An asymptotically sharp coefficients estimate for harmonic K-quasiconformal mappings

Abstract

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.

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

$$ h(z)=\sum_{n=0}^{\infty}a_{n}z^{n} \quad \mbox{and} \quad g(z)=\sum_{n=1}^{\infty}b_{n}z^{n} $$

(1)

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

$$ \bigl\vert w_{z}(0)\bigr\vert ^{2}+\bigl\vert w_{\bar{z}}(0)\bigr\vert ^{2}\geq c $$

(2)

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

$$ |a_{1}|^{2}+|b_{1}|^{2} \geq\frac{27}{4\pi^{2}}. $$

(3)

The lower bound $\frac{27}{4\pi^{2}}$ is the best possible.

Let

$$p(r,x-\varphi)=\frac{1-r^{2}}{2\pi(1-2r\cos(x-\varphi)+r^{2})} $$

denote the Poisson kernel, then every bounded harmonic mapping w defined in ${\mathbb{D}}$ has the following representation:

$$ w(z)=P[f](z)= \int _{0}^{2\pi}p(r,x-\varphi)f\bigl(e^{ix} \bigr)\, dx, $$

(4)

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

$$ \bigl\vert \partial_{z} w(z)\bigr\vert ^{2}+\bigl\vert \partial_{\bar{z}}w(z)\bigr\vert ^{2}\geq\frac{1}{4} \biggl(1+\frac{1}{K} \biggr)^{2} \max \biggl\{ \frac{4}{\pi^{2}}, L_{K}^{2} \biggr\} $$

(5)

holds for every $z\in{\mathbb{D}}$, where

$$ L_{K}:=\frac{2}{\pi} \int _{0}^{\frac{1}{\sqrt{2}}} \frac{d (\Phi_{1/K}(s)^{2} )}{s\sqrt{1-s^{2}}} $$

(6)

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 Grötzsch’s 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

$$ 4^{1-K}s^{K}\leq\Phi_{1/K}(s)< 4^{D(s)(1-K)}s^{K} $$

(7)

and

$$ s^{1/K}\leq\Phi_{K}(s)< 4^{(1-s^{2})^{\frac{3}{4}}(1-1/K)}s^{1/K}, $$

(8)

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:

$$ |a_{1}|^{2}+|b_{1}|^{2} \geq B_{1}(K):=2-2^{2(1-1/K)(2+2^{5/4})}\frac{2K^{2}\Gamma (\frac{2}{K})}{(K+1)\Gamma^{2}(\frac{1}{K})} $$

(9)

which satisfies $\lim_{K\rightarrow1^{+}}B_{1}(K)=1$, where Γ is the gamma function.

For $n\geq2$ we have

$$ |a_{n}|^{2}+|b_{n}|^{2} \geq B_{n}(K), $$

(10)

where

$$\begin{aligned}& B_{n}(K):=\chi(K)+\frac{2^{2(1-1/K)(2+2^{5/4})}\Gamma(1+\frac {2}{K})(n-1-\frac{1}{K})!}{ \Gamma^{2}(\frac{1}{K})(n+\frac{1}{K})\frac{1}{K}(n-1+\frac{1}{K})!}, \\& \biggl(n-1-\frac{1}{K}\biggr)!:=\biggl(n-1-\frac{1}{K}\biggr) \biggl(n-2-\frac{1}{K}\biggr)\cdots \biggl(1-\frac{1}{K}\biggr), \\& \biggl(n-1+\frac{1}{K}\biggr)!:=\biggl(n-1+\frac{1}{K}\biggr) \biggl(n-2+\frac{1}{K}\biggr)\cdots \biggl(1+\frac{1}{K}\biggr), \end{aligned}$$

(11)

and

$$ \chi(K):=2-\frac{2^{2(1-1/K)(2+2^{5/4})}\Gamma(1+\frac {2}{K})}{\Gamma^{2}(1+\frac{1}{K})} $$

(12)

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:

$$ \frac{2^{5(1-K^{2})/2}}{(K^{2}+K-1)^{K}}\leq\bigl\vert f '(z)\bigr\vert \leq K^{3K}2^{5(K-\frac{1}{K})/2} $$

(13)

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:

$$ |a_{n}|^{2}+|b_{n}|^{2} \leq A_{n}(K):=\frac{16}{n^{2}\pi^{2}}K^{6K}2^{5(K-1/K)}. $$

(14)

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

$$ \int_{0}^{\pi}\sin^{\frac{2}{K}}(t)\cos(2nt)\, dt= \frac{\pi}{4^{\frac {1}{K}}}\frac{(-1)^{n}\Gamma (1+\frac{2}{K} )}{\Gamma (1+\frac{1}{K}-n )\Gamma (1+\frac{1}{K}+n )} $$

(15)

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

$$ \max_{0\leq t\leq2\pi}\varphi(t)=\varphi \biggl( \frac{\pi }{2} \biggr)=\sqrt[4]{2}. $$

(16)

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

$$ 2^{10-10K}\sin^{2K}(t)\leq\sin^{2} \biggl(\frac{\theta}{2} \biggr)\leq2^{2(1-1/K)(1+2^{5/4})}\sin^{2/K}(t) $$

(17)

hold for every $0\leq s<2\pi$, $0\leq t\leq\pi$.

