}. Given z ∈ D we see that Hence and by the formula (2.1) we see that Hence and by (2.17), and consequently, Combining this with (2.18) we obtain Combining (2.9) with (2.17) we have (2.21) Hence and by (2.20), From (2.9) it follows that This together with (2.22), (2.14) and (2.15) leads to Hence and by the equalities in (2.20) we conclude that This means that f 0 is an extremal function in Theorem 2.1, and so the estimation (2.5) is sharp, which completes the proof.
Now we are ready to prove our main result.
where (F) is the set of all z ∈ T such that the limit exists. Then Setting we see that the function f is measurable and bounded. Since F(D) ⊂ D, we can apply the dominated convergence theorem to show that F = P[ f ]. From the condition (2.28) it follows that f (z) ∈ D k for a.e. z ∈ T k and every k ∈ {0, 1, 2}. Theorem 2.1 now implies the estimation (2.29). From the properties of the Poisson integral it follows that the function This means that F 0 is an extremal function in Corollary 2.2, and so the estimation (2.29) is sharp, which completes the proof. This property can be also deduced from a more general result; cf. [6] or [4,Sect. 3.3]. Moreover, from the boundary normalization condition (2.32) it follows, that Hence f (e 0 ) = e 2 or f (e 1 ) = e 2 , which contradicts the equalities (2.33). Thus e 2 / ∈ f (T 0 ). Moreover, by the continuity of f and by (2.33) we see that f (T 0 ) is a closed arc of T containing the points e 0 and e 1 . Consequently, is an open arc of T, and so is the set . Therefore, f (T 0 ) ⊂ T 0 . The same conclusion can be drawn for the arcs T 1 and T 2 in place of T 0 , which means that It follows that for every k ∈ {0, 1, 2}, and consequently the condition (2.28) holds. Then Corollary 2.2 yields the estimation (2.29), which proves the corollary.

Remark 2.5 Using the trigonometric identity
we can transform the right hand side of the inequality in (2.29) as follows Let us recall that the hyperbolic metric ρ h in D is given by the formula cf. e.g. [1]. Hence which together with (2.35) leads to (2.36) Therefore the right hand side of the inequalities in (2.5) and (2.29) can be replaced by the right hand side of the equality in (2.35) or (2.36).

The Extremal Functions
As shown in the proof of Theorem 2.1 the function f 0 , given by the formula (2.25), yields the equality in (2.5). Consequently, the function F 0 , given by the formula (2.30), is extremal in Corollary 2.2. Since for every z = r e iθ ∈ D the equalities (2.26) hold, we conclude from (2.18) that Im log(z−e 2 )−log(z−e 1 ) , z = r e iθ ∈ D. (3.1) We want to find all extremal functions in Theorem 2.1 and Corollary 2.2. To this end we need the following lemma.
where C ζ → λ 0 (ζ ) := e −iα 0 ζ and C ζ → λ p (ζ ) := 1 − ζ / p for p ∈ T. From (3.4) it follows that Reλ b ( f (z)) ≥ 0 for a.e. z ∈ T . By the equality (3.3) we have Hence the real-valued harmonic function P[Re((λ b • f ) · χ T )] attains minimum at a, and so, by the minimum principle for real-valued harmonic functions, Then for a.e. z ∈ T, we see that the set {S 0 , S 1 , S 2 } is closed with respect to the composition operation of mappings •, i.e., ({S 0 , S 1 , S 2 }, •) is a group. Let F be the class of all integrable functions f : T → C satisfying the condition (2.4). It is easily seen that By the definition of the class H (see Remark 2.3) we also have From the properties of the Poisson integral it may be also concluded that

Theorem 3.3 Let f : T → C be an integrable function which satisfies the condition (2.4).
Then

9)
if and only if there exists k ∈ {0, 1, 2} such that one of the two following conditions holds:

10)
where f 0 is the function given by the formula (2.25); (ii) z = 0 and Proof Fix z ∈ D and f ∈ F. As shown in the final part of the proof of Theorem 2.1 the function f 0 satisfies (2.27). By the formula (3.12) we have Applying now the property (3.8) we deduce the equality (3.9) from the condition (i), provided z = 0, and from the condition (ii), provided z = 0. Conversely, assume that the equality (3.9) holds. Analyzing respective parts in the proof of Theorem 2.1 we conclude from (2.12), (2.24) and the second equality in (2.27) that Finally, (3.18), (3.17), (2.25) and (3.5) lead to a.e. on T.
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