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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 inequality Hübner inequalities coefficients estimate harmonic quasiconformal mapping


30C62 30C20 30F15

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.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

School of Mathematical Sciences, Huaqiao University


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