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Several properties of a class of generalized harmonic mappings

Bo-Yong Long  Qi-Han Wang

School of Mathematical Sciences, Anhui University, Hefei 230601, China

footnotetext: Supported by NSFC (No.12271001) and Natural Science Foundation of Anhui Province (2308085MA03), China. E-mail: [email protected][email protected]

Abstract:  We call the solution of a kind of second order homogeneous partial differential equation as real kernel α\alpha-harmonic mappings. In this paper, the representation theorem, the Lipschitz continuity, the univalency and the related problems of the real kernel α\alpha-harmonic mappings are explored.
Keywords: Weighted Laplacian operator; univalency; Polyharmonic mappings; Lipschitz continuity; Gauss hypergeometric function
2010 Mathematics Subject Classification: Primary 30C45 Secondary 33C05, 30B10

1 Introduction

Let 𝔻\mathbb{D} be the open unit disk and 𝕋\mathbb{T} the unit circle. For α\alpha\in\mathbb{R} and z𝔻z\in\mathbb{D}, let

Tα=α24(1|z|2)α1+α2(1|z|2)α1(zz+z¯z¯)+(1|z|2)αT_{\alpha}=-\frac{\alpha^{2}}{4}(1-|z|^{2})^{-\alpha-1}+\frac{\alpha}{2}(1-|z|^{2})^{-\alpha-1}\left(z\frac{\partial}{\partial z}+\bar{z}\frac{\partial}{\partial\bar{z}}\right)+(1-|z|^{2})^{-\alpha}\triangle

be the second order elliptic partial differential operator, where \triangle is the usual complex Laplacian operator

:=42zz¯=2x2+2y2.\triangle:=4\frac{\partial^{2}}{\partial z\partial\bar{z}}=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}.

The corresponding partial differential equation is

Tα(u)=0in𝔻.\displaystyle T_{\alpha}(u)=0\quad\mbox{in}\,\mathbb{D}. (1.1)

The associated Dirichlet boundary value problem is

{Tα(u)=0in𝔻,u=uon𝕋.\;\left\{\begin{array}[]{rr}T_{\alpha}(u)=0&\quad\mbox{in}\,\mathbb{D},\\ u=u^{*}&\quad\mbox{on}\,\mathbb{T}.\end{array}\right. (1.2)

Here, the boundary data u𝔇(𝕋)u^{*}\in\mathfrak{D}^{\prime}(\mathbb{T}) is a distribution on the boundary of 𝔻\mathbb{D}, and the boundary condition in (1.2) is interpreted in the distributional sense that uruu_{r}\rightarrow u^{*} in 𝔇(𝕋)\mathfrak{D}^{\prime}(\mathbb{T}) as r1r\rightarrow 1^{-}, where

ur(eiθ)=u(reiθ),eiθ𝕋,u_{r}(e^{i\theta})=u(re^{i\theta}),\quad e^{i\theta}\in\mathbb{T},

for r[0,1)r\in[0,1). In [24], Olofsson proved that, for the parameter α>1\alpha>-1, if a function u𝒞2(𝔻)u\in\mathcal{C}^{2}(\mathbb{D}) satisfies (1.1) with limr1ur=u𝔇(𝕋)\lim_{r\rightarrow 1^{-}}u_{r}=u^{*}\in\mathfrak{D}^{\prime}(\mathbb{T}) , then it has the form of Poisson type integral

u(z)=12π02πKα(zeiτ))u(eiτ)𝑑τ,forz𝔻,u(z)=\frac{1}{2\pi}\int_{0}^{2\pi}K_{\alpha}(ze^{-i\tau)})u^{*}(e^{i\tau})d\tau,\quad\mbox{for}\,\,z\in\mathbb{D}, (1.3)

where

Kα(z)=cα(1|z|2)α+1|1z|α+2,K_{\alpha}(z)=c_{\alpha}\frac{(1-|z|^{2})^{\alpha+1}}{|1-z|^{\alpha+2}}, (1.4)

cα=Γ2(α/2+1)/Γ(1+α)c_{\alpha}=\Gamma^{2}(\alpha/2+1)/\Gamma(1+\alpha) and Γ(s)=0ts1et𝑑t\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}dt for s>0s>0 is the standard Gamma function. If α1\alpha\leq-1, u𝒞2(𝔻)u\in\mathcal{C}^{2}(\mathbb{D}) satisfies (1.1), and the boundary limit u=limr1uru^{*}=\lim_{r\rightarrow 1^{-}}u_{r} exists in 𝔇(𝕋)\mathfrak{D}^{\prime}(\mathbb{T}), then u(z)=0u(z)=0 for all z𝔻z\in\mathbb{D}. So, in the following of this paper, we always assume that α>1\alpha>-1.

For c0,1,2,c\neq 0,\,-1,\,-2,..., the Gauss hypergeometric function is defined by the series

F(a,b;c;x)=n=0(a)n(b)n(c)nxnn!F(a,b;c;x)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{x^{n}}{n!}

for |x|<1|x|<1, and has a continuation to the complex plane with branch points at 11 and \infty, where (a)0=1(a)_{0}=1 and (a)n=a(a+1)(a+n1)(a)_{n}=a(a+1)\cdots(a+n-1) for n=1,2,n=1,2,... are the Pochhammer symbols. Obviously, for n=0,1,2,n=0,1,2,..., (a)n=Γ(a+n)/Γ(a)(a)_{n}=\Gamma(a+n)/\Gamma(a). It is easily to verified that

ddxF(a,b;c;x)=abcF(a+1,b+1;c+1;x).\frac{d}{dx}F(a,b;c;x)=\frac{ab}{c}F(a+1,b+1;c+1;x). (1.5)

Furthermore, it holds that (cf.[3] )

limx1F(a,b;c;x)=Γ(c)Γ(cab)Γ(ca)Γ(cb)\lim_{x\rightarrow 1}F(a,b;c;x)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} (1.6)

if Re(cab)>0Re(c-a-b)>0.

The following Lemma 1.1 involves the determination of monotonicity of Gauss hypergeometric functions.

Lemma 1.1.

[24] Let c>0c>0, aca\leq c, bcb\leq c and ab0ab\leq 0 (ab0)(ab\geq 0). Then the function F(a,b;c;x)F(a,b;c;x) is decreasing (increasing) on x(0,1)x\in(0,1).

The following result of [24] is the homogeneous expansion of solutions of (1.1).

Theorem 1.2.

[24] Let α\alpha\in\mathbb{R} and u𝒞2(𝔻)u\in\mathcal{C}^{2}(\mathbb{D}). Then uu satisfies (1.1) if and only if it has a series expansion of the form

u(z)=k=0ckF(α2,kα2;k+1;|z|2)zk+k=1ckF(α2,kα2;k+1;|z|2)z¯k,z𝔻,u(z)=\sum_{k=0}^{\infty}c_{k}F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;|z|^{2})z^{k}+\sum_{k=1}^{\infty}c_{-k}F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;|z|^{2})\bar{z}^{k},\quad z\in\mathbb{D}, (1.7)

for some sequence {ck}\{c_{k}\}_{-\infty}^{\infty} of complex number satisfying

lim|k|sup|ck|1|k|1.\lim_{|k|\rightarrow\infty}\sup|c_{k}|^{\frac{1}{|k|}}\leq 1. (1.8)

In particular, the expansion (1.7), subject to (1.8), converges in 𝒞(𝔻)\mathcal{C}^{\infty}(\mathbb{D}), and every solution uu of (1.1) is 𝒞\mathcal{C}^{\infty}-smooth in 𝔻\mathbb{D}.

