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An asymptotically sharp coefficients estimate for harmonic K-quasiconformal mappings

Journal of Inequalities and Applications20162016:84

Received: 17 November 2015

Accepted: 24 February 2016

Published: 2 March 2016


By using the improved Hübner inequalities, in this paper we obtain an asymptotically sharp lower bound estimate for the coefficients of harmonic K-quasiconformal self-mappings of the unit disk \({\mathbb{D}}\) which keep the origin fixed. The result partly improves the former results given by (Partyka and Sakan in Ann. Acad. Sci. Fenn., Math. 30:167-182, 2005) and (Zhu and Zeng in J. Comput. Anal. Appl. 13:1081-1087, 2011). Furthermore, using some estimate for the derivative of the boundary function of a harmonic K-quasiconformal self-mapping w of \({\mathbb{D}}\) which keeps the origin fixed, we obtain an upper bound estimate for the coefficients of w.


Heinz inequalityHübner inequalitiescoefficients estimateharmonic quasiconformal mapping



1 Introduction

Let \({\mathbb{D}}=\{z: |z|<1\}\) denote the unit disk, \(w(z)\) be a harmonic mapping defined in \({\mathbb{D}}\). Then \(w(z)\) can be presented as \(w(z)=h(z)+\overline{g(z)}\), where
$$ h(z)=\sum_{n=0}^{\infty}a_{n}z^{n} \quad \mbox{and} \quad g(z)=\sum_{n=1}^{\infty}b_{n}z^{n} $$
are both analytic in \({\mathbb{D}}\). By Lewy’s theorem [3], we know that \(w(z)\) is locally univalent and sense-preserving in \({\mathbb{D}}\) if and only if its Jacobian satisfies the following inequality:
$$J_{f}(z)=\bigl\vert w_{z}(z)\bigr\vert ^{2}- \bigl\vert w_{\bar{z}}(z)\bigr\vert ^{2}=\bigl\vert h'(z)\bigr\vert ^{2}-\bigl\vert g'(z) \bigr\vert ^{2}>0 $$
for all \(z\in{\mathbb{D}}\). One of the basic properties for harmonic self-mappings of \({\mathbb {D}}\) is the Heinz inequality [4].

Lemma A

Let w map the unit disk harmonically onto itself with \(w(0)=0\). Then
$$ \bigl\vert w_{z}(0)\bigr\vert ^{2}+\bigl\vert w_{\bar{z}}(0)\bigr\vert ^{2}\geq c $$
for some absolute constant \(c>0\).

Subsequently, in 1982, Hall [5] obtained the sharp lower bound of c.

Theorem B

Let \(w(z)=h(z)+\overline{g(z)}=\sum_{n=1}^{\infty }a_{n}z^{n}+\overline{\sum_{n=1}^{\infty}b_{n}z^{n}}\) be a univalent harmonic mapping of the unit disk onto itself, then its coefficients satisfy the inequality
$$ |a_{1}|^{2}+|b_{1}|^{2} \geq\frac{27}{4\pi^{2}}. $$
The lower bound \(\frac{27}{4\pi^{2}}\) is the best possible.
$$p(r,x-\varphi)=\frac{1-r^{2}}{2\pi(1-2r\cos(x-\varphi)+r^{2})} $$
denote the Poisson kernel, then every bounded harmonic mapping w defined in \({\mathbb{D}}\) has the following representation:
$$ w(z)=P[f](z)= \int _{0}^{2\pi}p(r,x-\varphi)f\bigl(e^{ix} \bigr)\, dx, $$
where \(z=re^{i\varphi}\in{\mathbb{D}}\) and f is a bounded integrable function defined on the unit circle \(\mathbf{T}:=\partial{\mathbb{D}}\).
Suppose that \(w(z)\) is a sense-preserving univalent harmonic mapping of \({\mathbb{D}}\) onto a domain \(\Omega\subseteq\mathbb{C}\). Then \(w(z)\) is a harmonic K-quasiconformal mapping if and only if
$$K(w):=\sup_{z\in{\mathbb{D}}} \frac{|w_{z}(z)|+|w_{\bar{z}}(z)|}{|w_{z}(z)|-|w_{\bar{z}}(z)|}\leq K. $$

Under the additional assumption that \(w(z)\) is a K-quasiconformal mapping, in 2005 Partyka and Sakan [1] obtained an asymptotically sharp variant of Heinz’s inequality as follows (see also [2]).

