Lipschitz continuity for weighted harmonic functions in the unit disc

Abstract

We study membership in Lipschitz classes ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ for a class of α-harmonic functions in the open unit disc $\mathbb{D}$ in the complex plane. From earlier work by Olofsson and Wittsten we know that such an α-harmonic function u is the α-harmonic Poisson integral $P_{\alpha }\left[f\right]$ of its boundary value function f on the unit circle $\mathbb{T}$. We determine when the Poisson integral $P_{\alpha }\left[f\right]$ belongs to a Lipschitz class ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ for the unit disc.

Keywords:

• Lipschitz continuity
• harmonic function
• Poisson integral
• Fourier multiplier

AMS SUBJECT CLASSIFICATIONS:

• Primary: 31A05
• Secondary: 35J25

1. Introduction

Let $\mathbb{D}$ be the open unit disc in the complex plane, let $\mathbb{T}=\mathrm{\partial }\mathbb{D}$ be the unit circle, and denote by ${\mathrm{\partial }}_z=\frac{1}{2}\left(\frac{\mathrm{\partial }}{\mathrm{\partial }x}+\frac{1}{i}\frac{\mathrm{\partial }}{\mathrm{\partial }y}\right)\mathrm{a}\mathrm{n}\mathrm{d}{\overline{\mathrm{\partial }}}_z=\frac{1}{2}\left(\frac{\mathrm{\partial }}{\mathrm{\partial }x}-\frac{1}{i}\frac{\mathrm{\partial }}{\mathrm{\partial }y}\right)$ the standard complex partial derivatives, where z=x + iy is the complex coordinate. A chief concern of this paper is the homogeneous second order partial differential equation (1) ${\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},$(1) where (2) ${\mathrm{\Delta }}_{\alpha ,z}={\mathrm{\partial }}_z(1-\mid z{\mid }^2{)}^{-\alpha }{\overline{\mathrm{\partial }}}_z,z\in \mathbb{D},$(2) is a certain weighted Laplacian with weight parameter $\alpha >-1$. The restriction $\alpha >-1$ on the weight parameter is needed for Equation (1(1) ${\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},$(1) ) to have a sufficiently rich supply of solutions with well-behaved boundary behaviour.

A solution u of Equation (1(1) ${\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},$(1) ) is called an α-harmonic function. Such a function u is smooth in $\mathbb{D}$ but can be rather wild near the boundary $\mathbb{T}=\mathrm{\partial }\mathbb{D}$. Notice that ${\mathrm{\Delta }}_0=\mathrm{\partial }\overline{\mathrm{\partial }}$ is one-quarter of the usual Laplacian and that a 0-harmonic function is a harmonic function in $\mathbb{D}$ in the usual sense. Notice also that the differential operator ${\mathrm{\Delta }}_{\alpha }$ has a certain singular or degenerate behaviour on the unit circle $\mathbb{T}$ when $\alpha \ne 0$. The study of weighted Laplacians of the above form (2(2) ${\mathrm{\Delta }}_{\alpha ,z}={\mathrm{\partial }}_z(1-\mid z{\mid }^2{)}^{-\alpha }{\overline{\mathrm{\partial }}}_z,z\in \mathbb{D},$(2) ) goes back to Garabedian [1] and has attracted quite some interest recently.

Of particular interest is the Dirichlet boundary value problem (3) $\left\{\begin{array}{l}{\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},\\ u=f\mathrm{o}\mathrm{n}\mathbb{T},\end{array}\right.$(3) for α-harmonic functions. Here $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ is a distribution on $\mathbb{T}$ and the boundary condition in (3(3) $\left\{\begin{array}{l}{\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},\\ u=f\mathrm{o}\mathrm{n}\mathbb{T},\end{array}\right.$(3) ) is interpreted in a distributional sense that $\underset{r\to 1}{lim}{u}_r=f$ in ${\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$, where (4) $u_r\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=u\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right),{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T},$(4) for $0\le r<1$. From Olofsson and Wittsten [2] it is known that a twice continuously differentiable function $u\in C^2\left(\mathbb{D}\right)$ in $\mathbb{D}$ solves the Dirichlet problem (3(3) $\left\{\begin{array}{l}{\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},\\ u=f\mathrm{o}\mathrm{n}\mathbb{T},\end{array}\right.$(3) ) if and only if it has the form of a Poisson integral $u\left(z\right)=P_{\alpha }\left[f\right]\left(z\right)=P_{\alpha ,r}\ast f\left({\mathrm{e}}^{\mathrm{i}\theta }\right),z=r{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{D},$ where (5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) is the α-harmonic Poisson kernel, $P_{\alpha ,r}=(P_{\alpha }{)}_r$ for $0\le r<1$ in accordance with (4(4) $u_r\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=u\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right),{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T},$(4) ) and $\ast$ is convolution in ${\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$. For an integrable function $f\in L^1\left(\mathbb{T}\right)$ on $\mathbb{T}$ the above Poisson integral has the form (6) $P_{\alpha }\left[f\right]\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right)f\left({\mathrm{e}}^{\mathrm{i}\tau }\right)\mathrm{d}\tau ,z\in \mathbb{D},$(6) where the function $P_{\alpha }$ is as in (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ). Notice that $u=P_0\left[f\right]$ is the usual Poisson integral representation of a harmonic function in $\mathbb{D}$. The Poisson integral representation $u=P_{\alpha }\left[f\right]$ leads to improved convergence to boundary values (in norm as well as non-tangentially) of solutions of (3(3) $\left\{\begin{array}{l}{\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},\\ u=f\mathrm{o}\mathrm{n}\mathbb{T},\end{array}\right.$(3) ) provided the boundary data $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ has some appropriate regularity, see [2–4] for results.