Let

v(z)=k=0ckzk+k=1ckz¯k,z𝔻.v(z)=\sum_{k=0}^{\infty}c_{k}z^{k}+\sum_{k=1}^{\infty}c_{-k}\bar{z}^{k},\quad z\in\mathbb{D}. (1.9)

It is obvious that v(z)v(z) is a harmonic mapping, i.e., v=0\triangle v=0. We observe that u(z)u(z) of (1.7) and v(z)v(z) have same coefficient sequence {ck}\{c_{k}\}_{-\infty}^{\infty}. Actually, if α=0\alpha=0, then u(z)=v(z)u(z)=v(z).

Observe that the kernel KαK_{\alpha} in (1.4) is real. We call uu of (1.3) or (1.7) as real kernel α\alpha-harmonic mappings. Furthermore, suppose u(z)u(z) and v(z)v(z) have the expansions of (1.7) and (1.9), respectively. We call v(z)v(z) as the corresponding harmonic mapping of u(z)u(z). Conversely, we call u(z)u(z) as the corresponding real kernel α\alpha-harmonic mapping of v(z)v(z).

If we take α=2(p1)\alpha=2(p-1), then a real kernel α\alpha-harmonic mapping uu is polyharmonic (or pp-harmonic), where p{1,2,}p\in\{1,2,...\} (cf.[1, 6, 13, 5, 27, 11, 2, 15]). In particular, if α=0\alpha=0, then uu is harmonic (cf.[10, 18, 19, 20]). Thus, the real kernel α\alpha-harmonic mapping is a kind of generalization of classical harmonic mapping. Furthermore, by [25], we know that it is related to standard weighted harmonic mappings. For the related discussion on standard weighted harmonic mappings, see [8, 17, 16, 23].

For the real kernel α\alpha-harmonic mappings, the Schwarz-Pick type estimates and coefficient estimates are obtained in [7]; the starlikeness, convexity and Landau type theorem are studied in [22]; the sharp Heinz type inequality is established and the extremal functions of Schwartz type lemma are explored in [21]; the Lipschitz continuity with respect to the distance ratio metric is proved in [14]. In [12], using the properties of the real kernel α\alpha-harmonic mappings, the authors established some Schwarz type lemmas for mappings satisfying a class of inhomogeneous biharmonic Dirichlet problem.

In this paper, we continue to study the properties of the real kernel α\alpha-harmonic mappings. The main idea of this paper is that by establishing the relationship between harmonic mapping and the corresponding real kernel α\alpha-harmonic mapping, we use the harmonic mapping to characterize the corresponding real kernel α\alpha-harmonic mapping. In section 2, for a nonnegative even number α\alpha, we get an explicit representation theorem which determines the relation between the real kernel α\alpha-harmonic mapping and the corresponding harmonic mapping. As its application, in section 3, we show that the Lipschitz continuity of a real kernel α\alpha-harmonic mapping is determined by the corresponding harmonic mapping. In section 4, for a subclass of the real kernel α\alpha-harmonic mappings, we discuss its univalency and explore its Radó-Kneser-Choquet type theorem. In section 5, we explore the influence of parameters α\alpha on the image area of the real kernel α\alpha-harmonic mappings.

2 Representation theorem

Theorem 2.1.

Let v(z)=h(z)+g(z)¯=k=0ckzk+k=1ckz¯kv(z)=h(z)+\overline{g(z)}=\sum_{k=0}^{\infty}c_{k}z^{k}+\sum_{k=1}^{\infty}c_{-k}\bar{z}^{k} be a harmonic mapping defined on the unit disk 𝔻\mathbb{D}. If α2=p1\frac{\alpha}{2}=p-1 is a nonnegative integer, then the corresponding real kernel α\alpha-harmonic mapping of v(z)v(z) can be represented by

u(z)=n=0p1|z|2n(1p)nn!(In+Jn¯),\displaystyle u(z)=\sum_{n=0}^{p-1}|z|^{2n}\frac{(1-p)_{n}}{n!}(I_{n}+\overline{J_{n}}), (2.1)

where InI_{n} and JnJ_{n} satisfy the recurrence formulas

In=In1p0zzn1In1𝑑zzn,\displaystyle I_{n}=I_{n-1}-p\frac{\int_{0}^{z}z^{n-1}I_{n-1}dz}{z^{n}}, (2.2)
Jn=Jn1p0zzn1Jn1𝑑zznn=1,2,,p1,\displaystyle J_{n}=J_{n-1}-p\frac{\int_{0}^{z}z^{n-1}J_{n-1}dz}{z^{n}}\qquad n=1,2,...,p-1, (2.3)

I0=h(z)I_{0}=h(z), and J0=g(z)J_{0}=g(z).

Proof.

Let H(z)=k=0ckF(α2,kα2;k+1;|z|2)zkH(z)=\sum_{k=0}^{\infty}c_{k}F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;|z|^{2})z^{k} and G(z)=k=1ck¯F(α2,kα2;k+1;|z|2)zkG(z)=\sum_{k=1}^{\infty}\overline{c_{-k}}F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;|z|^{2})z^{k}. Then by the assumption and (1.7), we have

u(z)=H(z)+G(z)¯.\displaystyle u(z)=H(z)+\overline{G(z)}. (2.4)

When α2=p1\frac{\alpha}{2}=p-1, rewrite H(z)H(z) as

H(z)=k=0ckF(1p,k+1p;k+1;|z|2)zk\displaystyle H(z)=\sum_{k=0}^{\infty}c_{k}F(1-p,k+1-p;k+1;|z|^{2})z^{k}
=k=0ckzk(n=0(1p)n(k+1p)n(k+1)n|z|2nn!)\displaystyle\qquad\,=\sum_{k=0}^{\infty}c_{k}z^{k}\left(\sum_{n=0}^{\infty}\frac{(1-p)_{n}(k+1-p)_{n}}{(k+1)_{n}}\frac{|z|^{2n}}{n!}\right)
=k=0ckzk(n=0p1(1p)n(k+1p)n(k+1)n|z|2nn!)\displaystyle\qquad\,=\sum_{k=0}^{\infty}c_{k}z^{k}\left(\sum_{n=0}^{p-1}\frac{(1-p)_{n}(k+1-p)_{n}}{(k+1)_{n}}\frac{|z|^{2n}}{n!}\right)
=n=0p1|z|2n(1p)nn!In,\displaystyle\qquad\,=\sum_{n=0}^{p-1}|z|^{2n}\frac{(1-p)_{n}}{n!}I_{n}, (2.5)

where

In=k=0(k+1p)n(k+1)nckzk.I_{n}=\sum_{k=0}^{\infty}\frac{(k+1-p)_{n}}{(k+1)_{n}}c_{k}z^{k}.