Theorem C

Let \(w(z)\) be a harmonic K-quasiconformal mapping of \({\mathbb{D}}\) onto itself satisfying \(w(0)=0\). Then the inequality
$$ \bigl\vert \partial_{z} w(z)\bigr\vert ^{2}+\bigl\vert \partial_{\bar{z}}w(z)\bigr\vert ^{2}\geq\frac{1}{4} \biggl(1+\frac{1}{K} \biggr)^{2} \max \biggl\{ \frac{4}{\pi^{2}}, L_{K}^{2} \biggr\} $$
holds for every \(z\in{\mathbb{D}}\), where
$$ L_{K}:=\frac{2}{\pi} \int _{0}^{\frac{1}{\sqrt{2}}} \frac{d (\Phi_{1/K}(s)^{2} )}{s\sqrt{1-s^{2}}} $$
is a strictly decreasing function of K. For \(L>0\), \(\Phi_{L}(s)\) is the Hersch-Pfluger distortion function defined by the equalities \(\Phi_{L}(s):=\mu^{-1}(\mu(s)/L) \), \(0< s<1\); \(\Phi_{L}(0):=0 \), \(\Phi_{L}(1):=1\), where \(\mu(s)\) stands for the module of Grötzschs extremal domain \({\mathbb{D}}\backslash[0,s]\).

In 2010, Qiu and Ren [6] improved the Hübner inequalities as follows.

Theorem D

For all \(s\in(0,1)\) and \(K\in(1, \infty)\), we have
$$ 4^{1-K}s^{K}\leq\Phi_{1/K}(s)< 4^{D(s)(1-K)}s^{K} $$
$$ s^{1/K}\leq\Phi_{K}(s)< 4^{(1-s^{2})^{\frac{3}{4}}(1-1/K)}s^{1/K}, $$
where \(D(s)=(1-s)(1+s)^{1/\ln4}\).

A sense-preserving harmonic mapping of \({\mathbb{D}}\) onto itself can be represented as the Poisson extension of the boundary function \(f(e^{it})=e^{i\gamma(t)}\), where \(\gamma(t)\) is a continuous non-decreasing function with \(\gamma(2\pi )-\gamma(0)=2\pi\) and \(\gamma(t+2\pi)=\gamma(t)+2\pi\) (cf. [7, 8]). The coefficients \(a_{n}\) and \(b_{n}\) have an alternative interpretation as Fourier coefficients of the periodic function \(e^{i\gamma(t)}\), and so Heinz’s lemma can be viewed as a statement about Fourier series.