Let us recall the notion of Lipschitz continuity. For a bounded subset E of the complex plane and parameter $0<\text{β}\le 1$ we denote by ${\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ the set of complex-valued functions f on E such that $\mid f\left(z_1\right)-f\left(z_2\right)|\le C\mid z_1-z_2{\mid }^{\text{β}},z_1,z_2\in E,$ for some finite constant $C=C\left(f\right)$. The class ${\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ is commonly referred to as the Lipschitz (or Hölder) class for the set E of order β. The space ${\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ is often equipped with the norm $\parallel f{\parallel }_{{\mathrm{\Lambda }}_{\text{β}}\left(E\right)}=\underset{z\in E}{sup}\mid f\left(z\right)\mid +\underset{\begin{array}{c}{z}_1,z_2\in E\\ z_1\ne z_2\end{array}}{sup}\frac{\mid f\left(z_1\right)-f\left(z_2\right)\mid }{\mid z_1-z_2{\mid }^{\text{β}}},f\in {\mathrm{\Lambda }}_{\text{β}}\left(E\right),$ and is then a commutative Banach algebra under pointwise multiplication of functions (see for instance Sherbert [5]). It is well-known that every function f in ${\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ extends uniquely to a continuous function on the closure $\overline{E}$ of E in such a way that $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\overline{E}\right)$. It is also well-known that ${\mathrm{\Lambda }}_{{\text{β}}_1}\left(E\right)\supset {\mathrm{\Lambda }}_{{\text{β}}_2}\left(E\right)$ if $0<{\text{β}}_1<{\text{β}}_2\le 1$. The family of spaces ${\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ forms a much natural scale to measure smoothness of continuous functions.

A main purpose of this paper is to study membership in Lipschitz classes for α-harmonic functions. Observe that $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ implies that $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$, where $f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=\underset{r\to 1}{lim}u\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right),{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}.$ A fundamental question is to determine when the Poisson integral $u=P_{\alpha }\left[f\right]$ of a function $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ belongs to the space ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Notice that this latter question is an extension problem in the sense that we ask when the function $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ extends to an α-harmonic function u in the class ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. We shall here solve this problem in full generality. The solution we propose depends on the individual parameters $\alpha >-1$ and $0<\text{β}\le 1$ as well as on spectral properties of the function $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$.

Our results contrast the situation in the category of continuous functions where one has that the Poisson integral $u=P_{\alpha }\left[f\right]$ extends to a continuous function on the closed disc $\overline{\mathbb{D}}$ if $f\in C\left(\mathbb{T}\right)$ is a continuous function on $\mathbb{T}$ and $\alpha >-1$ (see Olofsson and Wittsten [2, Section 5]). The study of Lipschitz conditions $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ goes back to classical work of Hardy and Littlewood [6, Section 5] in the case of analytic functions.