Because

(k+1p)n(k+1)n=(k+1p)n1(k+np)(k+1)n1(k+n)\displaystyle\frac{(k+1-p)_{n}}{(k+1)_{n}}=\frac{(k+1-p)_{n-1}(k+n-p)}{(k+1)_{n-1}(k+n)}
=(k+1p)n1(k+1)n1pk+n(k+1p)n1(k+1)n1,\displaystyle\qquad\qquad\quad\,\,\,=\frac{(k+1-p)_{n-1}}{(k+1)_{n-1}}-\frac{p}{k+n}\frac{(k+1-p)_{n-1}}{(k+1)_{n-1}},

we can get

In=k=0(k+1p)n1(k+1)n1ckzkk=0pk+n(k+1p)n1(k+1)n1ckzk\displaystyle I_{n}=\sum_{k=0}^{\infty}\frac{(k+1-p)_{n-1}}{(k+1)_{n-1}}c_{k}z^{k}-\sum_{k=0}^{\infty}\frac{p}{k+n}\frac{(k+1-p)_{n-1}}{(k+1)_{n-1}}c_{k}z^{k}
=In1p0zzn1In1𝑑zzn.\displaystyle\quad=I_{n-1}-p\frac{\int_{0}^{z}z^{n-1}I_{n-1}dz}{z^{n}}.

This is (2.2).

Similarly, we can get

G(z)=n=0p1|z|2n(1p)nn!Jn,\displaystyle G(z)=\sum_{n=0}^{p-1}|z|^{2n}\frac{(1-p)_{n}}{n!}J_{n}, (2.6)

where JnJ_{n} is defined as in (2.3). Therefore, equation (2.1) follows from equations (2.4)-(2.6). ∎

Example 2.1.

From the recurrence formula (2.1), we have the following:

(i)  When α=0\alpha=0, i.e. p=1p=1,

u(z)=v(z);\displaystyle u(z)=v(z);

(ii)  When α=2\alpha=2, i.e. p=2p=2,

u(z)=h+g¯|z|2(h20zh(z)𝑑zz+g20zg(z)𝑑zz¯)\displaystyle u(z)=h+\bar{g}-|z|^{2}\left(h-2\frac{\int_{0}^{z}h(z)dz}{z}+\overline{g-2\frac{\int_{0}^{z}g(z)dz}{z}}\right)
=k=0ckzk+k=1ckz¯k|z|2(k=0ckk1k+1zk+k=1ckk1k+1z¯k);\displaystyle\qquad=\sum_{k=0}^{\infty}c_{k}z^{k}+\sum_{k=1}^{\infty}c_{-k}\bar{z}^{k}-|z|^{2}\left(\sum_{k=0}^{\infty}c_{k}\frac{k-1}{k+1}z^{k}+\sum_{k=1}^{\infty}c_{-k}\frac{k-1}{k+1}\bar{z}^{k}\right); (2.7)

(iii)  When α=4\alpha=4, i.e. p=3p=3,

u(z)=h+g¯2|z|2(h30zh(z)𝑑zz+g30zg(z)𝑑zz¯)\displaystyle u(z)=h+\bar{g}-2|z|^{2}\left(h-3\frac{\int_{0}^{z}h(z)dz}{z}+\overline{g-3\frac{\int_{0}^{z}g(z)dz}{z}}\right)
+|z|4(h30zh(z)𝑑zz30zzh(z)𝑑zz2+90z0zh(z)𝑑z𝑑zz2\displaystyle\qquad\quad+|z|^{4}\left(h-3\frac{\int_{0}^{z}h(z)dz}{z}-3\frac{\int_{0}^{z}zh(z)dz}{z^{2}}+9\frac{\int_{0}^{z}\int_{0}^{z}h(z)dzdz}{z^{2}}\right.
+g30zg(z)𝑑zz30zzg(z)𝑑zz2+90z0zg(z)𝑑z𝑑zz2¯)\displaystyle\qquad\quad\left.+\overline{g-3\frac{\int_{0}^{z}g(z)dz}{z}-3\frac{\int_{0}^{z}zg(z)dz}{z^{2}}+9\frac{\int_{0}^{z}\int_{0}^{z}g(z)dzdz}{z^{2}}}\right)
=k=0ckzk+k=1ckz¯k2|z|2(k=0ckk2k+1zk+k=1ckk2k+1z¯k)\displaystyle\qquad=\sum_{k=0}^{\infty}c_{k}z^{k}+\sum_{k=1}^{\infty}c_{-k}\bar{z}^{k}-2|z|^{2}\left(\sum_{k=0}^{\infty}c_{k}\frac{k-2}{k+1}z^{k}+\sum_{k=1}^{\infty}c_{-k}\frac{k-2}{k+1}\bar{z}^{k}\right)
+|z|4(k=0ck(k1)(k2)(k+1)(k+2)zk+k=1ck(k1)(k2)(k+1)(k+2)z¯k).\displaystyle\quad\qquad+|z|^{4}\left(\sum_{k=0}^{\infty}c_{k}\frac{(k-1)(k-2)}{(k+1)(k+2)}z^{k}+\sum_{k=1}^{\infty}c_{-k}\frac{(k-1)(k-2)}{(k+1)(k+2)}\bar{z}^{k}\right).

3 Lipschitz continuity

Theorem 3.1.

Let u(z)u(z) be the corresponding real kernel α\alpha-harmonic mapping of v(z)=h+g¯v(z)=h+\bar{g} on the unit disk 𝔻\mathbb{D}. If v(z)v(z) is Lipschitz continuous on the unit disk 𝔻\mathbb{D} and α2=p1\frac{\alpha}{2}=p-1 is a nonnegative integer, then uu is Lipschitz continuous on the unit disk 𝔻\mathbb{D} as well.

Proof.

By the assumption and (2.1), it is sufficient to prove that InI_{n} and JnJ_{n} are Lipschitz continuous on the unit disk 𝔻\mathbb{D} for n=0,1,2,,p1n=0,1,2,...,p-1. In the following, we just prove the Lipschitz continuity of InI_{n}. The case of JnJ_{n} is similar.