In this paper, assuming that \(w(z)\) is a harmonic K-quasiconformal mapping of \({\mathbb{D}}\) onto itself satisfying \(w(0)=0\), by using Theorem D we obtain a sharp lower bound for its coefficients as follows:
$$ |a_{1}|^{2}+|b_{1}|^{2} \geq B_{1}(K):=2-2^{2(1-1/K)(2+2^{5/4})}\frac{2K^{2}\Gamma (\frac{2}{K})}{(K+1)\Gamma^{2}(\frac{1}{K})} $$
which satisfies \(\lim_{K\rightarrow1^{+}}B_{1}(K)=1\), where Γ is the gamma function.
For \(n\geq2\) we have
$$ |a_{n}|^{2}+|b_{n}|^{2} \geq B_{n}(K), $$
$$\begin{aligned}& B_{n}(K):=\chi(K)+\frac{2^{2(1-1/K)(2+2^{5/4})}\Gamma(1+\frac {2}{K})(n-1-\frac{1}{K})!}{ \Gamma^{2}(\frac{1}{K})(n+\frac{1}{K})\frac{1}{K}(n-1+\frac{1}{K})!}, \\& \biggl(n-1-\frac{1}{K}\biggr)!:=\biggl(n-1-\frac{1}{K}\biggr) \biggl(n-2-\frac{1}{K}\biggr)\cdots \biggl(1-\frac{1}{K}\biggr), \\& \biggl(n-1+\frac{1}{K}\biggr)!:=\biggl(n-1+\frac{1}{K}\biggr) \biggl(n-2+\frac{1}{K}\biggr)\cdots \biggl(1+\frac{1}{K}\biggr), \end{aligned}$$
$$ \chi(K):=2-\frac{2^{2(1-1/K)(2+2^{5/4})}\Gamma(1+\frac {2}{K})}{\Gamma^{2}(1+\frac{1}{K})} $$
is a decreasing function of K with \(\chi(1)=0\).
Assume that \(w(z)=P[f](z)\) is a harmonic K-quasiconformal mapping of \({\mathbb{D}}\) onto itself with the boundary function \(f(e^{it})=e^{i\gamma(t)}\), satisfying \(w(0)=0\). In Theorem 3.2 of [9], Partyka and Sakan proved that the following inequalities:
$$ \frac{2^{5(1-K^{2})/2}}{(K^{2}+K-1)^{K}}\leq\bigl\vert f '(z)\bigr\vert \leq K^{3K}2^{5(K-\frac{1}{K})/2} $$
hold for a.e. \(z=e^{it}\in\mathbf{T}\). Applying the above inequalities we obtain an upper bound for the coefficients of a harmonic K-quasiconformal self-mapping \(w(z)\) of \({\mathbb{D}}\) satisfying \(w(0)=0\) as follows:
$$ |a_{n}|^{2}+|b_{n}|^{2} \leq A_{n}(K):=\frac{16}{n^{2}\pi^{2}}K^{6K}2^{5(K-1/K)}. $$
Furthermore we show that (9) and (10) are sharp as \(K\rightarrow1\).

2 Auxiliary results

Lemma 1

Let \(K>1\) be a constant. Then the equality
$$ \int_{0}^{\pi}\sin^{\frac{2}{K}}(t)\cos(2nt)\, dt= \frac{\pi}{4^{\frac {1}{K}}}\frac{(-1)^{n}\Gamma (1+\frac{2}{K} )}{\Gamma (1+\frac{1}{K}-n )\Gamma (1+\frac{1}{K}+n )} $$
holds for all nonnegative integer numbers \(n=0,1,2,\ldots \) .

Lemma 2

Let \(\varphi(t):=\vert \cos\frac{t}{2}\vert ^{\frac{3}{2}}+ \vert \sin\frac{t}{2}\vert ^{\frac{3}{2}}\), for any \(t\in[0, 2\pi]\). Then
$$ \max_{0\leq t\leq2\pi}\varphi(t)=\varphi \biggl( \frac{\pi }{2} \biggr)=\sqrt[4]{2}. $$

Lemma 3

Let \(w=P[f](z)\) be a harmonic K-quasiconformal self-mapping of \({\mathbb{D}}\) with the boundary function \(f(e^{it})=e^{i\gamma(t)}\). For every \(z_{1}=e^{i(s+t)}, z_{2}=e^{i(s-t)}\in\mathbf{T}\), let \(\theta=\gamma(s+t)-\gamma(s-t)\). Then \(f(z_{1})=e^{i\theta}f(z_{2})\) and the inequalities
$$ 2^{10-10K}\sin^{2K}(t)\leq\sin^{2} \biggl(\frac{\theta}{2} \biggr)\leq2^{2(1-1/K)(1+2^{5/4})}\sin^{2/K}(t) $$
hold for every \(0\leq s<2\pi\), \(0\leq t\leq\pi\).