An integrable function $f\in L^1\left(\mathbb{T}\right)$ on $\mathbb{T}$ is identified with the distribution $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ on $\mathbb{T}$ defined by $⟨f,\varphi ⟩=\frac{1}{2\pi }{\int }_{\mathbb{T}}\varphi \left({\mathrm{e}}^{\mathrm{i}\theta }\right)f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)\mathrm{d}\theta ,\varphi \in C^{\mathrm{\infty }}\left(\mathbb{T}\right),$ where $⟨\cdot ,\cdot ⟩$ is the distributional pairing and $C^{\mathrm{\infty }}\left(\mathbb{T}\right)$ is the space of indefinitely differentiable test functions on $\mathbb{T}$. The Fourier coefficients of a function $f\in L^1\left(\mathbb{T}\right)$ are defined by $\hat{f}\left(n\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}{\mathrm{e}}^{-\mathrm{i}n\theta }f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)\mathrm{d}\theta$ for integers $n\in \mathbb{Z}$, and similarly for distributions using the distributional pairing (see for instance [2] for details). The conjugate function of a distribution $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ is the distribution $\tilde{f}\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ defined by the condition that $\left(\tilde{f}\right)\stackrel{ˆ}{}\left(n\right)=-i\mathrm{s}\mathrm{g}\mathrm{n}\left(n\right)\hat{f}\left(n\right),n\in \mathbb{Z},$ where $\hat{f}\left(n\right)$ is the n-th Fourier coefficient of f, $\mathrm{s}\mathrm{g}\mathrm{n}\left(n\right)=n/\mid n\mid$ for $n\ne 0$ and $\mathrm{s}\mathrm{g}\mathrm{n}\left(0\right)=0$. A much influential introduction to Fourier analysis is Katznelson [7].

Let us first recall the situation in the classical case of usual harmonic functions in $\mathbb{D}$ ($\alpha =0$). If $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ and $0<\text{β}<1$, then $P_0\left[f\right]\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Furthermore, a harmonic function u in $\mathbb{D}$ belongs to the class ${\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$ if and only if it has the form $u=P_0\left[f\right]$ for some function $f\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$ such that $\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$ (see Theorem 6.7). These results are well-known and follow quite easily from the work of Hardy and Littlewood [6, Section 5] on analytic functions mentioned above. Recall here that the conjugate function $f\mapsto \tilde{f}$ is a bounded linear operator on the Lipschitz space ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ for $0<\text{β}<1$ (see Zygmund [8, Theorem III.13.29]). Results of this latter type often go under the name of Privalov's theorem.

Let us next consider the case when $\alpha >-1$ and $0<\text{β}\le 1$ are such that $\text{β}<\alpha +1$. For such parameters we show that $u=P_{\alpha }\left[f\right]$ belongs to ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ if $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ (see Theorem 3.2). This generalizes a classical result from the previous paragraph as well as a recent result of Li and Wang [9] for $\alpha >0$ to the bigger parameter range considered; a less precise result can be found inChen [10]. In fact, as we shall see later on, the inequality $\text{β}<\alpha +1$ defines the exact parameter range for the implication $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)⟹P_{\alpha }\left[f\right]\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ to hold true.

Lipschitz continuity $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ of a continuously differentiable function $u\in C^1\left(\mathbb{D}\right)$ in $\mathbb{D}$ is inferred from a gradient growth estimate of the form (7) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D}$(7) (see Theorem 3.1). This observation goes back to classical work of Hardy and Littlewood [6] and has been elaborated on by Gehring and Martio [11] and others. In the context of the previous paragraph we establish (7(7) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D}$(7) ) by careful estimations using the Poisson integral representation $u=P_{\alpha }\left[f\right]$ (see Theorems 2.5 and 2.10). Notably, we show that (8) $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(8) for $u=P_{\alpha }\left[f\right]$ with $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ provided $0<\text{β}\le 1$ (see Theorem 2.13).

It is an important observation that the combination of partial derivatives $(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u$ yields an estimate (8(8) $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(8) ) which is valid without any extra restriction on β more than what is previously stated. In fact, this more general validity of (8(8) $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(8) ) compared to (7(7) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D}$(7) ) is quite natural from the point of view of convolution structure in (6(6) $P_{\alpha }\left[f\right]\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right)f\left({\mathrm{e}}^{\mathrm{i}\tau }\right)\mathrm{d}\tau ,z\in \mathbb{D},$(6) ) and an interpretation of the differential operator $z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }}$ as angular derivative. The growth estimate (8(8) $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(8) ) is established for $u=P_{\alpha }\left[f\right]$ with a fairly economical constant $C=C\left(f\right)$ using a precise result on the $L^1$-means of the function $P_{\alpha }$ from Olofsson and Wittsten [2, Section 2] later refined in Olofsson [3] (see Theorem 2.4). We mention that inequality (8(8) $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(8) ) generalizes a recent result by Chen [10, Lemma 3.2].

Let us next consider the case when $-1<\alpha <0$ and $0<\text{β}\le 1$ are such that $\text{β}>\alpha +1$. For such parameters we show that if u is α-harmonic and $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$, then u is analytic in $\mathbb{D}$ (see Theorem 5.4). A closely related result is that, in this parameter range, the Poisson integral $u=P_{\alpha }\left[f\right]$ belongs to ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ if and only if $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ and $\hat{f}\left(n\right)=0$ for n<0 (see Theorem 5.5). This latter result puts earlier examples of Li and Wang [9] and Chen [10] in a precise context.

