EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 3, 2024, 1490-1496 ISSN 1307-5543 – ejpam.com Published by New York Business Global Riesz Inequality for Harmonic Quasiregular Mappings Elver Bajrami Department of Mathematics, University of Prishtina, Mother Teresa, No. 5, 10000, Prishtina, Kosovo Abstract. In this paper, we generalize the Riesz theorem for harmonic quasiregular mappings for a special case (when p = 2) in the unit disc. Our results improve similar results in this field and are proved with milder conditions. Moreover, we prove another variant forms of Riesz inequality for harmonic quasiregular functions. 2020 Mathematics Subject Classifications: 30H10, 30H05 Key Words and Phrases: Harmonic mappings, Quasiregular mappings, Riesz theorem 1. Introduction Let U = {z ∈ C : |z| < 1} be the unit disk and let T = {z ∈ C : |z| = 1} be the unit circle in plane. For p > 1 we define the Hardy class hp as the class of harmonic mappings f = g + h, where h and g are holomorphic mappings defined on unit disk U ⊂ C. Norm in this space is defined ||f ||p = ||f ||hp = sup 0 2, then following inequality holds ||f ||n ≤ c(k, n)||h||n, where c(k, n) = (2(1 + k2))n/2. 2. Proof of main results Proof. [Proof of Theorem 1] Let f = h+ ḡ = ∞∑ j=0 ajz j + ∞∑ j=1 bj z̄ j . We can assume that both of following integrals converge. If no, then we take the dilatation f(rz) for r < 1. If we integrate in the unit circle∫ T |g′(z)|2|dz| ≤ k2 ∫ U |h′(z)|2|dz| we get ∑ j=1 j2|bj |2 ≤ k2 ∑ j=1 j2|aj |2. Here k = K−1 K+1 . Let u = ℜ(g + h) and v = ℜ(i(h− g)). Then 4 1 π ∫ U |u|2dxdy = 4(ℜa0)2 + ∞∑ j=1 (|aj |2 + |bj |2 + 2ℜ(akbj)) and 4 1 π ∫ U |v|2dxdy = 4(ℑa0)2 + ∞∑ j=1 (|aj |2 + |bj |2 − 2ℜ(akbj)). Without loss of generality we assume that a0 = 0, because ℜ(a0) = 0 by assumption. We will find the best constant ck in the inequality 1 π ∫ U |u|2dxdy = ∞∑ j=0 (|aj |2 + |bj |2 + 2|aj ||bj |) ≤ ck 1 π ∫ U |u|2dxdy = ∞∑ j=0 (|aj |2 + |bj |2 − 2|aj ||bj |) E. Bajrami / Eur. J. Pure Appl. Math, 17 (3) (2024), 1490-1496 1494 under the condition ∑ j=1 j2|bj |2 ≤ k2 ∑ j=1 j2|aj |2. Let W (a, b) = ∑∞ j=0(|aj |2 + |bj |2 + 2|aj ||bj |)∑∞ j=0(|aj |2 + |bj |2 − 2|aj ||bj |) . We need to find the maximum of expression W under the condition H(a, b) = ∑ j=1 j2|bj |2 − k2 ∑ j=1 j2|aj |2 = 0. It is equivalent of finding the maximum of expression M = U V , where U = ∞∑ j=0 |aj |2, V = ∞∑ j=0 |bj |2 under the condition H(a, b) = 0. The Lagrangian is L = M − λH. Assume without loss of generality that aj ≥ 0 and bj ≥ 0. Also we have a0 = 0. Then Laj = 0 and Lbj = 0 imply that bj U = λj2bj , ajV U2 = k2λj2aj , j ≥ 1. λ cannot be zero, because in that case a = 0 and b = 0. If λ ̸= 0, then there exists j0 so that aj0 ̸= 0 and bj0 ̸= 0, aj = bj = 0 for j ̸= j0 and 1 U = k2V U2 . In this case we get M = k2. This implies that W ≤ 1+k2+2k 1+k2−2k = K2. In order to prove Theorem 2, we will need next result. Lemma 1. Let f be analytic function with condition f(0) = 0. Then∫ U ℜ(f(z))|dz| = 0. Proof. Since f(0) = 0, this function can be expressed as f(z) = ∑∞ k=1 akz k. Integrating last expression in unit circle, we get∫ U f(z)|dz| = ∞∑ k=1 ak ∫ 1 0 rkdr ∫ 2π 0 eiktdt. Since the integral ∫ 2π 0 eiktdt = 0, for each k ∈ N, we get∫ U f(z)|dz| = 0. Similarly ∫ U f(z)|dz| = 0. Now, identity ℜ(f(z)) = f(z)+f(z) 2 , follows that ∫ Uℜ(f(z))|dz| = 0. REFERENCES 1495 Proof. [Proof of Theorem 2] Let f(z) = h(z) + g(z) = ∑∞ k=0 akz k + ∑∞ k=1 bkz k, then∫ U |f(z)|2|dz| = ∫ U |h(z)|2|dz|+ ∫ U |g(z)|2|dz|+ 2 ∫ U ℜ(h(z)g(z))|dz|. As in the proof of Theorem 1, using Cauchy-Schwartz inequality, Lemma 1, we get∫ U |f(z)|2|dz| ≤ (1 + k2) ∫ T |h(z)|2|dz|+ 2 ∫ T ℜ(h(z)g(z))|dz| = ck ∫ T |h(z)|2|dz|. Which give required results. Proof. [Proof of Theorem 3] Let f(z) = h(z) + g(z) = ∑∞ k=0 akz k + ∑∞ k=1 bkz k be harmonic K-quasiregular, with g(0) = 0. We have∫ U |f(z)|n = ∫ U (|g + h|2) n 2 = ∫ U (|g|2 + |h|2 + 2ℜ(gh)) n 2 . Using Theorem 2 and inequality ℜ(gh) ≤ |hg| ≤ 1 2(|h| 2 + |g|2), we get∫ U |f(z)|n ≤ ∫ U (2(|h|2 + |g|2)) n 2 ≤ (2(1 + k2)) n 2 ∫ U |h|n. 3. Conclusion In this paper we generalize the Riesz theorem for harmonic quasiregular mappings for a special case in the unit disc. This result is given thought Theorem 1. In order to proving this result we use Lagrange multipliers. Probably this method can be used to prove also for other cases, to make a generalization of Riesz theorem. Moreover, thought Theorem 2 and 3, we prove another variant forms of Riesz inequality for harmonic quasiregular functions. References [1] E. F. Beckenbach. On a theorem of Fejér and Riesz. J. London Math. Soc., 13:82–86, 1938. [2] M. Stein C. Efferman. Hp spaces of several variables. Acta Math., 129:137–193, 1972. [3] T. Tao S. 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