According to the quasi-invariance of the harmonic measure (see (1.9) in [1]), we have
$$ \Phi_{1/K}\biggl(\cos\frac{t}{2}\biggr)\leq\cos \frac{\theta}{4}\leq\Phi _{K}\biggl(\cos\frac{t}{2}\biggr) $$
for every \(0\leq s<2\pi\), \(0\leq t\leq\pi\), and \(\theta=\gamma (s+t)-\gamma(s-t)\). Since \(\Phi^{2}_{K}(x)+\Phi^{2}_{1/K}(\sqrt{1-x^{2}})=1\) holds for every \(0\leq x\leq1\), this shows that
$$ \Phi_{1/K}\biggl(\sin\frac{t}{2}\biggr)\leq\sin \frac{\theta}{4}\leq\Phi _{K}\biggl(\sin\frac{t}{2}\biggr). $$
Using the Hübner inequalities, (7) and (8), we see that \(4^{1-K}s^{K}\leq\Phi_{1/K}(s)<4^{D(s)(1-K)}s^{K}\) and \(s^{1/K}\leq\Phi_{K}(s)<4^{(1-s^{2})^{\frac{3}{4}}(1-1/K)}s^{1/K}\). Applying (18), (19), and the above two inequalities, we have
$$\sin^{2} \biggl(\frac{\theta}{2} \biggr)\geq4\Phi_{1/K}^{2} \biggl(\sin \frac{t}{2} \biggr)\Phi_{1/K}^{2} \biggl( \cos\frac{t}{2} \biggr)\geq 2^{10(1-K)}\sin^{2K}t $$
$$ \sin^{2} \biggl(\frac{\theta}{2} \biggr)\leq4\Phi_{K}^{2} \biggl(\sin \frac{t}{2} \biggr)\Phi_{K}^{2} \biggl( \cos\frac{t}{2} \biggr) \leq2^{2(1-1/K) \{2 [\vert \cos(\frac{t}{2}) \vert ^{\frac{3}{2}}+\vert \sin(\frac{t}{2})\vert ^{\frac {3}{2}} ]+1 \}}\sin^{\frac{2}{K}}(t). $$
By using Lemma 2 we see that
$$\biggl\vert \cos\frac{t}{2}\biggr\vert ^{\frac{3}{2}}+\biggl\vert \sin\frac {t}{2}\biggr\vert ^{\frac{3}{2}}\leq\sqrt[4]{2}. $$
This implies that
$$2^{10-10K}\sin^{2K}(t)\leq\sin^{2} \biggl( \frac{\theta}{2} \biggr)\leq2^{2(1-1/K)(1+2^{5/4})}\sin^{2/K}(t) $$
hold for every \(0\leq s<2\pi\), \(0\leq t\leq\pi\), and \(\theta=\gamma (s+t)-\gamma(s-t)\).

This completes the proof. □

3 Main results

Theorem 1

Given \(K> 1\), let \(w(z)=P[f](z)=h(z)+\overline{g(z)}\) be a harmonic K-quasiconformal self-mapping of \({\mathbb{D}}\) satisfying \(w(0)=0\) with the boundary function \(f(e^{it})=e^{i\gamma (t)}\), where
$$ h(z)=\sum_{n=1}^{\infty}a_{n}z^{n} \quad \textit{and} \quad g(z)=\sum_{n=1}^{\infty}b_{n}z^{n} $$
are both analytic in \({\mathbb{D}}\). Then
$$|a_{1}|^{2}+|b_{1}|^{2}\geq B_{1}(K), $$
where \(B_{1}(K)\) is given by (9) and satisfies \(\lim_{K\rightarrow1^{+}}B_{1}(K)=1\). For \(n\geq2\),
$$|a_{n}|^{2}+|b_{n}|^{2}\geq B_{n}(K), $$
where \(B_{n}(K)\) is given by (11) and satisfies \(\lim_{n\rightarrow\infty}\lim_{K\rightarrow1^{+}}B_{n}(K)=0\).