The results in the previous paragraph are proved using properties of the homogeneous expansion of α-harmonic functions. Let $\alpha >-1$ and set $f_{\alpha ,n}\left(x\right)=\frac{\mathrm{\Gamma }(n+\alpha +1)}{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }(\alpha +1)}{\int }_0^1{t}^{n-1}(1-xt{)}^{\alpha }\mathrm{d}t,0\le x<1,$ for positive integers $n\in {\mathbb{Z}}^{+}$, where Γ is the standard Gamma function. We now introduce the functions $e_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D},$ for $n\in {\mathbb{Z}}^{+}$. From Olofsson and Wittsten [2, Section 5] we know that the Poisson integral $P_{\alpha }\left[f\right]$ has the series expansion (9) $P_{\alpha }\left[f\right]\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)e_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}\hat{f}\left(n\right)z^n,z\in \mathbb{D},$(9) for $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$.

We derive a system of two first-order partial differential equations in $\mathbb{D}$ satisfied by the function $e_{\alpha ,-n}$ (see Proposition 4.2). This system of partial differential equations is shown to characterize the function $e_{\alpha ,-n}$ in the punctured disc $\mathbb{D}\setminus \left\{0\right\}$ up to an additive term of the form $c/z^n$ (see Theorem 4.3). As an application of this development we show that $e_{\alpha ,-n}\left(z\right)=\frac{1}{z^n}-\frac{\mathrm{\Gamma }(n+\alpha +1)}{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }(\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha +1}{\int }_0^1(1-(1-\mid z{\mid }^2)t{)}^{n-1}{t}^{\alpha }\mathrm{d}t\frac{1}{z^n}$ for $z\in \mathbb{D}\setminus \left\{0\right\}$ (see Theorem 4.7). This latter formula refines an earlier result from Olofsson and Wittsten [2, Theorem 1.9]. As a consequence we have that $e_{\alpha ,-n}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ if and only if $\text{β}\le \alpha +1$ (see Lemma 5.1). Combined with some homogeneity type of arguments this leads to the results about Lipschitz continuity of α-harmonic functions in the case of big smoothness $\text{β}>\alpha +1$ discussed above.

Let us now consider the case when $-1<\alpha <0$ and $0<\text{β}\le 1$ are such that $\text{β}=\alpha +1$. The analysis of this case is the main technical contribution of this paper. We shall need the Fourier multiplier $M_{\alpha }$ defined by $\left(M_{\alpha }f\right)\stackrel{ˆ}{}\left(n\right)=\left\{\begin{array}{ll}\frac{\mathrm{\Gamma }(\mid n\mid +\alpha +1)}{\mathrm{\Gamma }(\mid n\mid )}\hat{f}\left(n\right)& \mathrm{f}\mathrm{o}\mathrm{r}n<0,\\ 0& \mathrm{f}\mathrm{o}\mathrm{r}n\ge 0,\end{array}\right.$ for $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$, where $\hat{f}\left(n\right)$ is the n-th Fourier coefficient of f. Denote also by $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ the space of essentially bounded measurable functions on $\mathbb{T}$ normed by the essential supremum $\parallel f{\parallel }_{\mathrm{\infty }}=\underset{{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}}{\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{p}}\mid f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)\mid ,f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right).$ In this parameter range, we show that the function $u=P_{\alpha }\left[f\right]$ belongs to the class ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ if and only if $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ is such that $M_{\alpha }f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ (see Theorems 6.3 and 6.5).

The constraint on a Lipschitz function $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ ($0<\text{β}<1$) that $M_{\text{β}-1}f$ belongs to $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ is quite demanding. An interesting example is provided by the Weierstrass function $f=f_{\text{β}}$ which has the properties that $f_{\text{β}}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ and $M_{\text{β}-1}{f}_{\text{β}}\notin L^1\left(\mathbb{T}\right)$ (see discussion following formula (49(49) $f_{\text{β}}\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=\sum _{k=1}^{\mathrm{\infty }}{b}^{-k\text{β}}{\mathrm{e}}^{-\mathrm{i}{b}^k\theta },{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T},$(49) )).

It is an important observation that the function $f_{\alpha ,n}$ above is a constant multiple of a hypergeometric function. Assuming that $u=P_{\alpha }\left[f\right]$ belongs to the class ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$, we prove necessity of the condition $M_{\alpha }f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ by a differentiation type argument using asymptotic properties of hypergeometric functions (see Theorem 6.3).