Observe that I0=h(z)I_{0}=h(z) is holomorphic on 𝔻\mathbb{D}. Then by the recurrence formula (2.2), it is easy to see that all InI_{n} are holomorphic on 𝔻\mathbb{D}. It follows that all InI^{\prime}_{n} are holomorphic on 𝔻\mathbb{D} too, where

In=In1pznIn1n0zzn1In1𝑑zzn+1,n=1,2,,p1.\displaystyle I^{\prime}_{n}=I^{\prime}_{n-1}-p\frac{z^{n}I_{n-1}-n\int_{0}^{z}z^{n-1}I_{n-1}dz}{z^{n+1}},\qquad n=1,2,...,p-1. (3.1)

Taking account of the maximum modulus principle of holomorphic functions, from equations (2.2) and (3.1), we get

supz𝔻|In|supz𝔻|In1|+psupz𝔻|In1|=(p+1)supz𝔻|In1|\displaystyle\sup_{z\in\mathbb{D}}|I_{n}|\leq\sup_{z\in\mathbb{D}}|I_{n-1}|+p\sup_{z\in\mathbb{D}}|I_{n-1}|=(p+1)\sup_{z\in\mathbb{D}}|I_{n-1}|

and

supz𝔻|In|supz𝔻|In1|+psupz𝔻|In1|+npsupz𝔻|In1|=supz𝔻|In1|+(n+1)psupz𝔻|In1|,\displaystyle\sup_{z\in\mathbb{D}}|I^{\prime}_{n}|\leq\sup_{z\in\mathbb{D}}|I^{\prime}_{n-1}|+p\sup_{z\in\mathbb{D}}|I_{n-1}|+np\sup_{z\in\mathbb{D}}|I_{n-1}|=\sup_{z\in\mathbb{D}}|I^{\prime}_{n-1}|+(n+1)p\sup_{z\in\mathbb{D}}|I_{n-1}|,

respectively. It follows that

supz𝔻|In|(p+1)nsupz𝔻|I0|\displaystyle\sup_{z\in\mathbb{D}}|I_{n}|\leq(p+1)^{n}\sup_{z\in\mathbb{D}}|I_{0}|

and

supz𝔻|In|supz𝔻|In1|+(n+1)psupz𝔻|In1|\displaystyle\sup_{z\in\mathbb{D}}|I^{\prime}_{n}|\leq\sup_{z\in\mathbb{D}}|I^{\prime}_{n-1}|+(n+1)p\sup_{z\in\mathbb{D}}|I_{n-1}|
supz𝔻|In2|+npsupz𝔻|In2|+(n+1)psupz𝔻|In1|\displaystyle\quad\qquad\leq\sup_{z\in\mathbb{D}}|I^{\prime}_{n-2}|+np\sup_{z\in\mathbb{D}}|I_{n-2}|+(n+1)p\sup_{z\in\mathbb{D}}|I_{n-1}|
\displaystyle\quad\qquad\leq...
supz𝔻|I0|+pi=1n(i+1)supz𝔻|Ii1|\displaystyle\quad\qquad\leq\sup_{z\in\mathbb{D}}|I^{\prime}_{0}|+p\sum_{i=1}^{n}(i+1)\sup_{z\in\mathbb{D}}|I_{i-1}|
supz𝔻|I0|+pi=1n(i+1)(p+1)i1supz𝔻|I0|.\displaystyle\quad\qquad\leq\sup_{z\in\mathbb{D}}|I^{\prime}_{0}|+p\sum_{i=1}^{n}(i+1)(p+1)^{i-1}\sup_{z\in\mathbb{D}}|I_{0}|. (3.2)

Because v=h+g¯v=h+\bar{g} is Lipschitz, there exists a constant MM such that

|h|=|I0|M\displaystyle|h^{\prime}|=|I^{\prime}_{0}|\leq M (3.3)

for z𝔻z\in\mathbb{D}. It follows that

supz𝔻|I0|=supz𝔻|h|M.\displaystyle\sup_{z\in\mathbb{D}}|I_{0}|=\sup_{z\in\mathbb{D}}|h|\leq M. (3.4)

Therefore, by inequalities (3.2)-(3.4), we get that there exists a constant C=C(M,p,n)C=C(M,p,n), such that

supz𝔻|In|(1+pi=1n(i+1)(p+1)i1)M=:C\displaystyle\sup_{z\in\mathbb{D}}|I^{\prime}_{n}|\leq(1+p\sum_{i=1}^{n}(i+1)(p+1)^{i-1})M=:C

for n=1,2,,p1n=1,2,...,p-1. It means that InI_{n} is Lipschitz continuous on 𝔻\mathbb{D}.

4 Univalency of a subclass of real kernel α\alpha-harmonic mappings

In the rest of this paper, we use the following notations. Let α>1\alpha>-1, z=reiθz=re^{i\theta}, and

t=|z|2=r2,\displaystyle t=|z|^{2}=r^{2},
F=Fk=F(α2,kα2;k+1;t),k=1,2,.,\displaystyle F=F_{k}=F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;t),\quad k=1,2,....,
Ft=Fk,t=Fk,t(α2,kα2;k+1;t)=dFkdt=dFdt.\displaystyle F_{t}=F_{k,t}=F_{k,t}(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;t)=\frac{dF_{k}}{dt}=\frac{dF}{dt}.

Furthermore, let

Fk(1)=limt1F(α2,kα2;k+1;t).\displaystyle F_{k}(1)=\lim_{t\rightarrow 1^{-}}F(-\frac{\alpha}{2},k-\frac{\alpha}{2};k+1;t).

Then by (1.6), we have

Fk(1)=Γ(k+1)Γ(1+α)Γ(k+1+α2)Γ(1+α2).\displaystyle F_{k}(1)=\frac{\Gamma(k+1)\Gamma(1+\alpha)}{\Gamma(k+1+\frac{\alpha}{2})\Gamma(1+\frac{\alpha}{2})}. (4.1)
Lemma 4.1.

Let rnr_{n} and sns_{n}(n=0,1,2,)(n=0,1,2,...) be real numbers, and let the power series

R(x)=n=0rnxnandS(x)=n=0snxnR(x)=\sum_{n=0}^{\infty}r_{n}x^{n}\quad\mbox{and}\quad S(x)=\sum_{n=0}^{\infty}s_{n}x^{n}

be convergent for |x|<r|x|<r, (r>0)(r>0) with sn>0s_{n}>0 for all nn. If the non-constant sequence {rn/sn}\{r_{n}/s_{n}\} is increasing (decreasing) for all nn, then the function xR(x)/S(x)x\mapsto R(x)/S(x) is strictly increasing (resp. decreasing) on (0,r)(0,r).

Lemma 4.1 is basically due to [4] (see also [28]) and in this form with a general setting was stated in [26] along with many applications which were later adopted by a number of researchers.

Lemma 4.2.

[22] Let α2(0,1]\frac{\alpha}{2}\in(0,1]. Then it holds that

(1)FkF11fork=1,2,3,andt[0,1);\displaystyle(1)\quad\frac{F_{k}}{F_{1}}\leq 1\,\,\mbox{for}\,\,k=1,2,3,...\,\,\mbox{and}\,\,t\in[0,1);
(2)|Fk,t|F1<(kα2)Γ(k+1)Γ(2+α2)2Γ(k+1+α2)fork=1,2,3,andt(0,1).\displaystyle(2)\quad\frac{|F_{k,t}|}{F_{1}}<\frac{(k-\frac{\alpha}{2})\Gamma(k+1)\Gamma(2+\frac{\alpha}{2})}{2\Gamma(k+1+\frac{\alpha}{2})}\,\,\mbox{for}\,\,k=1,2,3,...\mbox{and}\,\,t\in(0,1).
Theorem 4.3.