Since \(w(z)=P[f](z)=\sum_{n=1}^{\infty}a_{n}z^{n}+\overline{\sum_{n=1}^{\infty}b_{n}z^{n}}\), using Parseval’s relation (cf. [7]) we have
$$\frac{1}{2\pi} \int _{0}^{2\pi}e^{i[\gamma(s+t)-\gamma(s-t)]}\, ds= \sum _{n=1}^{\infty}\bigl(|a_{n}|^{2}e^{2int}+|b_{n}|^{2}e^{-2int} \bigr) $$
for arbitrary \(t\in R\). Taking real parts, we arrive at the formula
$$ 1-2J(t)=\sum_{n=1}^{\infty} \bigl(|a_{n}|^{2}+|b_{n}|^{2}\bigr) \cos(2nt), $$
$$J(t)=\frac{1}{2\pi} \int _{0}^{2\pi}\sin^{2} \biggl( \frac{\gamma(s+t)- \gamma(s-t)}{2} \biggr)\, ds. $$
Since \(w(z)\) is a harmonic K-quasiconformal mapping, by Lemma 3 we have
$$ 2^{10-10K}\sin^{2K}(t)\leq J(t) \leq2^{2(1-1/K)(1+2^{5/4})}\sin^{2/K}(t). $$
Hence \((|a_{n}|^{2}+|b_{n}|^{2})\int_{0}^{\pi}\cos(2nt)(1+\cos(2nt))\, dt= \int _{0}^{\pi}(1-2J(t))(1+\cos(2nt))\, dt\). Using (15) we also obtain
$$\begin{aligned} |a_{n}|^{2}+|b_{n}|^{2} =& \frac{2}{\pi} \biggl(\pi-2 \int_{0}^{\pi}J(t) \bigl(1+\cos (2nt)\bigr)\,dt \biggr) \\ \geq&\frac{2}{\pi} \biggl(\pi-2\cdot2^{2(1-1/K)(1+2^{5/4})} \int _{0}^{\pi}\sin^{\frac{2}{K}}(t) \bigl(1+ \cos(2nt)\bigr)\,dt \biggr) \\ =&2-2^{2(1-1/K)(2+2^{5/4})}\frac{\Gamma(1+\frac{2}{K})}{\Gamma ^{2}(1+\frac{1}{K})}- \frac{2^{2(1-1/K)(2+2^{5/4})}(-1)^{n}\Gamma(1+\frac{2}{K})}{\Gamma (1+\frac{1}{K}-n)\Gamma(1+\frac{1}{K}+n)} \\ :=&\chi(K)+\frac{2^{2(1-1/K)(2+2^{5/4})}(-1)^{n+1}\Gamma(1+\frac {2}{K})}{\Gamma(1+\frac{1}{K}-n)\Gamma(1+\frac{1}{K}+n)}. \end{aligned}$$
For \(n=1\), using the formula \(\Gamma(z+1)=z\Gamma(z)\) and simplifying the above result we obtain the following inequality:
$$|a_{1}|^{2}+|b_{1}|^{2}\geq B_{1}(K):=2-2^{2(1-1/K)(2+2^{5/4})}\frac{2K^{2}\Gamma (\frac{2}{K})}{(K+1)\Gamma^{2}(\frac{1}{K})}. $$
By computation we know that \(B_{1}(K)\) is a decreasing function of K and satisfies
$$\lim_{K\rightarrow1^{+}}B_{1}(K)=1. $$
The above estimate is sharp. Consider the conformal mapping \(w(z)=e^{ix}z\), where \(x\in\mathbb{R}\) is a real number. Then we have \(|a_{1}|+|b_{1}|=1\).
For \(n\geq2\), we have
$$\begin{aligned}& \Gamma\biggl(1+\frac{1}{K}-n\biggr)=\frac{\Gamma(\frac{1}{K})}{(1+\frac {1}{K}-n)(2+\frac{1}{K}-n)\cdots(\frac{1}{K}-1)} = \frac{(-1)^{n-1}\Gamma(\frac{1}{K})}{(n-1-\frac{1}{K})!}, \\& \Gamma\biggl(1+\frac{1}{K}+n\biggr)=\biggl(n+\frac{1}{K}\biggr) \biggl(n+\frac{1}{K}-1\biggr)\cdots \biggl(\frac{1}{K}\biggr)\Gamma \biggl(\frac{1}{K}\biggr)=\frac{1}{K}\Gamma\biggl(\frac {1}{K} \biggr) \biggl(n+\frac{1}{K}\biggr)!, \end{aligned}$$
$$|a_{n}|^{2}+|b_{n}|^{2}\geq\chi(K)+ \frac{2^{2(1-1/K)(2+2^{5/4})}\Gamma (1+\frac{2}{K})(n-1-\frac{1}{K})!}{ \Gamma^{2}(\frac{1}{K})(n+\frac{1}{K})\frac{1}{K}(n-1+\frac {1}{K})!}:=B_{n}(K). $$
By calculating we see that \(\chi(K)\) is a decreasing function of K with \(\chi(1)=0\). The function \(B_{n}(K)\) is a continuous function of K with \(\lim_{K\rightarrow1^{+}}B_{n}(K)=\frac{2}{(n+1)n(n-1)}\). This implies that \(B_{n}(K)>0\) holds for all \(n\geq2\) and some \(K>1\).