Differentiating (9(9) $P_{\alpha }\left[f\right]\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)e_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}\hat{f}\left(n\right)z^n,z\in \mathbb{D},$(9) ) using the above mentioned partial differential equations for the $e_{\alpha ,-n}$'s we show that $\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{\mathrm{\Gamma }(\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }{P}_0[M_{\alpha }f\left]\right(z)$ for $z\in \mathbb{D}$, where $u=P_{\alpha }\left[f\right]$ (see Lemma 6.4). This latter formula suggests an obvious estimation of $\overline{\mathrm{\partial }}u$. Combined with (8(8) $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(8) ) this yields that $u=P_{\alpha }\left[f\right]$ satisfies a growth estimate of the form (7(7) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D}$(7) ) if $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ is such that $M_{\alpha }f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ (see Theorem 6.5).

In conclusion we have shown that an inequality of the form (7(7) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D}$(7) ) is also necessary for Lipschitz continuity $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ of an α-harmonic function u. We point out that such an implication requires interior rigidity of the function u and is not true in general (see Proposition 3.5).

For further results on Lipschitz continuity of analytic functions in $\mathbb{D}$ we mention Dyakonov [12]. Recent papers dealing with aspects of α-harmonic functions not quoted above are [4, 13–17]. We mention also a related interesting paper by Borichev and Hedenmalm [18].

The author thanks Kevin Klein and Jens Wittsten for useful discussions.

2. Bounds for partial derivatives

In this section, we prove some growth bounds for partial derivatives of α-harmonic Poisson integrals. Applications of these bounds appear in later sections.

We first consider the $\overline{\mathrm{\partial }}$-derivative.

Lemma 2.1

Let $\alpha >-1$. Let $f\in L^1\left(\mathbb{T}\right)$ and set $u=P_{\alpha }\left[f\right]$. Then $\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}\overline{\mathrm{\partial }}{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right){\mathrm{e}}^{\mathrm{i}\tau }\left(f\right({\mathrm{e}}^{\mathrm{i}\tau })-f({\mathrm{e}}^{\mathrm{i}\theta }\left)\right)\mathrm{d}\tau$ for $z=r{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{D}$ with $0\le r<1$ and ${\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}$.

Proof.

The interpretation of the Poisson integral (6(6) $P_{\alpha }\left[f\right]\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right)f\left({\mathrm{e}}^{\mathrm{i}\tau }\right)\mathrm{d}\tau ,z\in \mathbb{D},$(6) ) as solution of the Dirichlet problem (3(3) $\left\{\begin{array}{l}{\mathrm{\Delta }}_{\alpha }u=0\mathrm{i}\mathrm{n}\mathbb{D},\\ u=f\mathrm{o}\mathrm{n}\mathbb{T},\end{array}\right.$(3) ) leads to the property that $\frac{1}{2\pi }{\int }_{\mathbb{T}}{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right)\mathrm{d}\tau =1$ for $z\in \mathbb{D}$. Let $c\in \mathbb{C}$ and notice that $u\left(z\right)=P_{\alpha }\left[f\right]\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right)\left(f\right({\mathrm{e}}^{\mathrm{i}\tau })-c)\mathrm{d}\tau +c$ for $z\in \mathbb{D}$. Differentiating we have that $\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\overline{\mathrm{\partial }}{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right){\mathrm{e}}^{\mathrm{i}\tau }\left(f\right({\mathrm{e}}^{\mathrm{i}\tau })-c)\mathrm{d}\tau$ for $z\in \mathbb{D}$. The choice $c=f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)$ gives the conclusion of the lemma.

We now calculate $\overline{\mathrm{\partial }}{P}_{\alpha }$.

Lemma 2.2

Let $\alpha >-1$ and let $P_{\alpha }$ be as in (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ). Then $\overline{\mathrm{\partial }}{P}_{\alpha }\left(z\right)=\frac{\alpha +1}{(1-\overline{z}{)}^2}{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }$ for $z\in \mathbb{D}$.

Proof.

From formula (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ) we have that $P_{\alpha }\left(z\right)=\frac{1}{1-z}{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha +1}$ for $z\in \mathbb{D}$. Differentiating we have that $\overline{\mathrm{\partial }}{P}_{\alpha }\left(z\right)=\frac{\alpha +1}{1-z}{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{(-z)(1-\overline{z})-(1-\mid z{\mid }^2)(-1)}{(1-\overline{z}{)}^2}$ for $z\in \mathbb{D}$. A simplification of terms now leads to the conclusion of the lemma.

From Lemma 2.2 we have that (10) $\mid \overline{\mathrm{\partial }}{P}_{\alpha }\left(z\right)\mid =(\alpha +1)\frac{(1-\mid z{\mid }^2{)}^{\alpha }}{\mid 1-z{\mid }^{\alpha +2}}$(10) for $z\in \mathbb{D}$.

The following lemma is well-known but included here for the sake of convenience.

Lemma 2.3

Let $0<\text{β}\le 1$. Then $(a+b{)}^{\text{β}}\le a^{\text{β}}+b^{\text{β}}$ for $a,b\ge 0$.