If α(0,2]\alpha\in(0,2], ck(N,N)c_{-k}\in(-N,N), where

N=α2((kα2)Γ(k+1)Γ(2+α2)Γ(k+1+α2)+k),\displaystyle N=\frac{\alpha}{2\left(\frac{(k-\frac{\alpha}{2})\Gamma(k+1)\Gamma(2+\frac{\alpha}{2})}{\Gamma(k+1+\frac{\alpha}{2})}+k\right)}, (4.2)

then the real kernel α\alpha-harmonic mapping

u(z)=F1z+ckFkz¯k,k=1,2,3,\displaystyle u(z)=F_{1}z+c_{-k}F_{k}\overline{z}^{k},\quad k=1,2,3..., (4.3)

is sense-preserving univalent in 𝔻\mathbb{D}.

Proof.

We divide the proof into two steps.

First step : Formula (4.3) implies that

uz=F1+F1,tt+ckFk,tz¯k+1,uz¯=F1,tz2+ck(Fk,tzz¯k+kFkz¯k1).\displaystyle u_{z}=F_{1}+F_{1,t}t+c_{-k}F_{k,t}\bar{z}^{k+1},\quad u_{\bar{z}}=F_{1,t}z^{2}+c_{-k}(F_{k,t}z\bar{z}^{k}+kF_{k}\bar{z}^{k-1}).

It follows that

|uz||uz¯|\displaystyle\quad|u_{z}|-|u_{\bar{z}}|
F1|F1,tt||ckFk,tz¯k+1||F1,tz2||ckFk,tzz¯k|k|ckFkz¯k1|\displaystyle\geq F_{1}-|F_{1,t}t|-|c_{-k}F_{k,t}\bar{z}^{k+1}|-|F_{1,t}z^{2}|-|c_{-k}F_{k,t}z\bar{z}^{k}|-k|c_{-k}F_{k}\bar{z}^{k-1}|
>F1|F1,t||ck||Fk,t||F1,t||ck||Fk,t|k|ck||Fk|\displaystyle>F_{1}-|F_{1,t}|-|c_{-k}||F_{k,t}|-|F_{1,t}|-|c_{-k}||F_{k,t}|-k|c_{-k}||F_{k}|
=F1[12|F1,t|F1|ck|(2|Fk,t|F1+k|Fk|F1)]\displaystyle=F_{1}\left[1-\frac{2|F_{1,t}|}{F_{1}}-|c_{-k}|\left(\frac{2|F_{k,t}|}{F_{1}}+k\frac{|F_{k}|}{F_{1}}\right)\right]
>F1[1(1α2)|ck|((kα2)Γ(k+1)Γ(2+α2)Γ(k+1+α2)+k)]\displaystyle>F_{1}\left[1-(1-\frac{\alpha}{2})-|c_{-k}|\left(\frac{(k-\frac{\alpha}{2})\Gamma(k+1)\Gamma(2+\frac{\alpha}{2})}{\Gamma(k+1+\frac{\alpha}{2})}+k\right)\right]
>0\displaystyle>0

for ck(N,N)c_{-k}\in(-N,N). The third inequality of the above holds because of Lemma 4.2. Therefore, u(z)u(z) is sense-preserving.

Second step : Let ck=|ck|eiβc_{-k}=|c_{-k}|e^{i\beta}. By assumption, we have β=0\beta=0 or π\pi. Let z=reiθz=re^{i\theta} and u(z)=Reiφu(z)=Re^{i\varphi}. Rewrite u(z)u(z) of (4.3) as

u(z)=F1reiθ+|ck|Fkrkei(βkθ)\displaystyle u(z)=F_{1}re^{i\theta}+|c_{-k}|F_{k}r^{k}e^{i(\beta-k\theta)}
=F1rcosθ+|ck|Fkrkcos(βkθ)+i(F1rsinθ+|ck|Fkrksin(βkθ)).\displaystyle\qquad=F_{1}r\cos\theta+|c_{-k}|F_{k}r^{k}\cos(\beta-k\theta)+i(F_{1}r\sin\theta+|c_{-k}|F_{k}r^{k}\sin(\beta-k\theta)). (4.4)

Then

tanφ=F1rsinθ+|ck|Fkrksin(βkθ)F1rcosθ+|ck|Fkrkcos(βkθ),\displaystyle\tan\varphi=\frac{F_{1}r\sin\theta+|c_{-k}|F_{k}r^{k}\sin(\beta-k\theta)}{F_{1}r\cos\theta+|c_{-k}|F_{k}r^{k}\cos(\beta-k\theta)}, (4.5)

where φ\varphi is the argument of u(z)u(z). It follows that

ddθ(tanφ)\displaystyle\frac{d}{d\theta}(\tan\varphi) =ddθ(F1rsinθ+|ck|Fkrksin(βkθ)F1rcosθ+|ck|Fkrkcos(βkθ))\displaystyle=\frac{d}{d\theta}\left(\frac{F_{1}r\sin\theta+|c_{-k}|F_{k}r^{k}\sin(\beta-k\theta)}{F_{1}r\cos\theta+|c_{-k}|F_{k}r^{k}\cos(\beta-k\theta)}\right)
=F12|ck|2Fk2r2(k1)k(k1)|ck|F1Fkrk1cos(β(k+1)θ)[F1cosθ+|ck|Fkrk1cos(βkθ)]2\displaystyle=\frac{F_{1}^{2}-|c_{-k}|^{2}F_{k}^{2}r^{2(k-1)}k-(k-1)|c_{-k}|F_{1}F_{k}r^{k-1}\cos(\beta-(k+1)\theta)}{[F_{1}\cos\theta+|c_{-k}|F_{k}r^{k-1}\cos(\beta-k\theta)]^{2}}
F12|ck|2Fk2r2(k1)k(k1)|ck|F1Fkrk1[F1cosθ+|ck|Fkrk1cos(βkθ)]2\displaystyle\geq\frac{F_{1}^{2}-|c_{-k}|^{2}F_{k}^{2}r^{2(k-1)}k-(k-1)|c_{-k}|F_{1}F_{k}r^{k-1}}{[F_{1}\cos\theta+|c_{-k}|F_{k}r^{k-1}\cos(\beta-k\theta)]^{2}}
=(F1+|ck|Fkrk1)(F1|ck|kFkrk1)[F1cosθ+|ck|Fkrk1cos(βkθ)]2\displaystyle=\frac{(F_{1}+|c_{-k}|F_{k}r^{k-1})(F_{1}-|c_{-k}|kF_{k}r^{k-1})}{[F_{1}\cos\theta+|c_{-k}|F_{k}r^{k-1}\cos(\beta-k\theta)]^{2}}
>0\displaystyle>0 (4.6)

for |ck|<1k|c_{-k}|<\frac{1}{k}. The last inequality of the above holds because of Lemma 4.2 (1). That is to say, tanφ\tan\varphi is strictly increasing with respect to θ\theta. So is φ\varphi, too.

In the following we divide into two cases to discuss.