The proof is completed. □

Remark 1

By computation we obtain
$$B_{1}(K)>\frac{27}{4\pi^{2}} $$
for all \(1\leq K\leq1.05174\). This shows that under the additional assumption that w is a K-quasiconformal mapping, the lower bound of the inequality (3) can be improved.
By the definition of the Gamma function we see that \(\Gamma(-n)=\infty \) holds for all nonnegative integer numbers n. According to the proof of Theorem 1 we know that for all \(n\geq2\), \(\lim_{K\rightarrow1^{+}}\Gamma(1+\frac{1}{K}-n)=\infty\). Therefore
$$\lim_{K\rightarrow1^{+}}B_{n}(K)=0 $$
holds for all \(n\geq2\).

Let \(t=0\) in equation (21). Then we have \(\sum_{n=1}^{\infty}(|a_{n}|^{2}+|b_{n}|^{2})=1\). The sharp coefficient estimate of \(a_{1}\) and \(b_{1}\) shows that if \(K\rightarrow1^{+}\) then \(|a_{1}|^{2}+|b_{1}|^{2}\geq B_{1}(K)\rightarrow1\). This shows that under the assumptions of Theorem 1 if additionally \(w(z)\) is a conformal self-mapping of \({\mathbb{D}}\) satisfying \(w(0)=0\), then all the coefficients \(b_{n}\) for \(n\geq1\) and \(a_{n}\) for \(n\geq2\) are zeros and \(|a_{1}|=1\), that is, \(w(z)=e^{i\theta}z\) for some \(\theta\in \mathbb{R}\).

Remark 2

In [1] the authors showed that an asymptotically sharp inequality holds for all z in \({\mathbb{D}}\). Our Theorem 1, however, gives an estimate at \(z=0\) only. In this sense, Theorem 1 partly improves the former results.

Theorem 2 shows that \(n^{2}(|a_{n}|^{2}+|b_{n}|^{2})\) is less than or equal to a positive number determined by K.