Proof

Sketch of proof

First show that $(1+x{)}^{\text{β}}\le 1+x^{\text{β}}$ for x>0. A homogeneity argument then leads to the conclusion of the lemma.

The interest in Lemma 2.3 comes from the fact that the power function operates on metrics: If d is a metric on a set X, then so is the function $d_{\text{β}}(x_1,x_2)=d(x_1,x_2{)}^{\text{β}},x_1,x_2\in X,$ for $0<\text{β}\le 1$.

Recall the standard Gamma function $\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{\mathrm{e}}^{-t}\mathrm{d}t,x>0.$ For the sake of easy reference we recall the following result from [2, Section 2].

Theorem 2.4

Let $\alpha >-1$ and set $M\left(\alpha \right)=\mathrm{\Gamma }(\alpha +1)/\mathrm{\Gamma }(\alpha /2+1{)}^2$. Then $\frac{1}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha +1}}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\theta }{\mid }^{\alpha +2}}\mathrm{d}\theta \le M\left(\alpha \right)$ for $0\le r<1$. Furthermore, $\underset{r\to 1}{lim}\frac{1}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha +1}}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\theta }{\mid }^{\alpha +2}}\mathrm{d}\theta =M\left(\alpha \right).$

For a detailed discussion of Theorem 2.4 and its relation to hypergeometry we refer to [3].

We next estimate the $\overline{\mathrm{\partial }}$-derivative of the Poisson integral. We denote by $C\left(\mathbb{T}\right)$ the space of continuous functions on $\mathbb{T}$.

Theorem 2.5

Let $\alpha >-1$ and $0<\text{β}\le 1$ be such that $\text{β}<\alpha +1$. Let $f\in C\left(\mathbb{T}\right)$ be such that $\mid f\left({\mathrm{e}}^{\mathrm{i}{\theta }_1}\right)-f\left({\mathrm{e}}^{\mathrm{i}{\theta }_2}\right)\mid \le C\mid {\mathrm{e}}^{\mathrm{i}{\theta }_1}-{\mathrm{e}}^{\mathrm{i}{\theta }_2}{\mid }^{\text{β}},{\mathrm{e}}^{\mathrm{i}{\theta }_1},{\mathrm{e}}^{\mathrm{i}{\theta }_2}\in \mathbb{T},$ where C is a finite positive constant. Set $u=P_{\alpha }\left[f\right]$. Then $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le 2C(\alpha +1)M(\alpha -\text{β})\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}}$ for $z\in \mathbb{D},$ where $M(\alpha -\text{β})$ is as in Theorem 2.4.

Proof.

Let $z=r{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{D}$ with $r\ge 0$ and ${\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}$. Recall Lemma 2.1. By the triangle inequality we have that $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le \frac{1}{2\pi }{\int }_{\mathbb{T}}\mid \overline{\mathrm{\partial }}{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}(\theta -\tau )}\right)\mid \mid f\left({\mathrm{e}}^{\mathrm{i}\tau }\right)-f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)\mid \mathrm{d}\tau .$ By the assumption on f we have that $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le \frac{C}{2\pi }{\int }_{\mathbb{T}}\mid \overline{\mathrm{\partial }}{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}(\theta -\tau )}\right)\mid \mid {\mathrm{e}}^{\mathrm{i}\tau }-{\mathrm{e}}^{\mathrm{i}\theta }{\mid }^{\text{β}}\mathrm{d}\tau =\frac{C}{2\pi }{\int }_{\mathbb{T}}\mid \overline{\mathrm{\partial }}{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}\tau }\right)\mid \mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau ,$ where the last equality follows by a change of variables. We next invoke Lemma 2.2 to the extent that (11) $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le (\alpha +1)\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha }}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +2}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(11) for $z\in \mathbb{D}$ with $r=\mid z\mid$.

We shall now estimate the factor $\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}$ in the integrand in (11(11) $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le (\alpha +1)\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha }}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +2}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(11) ). By Lemma 2.3 and the triangle inequality we have that $\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\le \mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}+\mid r{\mathrm{e}}^{\mathrm{i}\tau }-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}=\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}+(1-r{)}^{\text{β}}.$ By a geometric consideration we see that $\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\le 2\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}$ for 0<r<1 and ${\mathrm{e}}^{\mathrm{i}\tau }\in \mathbb{T}$.

We now return to the estimation of $\overline{\mathrm{\partial }}u$. By (11(11) $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le (\alpha +1)\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha }}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +2}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(11) ) and the result of the previous paragraph we have that $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le 2(\alpha +1)\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha }}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha -\text{β}+2}}\mathrm{d}\tau$ for $z\in \mathbb{D}$, where $r=\mid z\mid$. An application of Theorem 2.4 now yields the conclusion of the theorem.