Case 1 β=0\beta=0. It follows from (4.5) that

cotφ=cosθ+|ck|FkF1rk1coskθsinθ|ck|FkF1rk1sinkθ.\displaystyle\cot\varphi=\frac{\cos\theta+|c_{-k}|\frac{F_{k}}{F_{1}}r^{k-1}\cos k\theta}{\sin\theta-|c_{-k}|\frac{F_{k}}{F_{1}}r^{k-1}\sin k\theta}. (4.7)

Let’s take a close look at the changes in the value of the function cotφ\cot\varphi. Firstly, as is well-known, it is easy to verify by mathematical induction that

|sinkθsinθ|k\displaystyle\left|\frac{\sin k\theta}{\sin\theta}\right|\leq k (4.8)

for k=1,2,k=1,2,... and θ[0,2π)\theta\in[0,2\pi). If |ck|<1k|c_{-k}|<\frac{1}{k}, α(0,2]\alpha\in(0,2] and sinθ0\sin\theta\neq 0, then Lemma 4.2 (1) and inequality (4.8) imply that |sinθ|>|ck|FkF1rk1|sinkθ||\sin\theta|>|c_{-k}|\frac{F_{k}}{F_{1}}r^{k-1}|\sin k\theta|. So, sinθ0\sin\theta\neq 0 implies sinθ|ck|FkF1rk1sinkθ0\sin\theta-|c_{-k}|\frac{F_{k}}{F_{1}}r^{k-1}\sin k\theta\neq 0. In another words, the zero of the denominator of the right side of equation (4.7) comes only from the zero of sinθ\sin\theta. Secondly, sinθ\sin\theta only have two zeros in the intervals [0,2π)[0,2\pi). That is θ=0\theta=0 and π\pi. By (4.7), we have that if θ=0+\theta=0^{+}, then cotφ=+\cot\varphi=+\infty; if θ=π\theta=\pi^{-}, then cotφ=\cot\varphi=-\infty; if θ=π+\theta=\pi^{+}, then cotφ=+\cot\varphi=+\infty; if θ=2π\theta=2\pi^{-}, then cotφ=\cot\varphi=-\infty. Therefore, considering the continuity and monotonicity of cotφ\cot\varphi, we can get that the u(reiθ)u(re^{i\theta}) maps every circle |z|=r<1|z|=r<1 in a one-to-one manner onto a closed Jordan curve.

Case 2 β=π\beta=\pi. Considering (4.5), we have

cotφ=cosθ|ck|FkF1rk1coskθsinθ+|ck|FkF1rk1sinkθ.\displaystyle\cot\varphi=\frac{\cos\theta-|c_{-k}|\frac{F_{k}}{F_{1}}r^{k-1}\cos k\theta}{\sin\theta+|c_{-k}|\frac{F_{k}}{F_{1}}r^{k-1}\sin k\theta}.

Follow the discussion of Case 1. We omit the further details.

It is easy to see that N<1kN<\frac{1}{k}, where NN defined by (4.2). Therefore, considering the above two steps of the proof, by degree principle[9], we can get that u(z)u(z) is univalent in 𝔻\mathbb{D}.

The following is the well known Radó-Kneser-Choquet Theorem, which can be seen in the page 29 of [10].

Theorem 4.4.

If Ω\Omega\in\mathbb{C} is a bounded convex domain whose boundary is a Jordan curve γ\gamma and ff is a homeomorphism of the unit circle 𝕋\mathbb{T} onto γ\gamma, then its harmonic extension

u(z)=12π02π1|z|2|eitz|2f(eit)𝑑t\displaystyle u(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-|z|^{2}}{|e^{it}-z|^{2}}f(e^{it})dt

is univalent in 𝔻\mathbb{D} and defines a harmonic mapping of 𝔻\mathbb{D} onto Ω\Omega.

Next, we want to explore the Radó-Kneser-Choquet type theorem for real kernel α\alpha-harmonic mappings. We need the following Proposition at first.

Proposition 4.5.

Suppose α>1\alpha>-1 and ckc_{-k}\in\mathbb{R}. Let

f(eiθ)=F1(1)eiθ+ckFk(1)eikθ,k=1,2,3,.\displaystyle f(e^{i\theta})=F_{1}(1)e^{i\theta}+c_{-k}F_{k}(1)e^{-ik\theta},\quad k=1,2,3,.... (4.9)

Then ff maps the unit circle 𝕋\mathbb{T} onto a convex Jordan curve if and only if ck(M,M)c_{-k}\in(-M,M), where

M=Γ(k+1+α2)k2Γ(k+1)Γ(2+α2).\displaystyle M=\frac{\Gamma(k+1+\frac{\alpha}{2})}{k^{2}\Gamma(k+1)\Gamma(2+\frac{\alpha}{2})}. (4.10)
Proof.

Direct computation leads to

ddθ(f(eiθ))=F1(1)sinθkckFk(1)sinkθ+i(F1(1)cosθkckFk(1)coskθ).\displaystyle\frac{d}{d\theta}(f(e^{i\theta}))=-F_{1}(1)\sin\theta-kc_{-k}F_{k}(1)\sin k\theta+i(F_{1}(1)\cos\theta-kc_{-k}F_{k}(1)\cos k\theta).

Let ψ=ψ(θ)=arg{ddθf(eiθ)}\psi=\psi(\theta)=\arg\{\frac{d}{d\theta}f(e^{i\theta})\}. Then we have

ddθ(tanψ(θ))\displaystyle\frac{d}{d\theta}(\tan\psi(\theta)) =(F1(1))2k3(ckFk(1))2+k(k1)ckF1(1)Fk(1)cos((k+1)θ)(F1(1)sinθ+kckFk(1)sinkθ)2\displaystyle=\frac{(F_{1}(1))^{2}-k^{3}(c_{-k}F_{k}(1))^{2}+k(k-1)c_{-k}F_{1}(1)F_{k}(1)\cos((k+1)\theta)}{(F_{1}(1)\sin\theta+kc_{-k}F_{k}(1)\sin k\theta)^{2}}
(F1(1))2k3(ckFk(1))2k(k1)|ck|F1(1)Fk(1))(F1(1)sinθ+kckFk(1)sinkθ)2\displaystyle\geq\frac{(F_{1}(1))^{2}-k^{3}(c_{-k}F_{k}(1))^{2}-k(k-1)|c_{-k}|F_{1}(1)F_{k}(1))}{(F_{1}(1)\sin\theta+kc_{-k}F_{k}(1)\sin k\theta)^{2}}
=(F1(1)+k|ck|Fk(1))(F1(1)k2|ck|Fk(1))(F1(1)sinθ+kckFk(1)sinkθ)2\displaystyle=\frac{(F_{1}(1)+k|c_{-k}|F_{k}(1))(F_{1}(1)-k^{2}|c_{-k}|F_{k}(1))}{(F_{1}(1)\sin\theta+kc_{-k}F_{k}(1)\sin k\theta)^{2}}

Hence, ddθ(tanψ(θ))0\frac{d}{d\theta}(\tan\psi(\theta))\geq 0 if and only if |ck|F1(1)k2Fk(1)=Γ(k+1+α2)k2Γ(k+1)Γ(2+α2)|c_{-k}|\leq\frac{F_{1}(1)}{k^{2}F_{k}(1)}=\frac{\Gamma(k+1+\frac{\alpha}{2})}{k^{2}\Gamma(k+1)\Gamma(2+\frac{\alpha}{2})}. ∎

