Bohr radius for locally univalent harmonic mappings

Abstract

We consider the class of all sense-preserving harmonic mappings $f= h+\overline{g}$ of the unit disk $\ID$, where $h$ and $g$ are analytic with $g(0)=0$, and determine the Bohr radius if any one of the following conditions holds: \bee $h$ is bounded in $\ID$. $h$ satisfies the condition ${\rm Re}\, h(z)\leq 1$ in $\mathbb{D}$ with $h(0)>0$. both $h$ and $g$ are bounded in $\ID$. $h$ is bounded and $g'(0)=0$. \eee We also consider the problem of determining the Bohr radius when the supremum of th

Results & Lemmas (14)

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Theorem 1. Theorem 1. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving K–quasiconformal harmonic mapping of the disk…

Theorem 1. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving K–quasiconformal harmonic mapping of the disk D, where h is a bounded function in D. Then ∞ X n=0 |an|rn + ∞ X n=1 |bn|rn ≤||h||∞for r ≤K + 1 5K + 1. The constant (K + 1)/(5K + 1) is sharp.

Theorem 2. Theorem 2. Assume the hypothesis of Theorem 1. Then |a0|2 + ∞ X n=1 (|an| + |bn|)rn ≤||h||∞for r ≤K + 1 3K + 1. The constant (K + 1)/(3K +…

Theorem 2. Assume the hypothesis of Theorem 1. Then |a0|2 + ∞ X n=1 (|an| + |bn|)rn ≤||h||∞for r ≤K + 1 3K + 1. The constant (K + 1)/(3K + 1) is sharp. We would like to remark that the boundedness condition on h in Theorem 1 can be replaced by half-plane condition. However, the Bohr radius remains the same in this case too.

Theorem 3. Theorem 3. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving K–quasiconformal harmonic mapping of the disk…

Theorem 3. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving K–quasiconformal harmonic mapping of the disk D, where h satisﬁes the con- dition Re h(z) ≤1 in D and h(0) = a0 is positive. Then a0 + ∞ X n=1 |an|rn + ∞ X n=1 |bn|rn ≤1 for r ≤K + 1 5K + 1.

Corollary 1. Corollary 1. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving harmonic mapping of the disk D, where h is…

Corollary 1. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving harmonic mapping of the disk D, where h is a bounded function in D. Then (1) |a0| + ∞ X n=1 (|an| + |bn|)rn ≤||h||∞for r ≤1 5, and the number 1/5 is sharp. Moreover, either a0 = 0 or |a0| in (1) is replaced by |a0|2, then the constant 1/5 could be replaced by 1/3 which is also sharp.

Corollary 2. Corollary 2. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving harmonic mapping of the disk D, where h…

Corollary 2. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=1 bnzn is a sense- preserving harmonic mapping of the disk D, where h satisﬁes the condition Re h(z) ≤1 in D and h(0) = a0 is positive. Then a0 + ∞ X n=1 (|an| + |bn|)rn ≤1 for r ≤1 5, and the number 1/5 is sharp.

Theorem 4. Theorem 4. Suppose that either f = h + g or f = h + g, where h(z) = P∞ n=1 anzn and g(z) = P∞ n=1 bnzn are bounded analytic functions in D.…

Theorem 4. Suppose that either f = h + g or f = h + g, where h(z) = P∞ n=1 anzn and g(z) = P∞ n=1 bnzn are bounded analytic functions in D. Then ∞ X n=1 (|an| + |bn|)rn ≤max{||h||∞, ||g||∞} for r ≤ r 7 32. This number p 7/32 is sharp. As in the symmetric case of analytic functions (see [2, 11, 12]), we have the following

Theorem 5. Theorem 5. Let p ≥2. Suppose that f(z) = h(z)+g(z) = P∞ n=0 anzpn+1 +P∞ n=0 bnzpn+1 is a harmonic p–symmetric function in D, where h and g…

Theorem 5. Let p ≥2. Suppose that f(z) = h(z)+g(z) = P∞ n=0 anzpn+1 +P∞ n=0 bnzpn+1 is a harmonic p–symmetric function in D, where h and g are bounded functions in D. Then ∞ X n=0 (|an| + |bn|)rpn+1 ≤max{||h||∞, ||g||∞} for r ≤1 2. The number 1/2 is sharp. The proofs of these results will be given in Section 2. In Section 3, we extend fur- ther results for sense-preserving K–quasiconformal harmonic mappings of the disk D. In Section 4, we consider the problem of ﬁnding the Bohr radius for the sp