We now turn our attention to the ∂-derivative of the Poisson integral.

Lemma 2.6

Let $\alpha >-1$. Let $f\in L^1\left(\mathbb{T}\right)$ and set $u=P_{\alpha }\left[f\right]$. Then $\mathrm{\partial }u\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}\mathrm{\partial }{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right){\mathrm{e}}^{-\mathrm{i}\tau }\left(f\right({\mathrm{e}}^{\mathrm{i}\tau })-f({\mathrm{e}}^{\mathrm{i}\theta }\left)\right)\mathrm{d}\tau$ for $z\in \mathbb{D}$.

Proof.

The proof of the lemma is analogous to that of Lemma 2.1 and is therefore omitted.

Recall the partial fraction decomposition (12) $\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}=\frac{1}{1-z}+\frac{\overline{z}}{1-\overline{z}}$(12) of the classical Poisson kernel.

We now calculate $\mathrm{\partial }{P}_{\alpha }$.

Lemma 2.7

Let $\alpha >-1$ and let $P_{\alpha }$ be as in (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ). Then $\mathrm{\partial }{P}_{\alpha }\left(z\right)=\frac{1}{1-z}\left(\frac{1}{1-z}-\frac{\alpha \overline{z}}{1-\overline{z}}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }$ for $z\in \mathbb{D}$.

Proof.

Recall formulas (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ) and (12(12) $\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}=\frac{1}{1-z}+\frac{\overline{z}}{1-\overline{z}}$(12) ). Differentiating we have that $\begin{array}{rl}\mathrm{\partial }{P}_{\alpha }\left(z\right)& =\alpha {\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha -1}\frac{(-\overline{z})}{1-\overline{z}}\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}+{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1}{(1-z{)}^2}\\ & =\frac{1}{1-z}\left(\frac{1}{1-z}-\frac{\alpha \overline{z}}{1-\overline{z}}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\end{array}$ for $z\in \mathbb{D}$, where the last equality is straightforward to check.

It is evident that the quotient $(1-\overline{z})/(1-z)$ belongs to $\mathbb{T}$ for every $z\in \mathbb{C}$ with $z\ne 1$.

Lemma 2.8

We have that $\left\{\frac{1-\overline{z}}{1-z}:z\in \mathbb{D}\mathrm{a}\mathrm{n}\mathrm{d}\mid 1-z\mid <\epsilon \right\}=\mathbb{T}\setminus \{-1\}$ for every $\epsilon >0$.

Proof.

Let ${\mathrm{e}}^{\mathrm{i}\tau }\in \mathbb{T}\setminus \left\{1\right\}$. The line segment ${\gamma }_{{\mathrm{e}}^{\mathrm{i}\tau }}$ between ${\mathrm{e}}^{\mathrm{i}\tau }$ and the point 1 is parametrized by $z\left(t\right)=t+(1-t){\mathrm{e}}^{\mathrm{i}\tau },0<t<1.$ We have that $\frac{1-\overline{z\left(t\right)}}{1-z\left(t\right)}=\frac{1-t-(1-t){\mathrm{e}}^{-\mathrm{i}\tau }}{1-t-(1-t){\mathrm{e}}^{\mathrm{i}\tau }}=\frac{1-{\mathrm{e}}^{-\mathrm{i}\tau }}{1-{\mathrm{e}}^{\mathrm{i}\tau }}=\frac{{\mathrm{e}}^{-\mathrm{i}\tau /2}({\mathrm{e}}^{\mathrm{i}\tau /2}-{\mathrm{e}}^{-\mathrm{i}\tau /2})}{{\mathrm{e}}^{\mathrm{i}\tau /2}({\mathrm{e}}^{-\mathrm{i}\tau /2}-{\mathrm{e}}^{\mathrm{i}\tau /2})}=-{\mathrm{e}}^{-\mathrm{i}\tau }.$ Observe that $\mathbb{D}$ is the disjoint union of line segments ${\gamma }_{{\mathrm{e}}^{\mathrm{i}\tau }}$ as above. Thus $\left\{\frac{1-\overline{z}}{1-z}:z\in \mathbb{D}\mathrm{a}\mathrm{n}\mathrm{d}\mid 1-z\mid <\epsilon \right\}=\{-{\mathrm{e}}^{-\mathrm{i}\tau }:{\mathrm{e}}^{\mathrm{i}\tau }\in \mathbb{T}\setminus \{1\left\}\right\}=\mathbb{T}\setminus \{-1\}$ for $\epsilon >0$. This completes the proof of the lemma.