Theorem 2

Under the assumption of Theorem  1, the coefficients of \(w(z)\) satisfy the following inequality:
$$|a_{n}|^{2}+|b_{n}|^{2}\leq \frac{16}{n^{2}\pi^{2}}K^{6K}2^{5(K-1/K)},\quad n=1,2,\ldots. $$


For every \(z=re^{i\theta}\in{\mathbb{D}}\),
$$w\bigl(re^{i\theta}\bigr)=\sum_{n=1}^{\infty}a_{n}r^{n}e^{in\theta}+ \sum_{n=1}^{\infty}\bar{b}_{n}r^{n}e^{-in\theta}, $$
$$\begin{aligned}& a_{n}r^{n}=\frac{1}{2\pi} \int _{0}^{{2\pi}}w\bigl(re^{i\theta } \bigr)e^{-in\theta}\, d\theta,\quad n=1,2,\ldots, \\& \bar{b}_{n} r^{n}=\frac{1}{2\pi} \int _{0}^{{2\pi }}w\bigl(re^{i\theta} \bigr)e^{in\theta}\, d\theta, \quad n=1,2,\ldots. \end{aligned}$$
For every n we set \(a_{n}=|a_{n}|e^{i\alpha_{n}}\), \(b_{n}=|b_{n}|e^{i\beta _{n}}\), and \(\theta_{n}=\frac{\alpha_{n}+\beta_{n}}{2n}\). Then
$$\begin{aligned} \begin{aligned} \bigl(\vert a_{n}\vert +|b_{n}|\bigr)r^{n}&= \biggl\vert \frac{1}{2\pi} \int _{0}^{{2\pi}}w\bigl(re^{i\theta }\bigr) \bigl[e^{-i\alpha_{n}}e^{-in\theta}+ e^{i\beta_{n}}e^{in\theta}\bigr]\, d \theta\biggr\vert \\ &=\biggl\vert \frac{1}{2\pi} \int _{0}^{{2\pi}}w\bigl(re^{i\theta }\bigr) \bigl[e^{-in(\theta+\theta_{n})}+ e^{in(\theta+\theta_{n})}\bigr]\, d\theta\biggr\vert \\ &=\biggl\vert \frac{1}{\pi} \int _{0}^{{2\pi}}w\bigl(re^{i\theta}\bigr) \cos n( \theta+\theta_{n})\, d\theta\biggr\vert . \end{aligned} \end{aligned}$$
Integrating by parts we have
$$ \bigl(\vert a_{n}\vert +|b_{n}| \bigr)r^{n}=\biggl\vert \frac{1}{n\pi} \int _{0}^{{2\pi }}w_{\theta}\bigl(re^{i\theta} \bigr) \sin n(\theta+\theta_{n})\, d\theta\biggr\vert . $$
In Theorem 2.8 of [10], Kalaj proved that the radial limits of \(w_{\theta}\) and \(w_{r}\) exist almost everywhere and
$$\lim_{r\rightarrow1^{-}}\partial_{\theta}w\bigl(re^{i\theta } \bigr)=\frac{df(e^{i\theta})}{d\theta} $$
for almost every \(z=e^{i\theta}\in\mathbf{T}\). Here f is the boundary function of w. Hence, letting \(r\rightarrow1^{-}\) and using (13), (23) we see that
$$\vert a_{n}\vert +\vert b_{n}\vert \leq \frac{1}{n\pi} \int_{0}^{2\pi}\bigl\vert f' \bigl(e^{i\theta}\bigr)\bigr\vert \bigl\vert \sin n(\theta+ \theta_{n})\bigr\vert \, d\theta=\frac{4K^{3K}2^{5(K-1/K)/2}}{n\pi}. $$
It shows that \(|a_{n}|^{2}+|b_{n}|^{2}\leq(|a_{n}|+|b_{n}|)^{2}\leq\frac {16K^{6K}2^{5(K-1/K)}}{n^{2}\pi^{2}}:=A_{n}(K)\).

The proof is completed. □



The author of this work is supported by NNSF of China (Nos. 11301195, 11501220) and the China Scholarship Council and a research foundation of Huaqiao University (Project No. 2014KJTD14). The author would like to express her appreciation to Professor Jian-Feng Zhu and the referee for their helpful advice.

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Authors’ Affiliations

School of Mathematical Sciences, Huaqiao University, Quanzhou, China


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