Lemma 1. Lemma 1. Suppose that h(z) = P∞ n=0 anzn and g(z) = P∞ n=0 bnzn are two analytic functions in the unit disk D such that |g′(z)| ≤k|h′(z)|…

Lemma 1. Suppose that h(z) = P∞ n=0 anzn and g(z) = P∞ n=0 bnzn are two analytic functions in the unit disk D such that |g′(z)| ≤k|h′(z)| in D and for some k ∈[0, 1]. Then ∞ X n=1 |bn|2rn ≤k2 ∞ X n=1 |an|2rn for |z| = r < 1.

Lemma 2. Lemma 2. Suppose p is a natural number and 2r2p < 1. If h(z) = P∞ n=0 apn+1zpn+1 is analytic and |h(z)| ≤1 for z ∈D, then the following…

Lemma 2. Suppose p is a natural number and 2r2p < 1. If h(z) = P∞ n=0 apn+1zpn+1 is analytic and |h(z)| ≤1 for z ∈D, then the following inequalities hold: (4) ∞ X n=0 |apn+1|rpn+1 ≤ 1 rp−1(3 −2 √ 2 √ 1 −r2p)

Theorem 6. Theorem 6. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=2 bnzn is a sense- preserving K–quasiconformal harmonic mapping of the disk…

Theorem 6. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=2 bnzn is a sense- preserving K–quasiconformal harmonic mapping of the disk D, where h is a bounded function in D. Then (5) ∞ X n=0 |an|rn + ∞ X n=2 |bn|rn ≤||h||∞for r ≤rK, where rK is the positive root of the equation MK(r) = 1/2 and

Corollary 3. Corollary 3. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=2 bnzn is a sense- preserving harmonic mapping of the disk D, where h is…

Corollary 3. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=2 bnzn is a sense- preserving harmonic mapping of the disk D, where h is a bounded function in D. Then (9) ∞ X n=0 |an|rn + ∞ X n=2 |bn|rn ≤||h||∞for r ≤0.2942.... The number 0.2942... cannot be replaced by a number greater than R = 0.299825..., where R is the positive root of the equation

Theorem 7. Theorem 7. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=0 bnzn is a locally uni- valent K–quasiconformal harmonic mapping of the…

Theorem 7. Suppose that f(z) = h(z) + g(z) = P∞ n=0 anzn + P∞ n=0 bnzn is a locally uni- valent K–quasiconformal harmonic mapping of the disk D, where h′ is a bounded function in D. Then K + 1 2K ∞ X n=1 (|an| + |bn|)nrn−1 ≤||h′||∞for r ≤1/3 and the number 1/3 is sharp. The proof of Theorem 7 easily follows easily if we use the above method and the classical proof of Bohr’s 1/3-Theorem. The case K →∞gives the corresponding result for sense-preserving harmonic mappings of the disk D.

Theorem 8. Theorem 8. Let f ∈B and ∥f∥B ≤1. Then ∞ X n=0 |an|rn ≤1 for r ≤R = 0.55356..., where R is the positive solution to the equation 1 −R + R…

Theorem 8. Let f ∈B and ∥f∥B ≤1. Then ∞ X n=0 |an|rn ≤1 for r ≤R = 0.55356..., where R is the positive solution to the equation 1 −R + R log(1 −R) = 0. The number R cannot be replaced by a number greater than 0.624162....

Theorem 9. Theorem 9. Suppose that f = h + g is harmonic in D, g(0) = 0 and ||f||BH ≤1, where ||f||BH = |f(0)| + sup z∈D (1 −|z|2)(|h′(z)| + |g′(z)|).…

Theorem 9. Suppose that f = h + g is harmonic in D, g(0) = 0 and ||f||BH ≤1, where ||f||BH = |f(0)| + sup z∈D (1 −|z|2)(|h′(z)| + |g′(z)|). Then |a0| + ∞ X n=1 p |an|2 + |bn|2rn ≤1 for r ≤R = 0.55356. This number 0.55356 cannot be replaced by a number greater than 0.624162....