Now let f(eiθ)f(e^{i\theta}) be defined as in (4.9) with α(0,2]\alpha\in(0,2], ck(L,L)c_{-k}\in(-L,L), where L=min{M,N}L=\min\{M,N\}. Observe that limr1u(z):=u(eiθ)=f(eiθ)\lim_{r\rightarrow 1}u(z):=u^{*}(e^{i\theta})=f(e^{i\theta}), where u(z)u(z) are defined by (4.3). Similar to the second step of the proof of Theorem 4.3, we can verify that f(eiθ)f(e^{i\theta}) maps unit circle 𝕋\mathbb{T} onto a closed Jordan curve in a one-to-one manner, too. Therefore, considering Theorem 3.3 of [24] and Theorem 4.3 of the above, we actually get a Radó-Kneser-Choquet type theorem as follows:

Proposition 4.6.

Let u(eiθ)=f(eiθ)u^{*}(e^{i\theta})=f(e^{i\theta}) be defined by (4.9) with k=1,2,3,k=1,2,3,..., α(0,2]\alpha\in(0,2], ck(L,L)c_{-k}\in(-L,L), where L=min{M,N}L=\min\{M,N\}, NN and MM are defined by (4.2) and (4.10), respectively . Then u(eiθ)u^{*}(e^{i\theta}) is a homeomorphism of the unit circle 𝕋\mathbb{T} onto a convex Jordan curve γ\gamma which is a boundary of a bounded convex domain Ω\Omega\subset\mathbb{C}. Furthermore, u(z)u(z) defined by (1.3) defines a univalent real kernel α\alpha-harmonic mapping of 𝔻\mathbb{D} onto Ω\Omega.

Let us have a look at some special cases of Theorem 4.3 or Proposition 4.6.

Example 4.1.

Let α=2\alpha=2. Then M=k+12k2M=\frac{k+1}{2k^{2}} and N=k+1k2+3k2N=\frac{k+1}{k^{2}+3k-2}. Formula (4.9) deduces to

f(eiθ)=eiθ+2k+1ckeikθ.\displaystyle f(e^{i\theta})=e^{i\theta}+\frac{2}{k+1}c_{-k}e^{-ik\theta}.

Furthermore, let u(eiθ)=f(eiθ)u^{*}(e^{i\theta})=f(e^{i\theta}). Then (1.3), or (2.1), implies that the corresponding real kernel α\alpha-harmonic mapping is

u(z)=F1z+ckFkz¯k=z+ck(1k1k+1|z|2)z¯k.\displaystyle u(z)=F_{1}z+c_{-k}F_{k}\bar{z}^{k}=z+c_{-k}\left(1-\frac{k-1}{k+1}|z|^{2}\right)\bar{z}^{k}. (4.11)

Actually, it is biharmonic.

(1) If k=1k=1 or k=2k=2, then L=M=NL=M=N. If ck(L,L)c_{-k}\in(-L,L), then Proposition 4.6 says that the u(z)u(z) given by (4.11) is univalent, and u(𝔻)=Ωu(\mathbb{D})=\Omega is a convex domain.

(2) If k=3,4,5,k=3,4,5,..., then a direct computation leads to N>MN>M. Taking ck(M,N)c_{-k}\in(M,N), Theorem 4.3 and Proposition 4.5 imply that the above u(z)u(z) is still univalent, but u(𝔻)=Ωu(\mathbb{D})=\Omega is not a convex domain.

5 Area SuS_{u}

Let SuS_{u} denote the area of the Riemann surface of uu. Then we have the following results.

Theorem 5.1.

Let uu be a sense-preserving real kernel α\alpha-harmonic mapping that has the series expansion of the form (1.7) with c0=0c_{0}=0, continuous on 𝔻¯\overline{\mathbb{D}}. Let vv be the corresponding sense-preserving harmonic mapping that has the series expansion of the form (1.9), continuous on 𝔻¯\overline{\mathbb{D}}. If |ck||ck||c_{k}|\geq|c_{-k}| for k=1,2,k=1,2,..., then

(1)   Su<SvS_{u}<S_{v} for α(0,2)\alpha\in(0,2) and Su>SvS_{u}>S_{v} for α(1,0)\alpha\in(-1,0);

(2)    Su(α)S_{u}(\alpha) is strictly decreasing with respect to α(1,0.8)\alpha\in(-1,0.8), where Su(α)=SuS_{u}(\alpha)=S_{u}.

Proof.

By (1.7), direct computation leads to

uz\displaystyle u_{z} =k=1ck[Ftz¯zk+kFzk1]+k=1ckFtz¯k+1\displaystyle=\sum_{k=1}^{\infty}c_{k}[F_{t}\bar{z}z^{k}+kFz^{k-1}]+\sum_{k=1}^{\infty}c_{-k}F_{t}\bar{z}^{k+1}
=k=1ck[Ftrk+1+kFrk1]ei(k1)θ+k=1ckFtrk+1ei(k+1)θ\displaystyle=\sum_{k=1}^{\infty}c_{k}[F_{t}r^{k+1}+kFr^{k-1}]e^{i(k-1)\theta}+\sum_{k=1}^{\infty}c_{-k}F_{t}r^{k+1}e^{-i(k+1)\theta}

and

uz¯\displaystyle u_{\bar{z}} =k=1ckFtzk+1+k=1ck[Ftzz¯k+kFz¯k1]\displaystyle=\sum_{k=1}^{\infty}c_{k}F_{t}z^{k+1}+\sum_{k=1}^{\infty}c_{-k}[F_{t}z\bar{z}^{k}+kF\bar{z}^{k-1}]
=k=1ck[Ftrk+1+kFrk1]ei(k1)θ+k=1ckFtrk+1ei(k+1)θ.\displaystyle=\sum_{k=1}^{\infty}c_{-k}[F_{t}r^{k+1}+kFr^{k-1}]e^{-i(k-1)\theta}+\sum_{k=1}^{\infty}c_{k}F_{t}r^{k+1}e^{i(k+1)\theta}.