Proof

According to the quasi-invariance of the harmonic measure (see (1.9) in [1]), we have

$$ \Phi_{1/K}\biggl(\cos\frac{t}{2}\biggr)\leq\cos \frac{\theta}{4}\leq\Phi _{K}\biggl(\cos\frac{t}{2}\biggr) $$

(18)

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

$$ \Phi_{1/K}\biggl(\sin\frac{t}{2}\biggr)\leq\sin \frac{\theta}{4}\leq\Phi _{K}\biggl(\sin\frac{t}{2}\biggr). $$

(19)

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

$$\sin^{2} \biggl(\frac{\theta}{2} \biggr)\geq4\Phi_{1/K}^{2} \biggl(\sin \frac{t}{2} \biggr)\Phi_{1/K}^{2} \biggl( \cos\frac{t}{2} \biggr)\geq 2^{10(1-K)}\sin^{2K}t $$

and

$$ \sin^{2} \biggl(\frac{\theta}{2} \biggr)\leq4\Phi_{K}^{2} \biggl(\sin \frac{t}{2} \biggr)\Phi_{K}^{2} \biggl( \cos\frac{t}{2} \biggr) \leq2^{2(1-1/K) \{2 [\vert \cos(\frac{t}{2}) \vert ^{\frac{3}{2}}+\vert \sin(\frac{t}{2})\vert ^{\frac {3}{2}} ]+1 \}}\sin^{\frac{2}{K}}(t). $$

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

$$ h(z)=\sum_{n=1}^{\infty}a_{n}z^{n} \quad \textit{and} \quad g(z)=\sum_{n=1}^{\infty}b_{n}z^{n} $$

(20)

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 $\lim_{n\rightarrow\infty}\lim_{K\rightarrow1^{+}}B_{n}(K)=0$.

Proof

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

$$ 1-2J(t)=\sum_{n=1}^{\infty} \bigl(|a_{n}|^{2}+|b_{n}|^{2}\bigr) \cos(2nt), $$

(21)

where

$$J(t)=\frac{1}{2\pi} \int _{0}^{2\pi}\sin^{2} \biggl( \frac{\gamma(s+t)- \gamma(s-t)}{2} \biggr)\, ds. $$

Since $w(z)$ is a harmonic K-quasiconformal mapping, by Lemma 3 we have

$$ 2^{10-10K}\sin^{2K}(t)\leq J(t) \leq2^{2(1-1/K)(1+2^{5/4})}\sin^{2/K}(t). $$

(22)

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

$$\begin{aligned}& \Gamma\biggl(1+\frac{1}{K}-n\biggr)=\frac{\Gamma(\frac{1}{K})}{(1+\frac {1}{K}-n)(2+\frac{1}{K}-n)\cdots(\frac{1}{K}-1)} = \frac{(-1)^{n-1}\Gamma(\frac{1}{K})}{(n-1-\frac{1}{K})!}, \\& \Gamma\biggl(1+\frac{1}{K}+n\biggr)=\biggl(n+\frac{1}{K}\biggr) \biggl(n+\frac{1}{K}-1\biggr)\cdots \biggl(\frac{1}{K}\biggr)\Gamma \biggl(\frac{1}{K}\biggr)=\frac{1}{K}\Gamma\biggl(\frac {1}{K} \biggr) \biggl(n+\frac{1}{K}\biggr)!, \end{aligned}$$

then

$$|a_{n}|^{2}+|b_{n}|^{2}\geq\chi(K)+ \frac{2^{2(1-1/K)(2+2^{5/4})}\Gamma (1+\frac{2}{K})(n-1-\frac{1}{K})!}{ \Gamma^{2}(\frac{1}{K})(n+\frac{1}{K})\frac{1}{K}(n-1+\frac {1}{K})!}:=B_{n}(K). $$

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:

$$|a_{n}|^{2}+|b_{n}|^{2}\leq \frac{16}{n^{2}\pi^{2}}K^{6K}2^{5(K-1/K)},\quad n=1,2,\ldots. $$

Proof

For every $z=re^{i\theta}\in{\mathbb{D}}$,

$$w\bigl(re^{i\theta}\bigr)=\sum_{n=1}^{\infty}a_{n}r^{n}e^{in\theta}+ \sum_{n=1}^{\infty}\bar{b}_{n}r^{n}e^{-in\theta}, $$

hence

$$\begin{aligned}& a_{n}r^{n}=\frac{1}{2\pi} \int _{0}^{{2\pi}}w\bigl(re^{i\theta } \bigr)e^{-in\theta}\, d\theta,\quad n=1,2,\ldots, \\& \bar{b}_{n} r^{n}=\frac{1}{2\pi} \int _{0}^{{2\pi }}w\bigl(re^{i\theta} \bigr)e^{in\theta}\, d\theta, \quad n=1,2,\ldots. \end{aligned}$$

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

$$ \bigl(\vert a_{n}\vert +|b_{n}| \bigr)r^{n}=\biggl\vert \frac{1}{n\pi} \int _{0}^{{2\pi }}w_{\theta}\bigl(re^{i\theta} \bigr) \sin n(\theta+\theta_{n})\, d\theta\biggr\vert . $$

(23)

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. □