So,

Su\displaystyle S_{u} =02π01Ju(z)r𝑑r𝑑θ\displaystyle=\int_{0}^{2\pi}\int_{0}^{1}J_{u}(z)rdrd\theta
=2π01k=1[|ck(Ftrk+1+kFrk1)|2+|ckFtrk+1|2|ck(Ftrk+1+kFrk1)|2|ckFtrk+1|2]rdr\displaystyle=2\pi\int_{0}^{1}\sum_{k=1}^{\infty}\left[|c_{k}(F_{t}r^{k+1}+kFr^{k-1})|^{2}+|c_{-k}F_{t}r^{k+1}|^{2}-|c_{-k}(F_{t}r^{k+1}+kFr^{k-1})|^{2}-|c_{k}F_{t}r^{k+1}|^{2}\right]rdr
=2π01[k=1(|ck|2|ck|2)(k2F2r2k1+2kFFtr2k+1)]𝑑r\displaystyle=2\pi\int_{0}^{1}\left[\sum_{k=1}^{\infty}(|c_{k}|^{2}-|c_{-k}|^{2})(k^{2}F^{2}r^{2k-1}+2kFF_{t}r^{2k+1})\right]dr
=2πk=1[(|ck|2|ck|2)k01(kF2r2k1+2FFtr2k+1)𝑑r]\displaystyle=2\pi\sum_{k=1}^{\infty}\left[(|c_{k}|^{2}-|c_{-k}|^{2})k\int_{0}^{1}(kF^{2}r^{2k-1}+2FF_{t}r^{2k+1})dr\right]
=πk=1[(|ck|2|ck|2)k01d(F2r2k)]\displaystyle=\pi\sum_{k=1}^{\infty}\left[(|c_{k}|^{2}-|c_{-k}|^{2})k\int_{0}^{1}d(F^{2}r^{2k})\right]
=πΓ2(1+α)Γ2(1+α2)k=1[k(|ck|2|ck|2)Γ2(k+1)Γ2(k+1+α2)].\displaystyle=\pi\frac{\Gamma^{2}(1+\alpha)}{\Gamma^{2}(1+\frac{\alpha}{2})}\sum_{k=1}^{\infty}\left[k(|c_{k}|^{2}-|c_{-k}|^{2})\frac{\Gamma^{2}(k+1)}{\Gamma^{2}(k+1+\frac{\alpha}{2})}\right]. (5.1)

The last equality holds because of (1.6).

Similarly, we have

Sv=πk=1k(|ck|2|ck|2).\displaystyle S_{v}=\pi\sum_{k=1}^{\infty}k(|c_{k}|^{2}-|c_{-k}|^{2}). (5.2)

(1) Let ψ(x)=Γ(x)/Γ(x)\psi(x)=\Gamma^{\prime}(x)/\Gamma(x) be the digamma function. Then it is well known that(cf. [3]) ψ(x)\psi(x) is strictly increasing on (0,+)(0,+\infty).

Let

f(x)=Γ(1+α)Γ(x+1)Γ(1+α2)Γ(x+1+α2).\displaystyle f(x)=\frac{\Gamma(1+\alpha)\Gamma(x+1)}{\Gamma(1+\frac{\alpha}{2})\Gamma(x+1+\frac{\alpha}{2})}.

Then we have

(logf(x))\displaystyle(\log f(x))^{\prime} =ψ(x+1)ψ(x+1+α2).\displaystyle=\psi(x+1)-\psi(x+1+\frac{\alpha}{2}).

It follows that (logf(x))<0(\log f(x))^{\prime}<0 provided α>0\alpha>0, and (logf(x))>0(\log f(x))^{\prime}>0 provided α<0\alpha<0. Observe that

f(α/2)=1.\displaystyle f(\alpha/2)=1.

Therefore, for k=1,2,k=1,2,..., we have f(k)<1f(k)<1 if α(0,2)\alpha\in(0,2) as well as f(k)>1f(k)>1 if α(1,0)\alpha\in(-1,0). Taking account of (5.1) and (5.2), we can get Theorem 5.1 (1).

(2) As to digamma function ψ(x)\psi(x), we have (cf. [3])

ψ(1+x)=1x+ψ(x),\displaystyle\psi(1+x)=\frac{1}{x}+\psi(x), (5.3)
ψ(x)=γ+n=0(1n+11n+x),\displaystyle\psi(x)=-\gamma+\sum^{\infty}_{n=0}\left(\frac{1}{n+1}-\frac{1}{n+x}\right), (5.4)

and

ψ(x)=n=01(x+n)2\displaystyle\psi^{\prime}(x)=\sum^{\infty}_{n=0}\frac{1}{(x+n)^{2}} (5.5)

for any x(0,+)x\in(0,+\infty), where γ\gamma is the Euler-Mascheroni constant.

Let

h(α)=ψ(1+α)ψ(1+α2)12+α.h(\alpha)=\psi(1+\alpha)-\psi(1+\frac{\alpha}{2})-\frac{1}{2+\alpha}.

Then (5.5) implies that

h(α)=ψ(1+α)12ψ(1+α2)+1(2+α)2\displaystyle h^{\prime}(\alpha)=\psi^{\prime}(1+\alpha)-\frac{1}{2}\psi^{\prime}(1+\frac{\alpha}{2})+\frac{1}{(2+\alpha)^{2}}
=n=0(1(1+α+n)212(1+α2+n)2)+1(2+α)2\displaystyle\qquad\,=\sum^{\infty}_{n=0}\left(\frac{1}{(1+\alpha+n)^{2}}-\frac{1}{2(1+\frac{\alpha}{2}+n)^{2}}\right)+\frac{1}{(2+\alpha)^{2}}
=n=0(n+1)2α222(1+α2+n)2(1+α+n)2+1(2+α)2\displaystyle\qquad\,=\sum^{\infty}_{n=0}\frac{(n+1)^{2}-\frac{\alpha^{2}}{2}}{2(1+\frac{\alpha}{2}+n)^{2}(1+\alpha+n)^{2}}+\frac{1}{(2+\alpha)^{2}}
>0\displaystyle\qquad\,>0

for α(1,1)\alpha\in(-1,1). Furthermore, using (5.4), direct numerical computation shows

h(0.8)=0.0108<0.\displaystyle h(0.8)=-0.0108<0.

Thus, h(α)<0h(\alpha)<0 for α(1,0.8)\alpha\in(-1,0.8). Let

g(α)=Γ(1+α)Γ(1+α2)Γ(k+1+α2),k=1,2,.g(\alpha)=\frac{\Gamma(1+\alpha)}{\Gamma(1+\frac{\alpha}{2})\Gamma(k+1+\frac{\alpha}{2})},\quad k=1,2,....

Then it follows that

dlogg(α)dα=ψ(1+α)12ψ(1+α2)12ψ(k+1+α2)\displaystyle\frac{d\log g(\alpha)}{d\alpha}=\psi(1+\alpha)-\frac{1}{2}\psi(1+\frac{\alpha}{2})-\frac{1}{2}\psi(k+1+\frac{\alpha}{2})
<ψ(1+α)12ψ(1+α2)12ψ(2+α2)\displaystyle\qquad\qquad\,<\psi(1+\alpha)-\frac{1}{2}\psi(1+\frac{\alpha}{2})-\frac{1}{2}\psi(2+\frac{\alpha}{2})
=h(α).\displaystyle\qquad\qquad\,=h(\alpha).

That is to say g(α)g(\alpha) is strictly decreasing on (1,0.8)(-1,0.8). Therefore, (5.1) implies that Su(α)S_{u}(\alpha) is strictly decreasing with respect to α(1,0.8)\alpha\in(-1,0.8).

Acknowledgements  The authors heartily thank the anonymous reviewers for their careful review and for their effective suggestions.

Declarations

Conflict of interests The authors declare that they have no conflict of interest.

Data availability statement My manuscript has no associated date.

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