We remark in passing that $\frac{1-\overline{z}}{1-z}=-1$ if and only if $Re\left(z\right)=1$.

We now estimate $\mathrm{\partial }{P}_{\alpha }$.

Proposition 2.9

Let $\alpha >-1$ and let $P_{\alpha }$ be as in (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ). Then $\mid \mathrm{\partial }{P}_{\alpha }\left(z\right)\mid \le (\mid \alpha \mid +1)\frac{(1-\mid z{\mid }^2{)}^{\alpha }}{\mid 1-z{\mid }^{\alpha +2}}$ for $z\in \mathbb{D}$. The constant $\mid \alpha \mid +1$ is best possible.

Proof.

From Lemma 2.7 we have that $\mathrm{\partial }{P}_{\alpha }\left(z\right)=q\left(z\right)\frac{(1-\mid z{\mid }^2{)}^{\alpha }}{(1-z)(1-\overline{z}{)}^{\alpha +1}},$ for $z\in \mathbb{D}$, where $q\left(z\right)=\frac{1-\overline{z}}{1-z}-\alpha \overline{z}.$ Thus (13) $\mid \mathrm{\partial }{P}_{\alpha }\left(z\right)\mid =\mid q\left(z\right)\mid \frac{(1-\mid z{\mid }^2{)}^{\alpha }}{\mid 1-z{\mid }^{\alpha +2}}$(13) for $z\in \mathbb{D}$.

It remains to calculate the supremum $\underset{z\in \mathbb{D}}{sup}\mid q\left(z\right)\mid$. By the triangle inequality we have that $\underset{z\in \mathbb{D}}{sup}\mid q\left(z\right)\mid \le 1+\mid \alpha \mid$. From Lemma 2.8 we have that $\underset{z\in \mathbb{D}}{sup}\mid q\left(z\right)\mid \ge \mid {\mathrm{e}}^{\mathrm{i}\theta }-\alpha \mid$ for every ${\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}$. Maximizing over ${\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}$ we conclude that $\underset{z\in \mathbb{D}}{sup}\mid q\left(z\right)\mid =1+\mid \alpha \mid .$ This completes the proof of the proposition.

We now return to the α-harmonic Poisson integral.

Theorem 2.10

Let $\alpha >-1$ and $0<\text{β}\le 1$ be such that $\text{β}<\alpha +1$. Let $f\in C\left(\mathbb{T}\right)$ be such that $\mid f\left({\mathrm{e}}^{\mathrm{i}{\theta }_1}\right)-f\left({\mathrm{e}}^{\mathrm{i}{\theta }_2}\right)\mid \le C\mid {\mathrm{e}}^{\mathrm{i}{\theta }_1}-{\mathrm{e}}^{\mathrm{i}{\theta }_2}{\mid }^{\text{β}},{\mathrm{e}}^{\mathrm{i}{\theta }_1},{\mathrm{e}}^{\mathrm{i}{\theta }_2}\in \mathbb{T},$ where C is a finite positive constant. Set $u=P_{\alpha }\left[f\right]$. Then $\mid \mathrm{\partial }u\left(z\right)\mid \le 2C(\mid \alpha \mid +1)M(\alpha -\text{β})\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$ where $M(\alpha -\text{β})$ is as in Theorem 2.4.

Proof.

The proof parallels that of Theorem 2.5. Let $z=r{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{D}$ with $r\ge 0$ and ${\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}$. Recall Lemma 2.6. By the triangle inequality we have that $\mid \mathrm{\partial }u\left(z\right)\mid \le \frac{1}{2\pi }{\int }_{\mathbb{T}}\mid \mathrm{\partial }{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}(\theta -\tau )}\right)\mid \mid f\left({\mathrm{e}}^{\mathrm{i}\tau }\right)-f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)\mid \mathrm{d}\tau .$ By the assumption on f we have that $\mid \mathrm{\partial }u\left(z\right)\mid \le \frac{C}{2\pi }{\int }_{\mathbb{T}}\mid \mathrm{\partial }{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}(\theta -\tau )}\right)\mid {\mathrm{e}}^{\mathrm{i}\tau }-{\mathrm{e}}^{\mathrm{i}\theta }{\mid }^{\text{β}}\mathrm{d}\tau =\frac{C}{2\pi }{\int }_{\mathbb{T}}\mid \mathrm{\partial }{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}\tau }\right)\mid \mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau ,$ where the last equality follows by a change of variables. We next invoke Proposition 2.9 to the extent that (14) $\mid \mathrm{\partial }u\left(z\right)\mid \le (\mid \alpha \mid +1)\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha }}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +2}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(14) for $z\in \mathbb{D}$ with $r=\mid z\mid$. Formula (14(14) $\mid \mathrm{\partial }u\left(z\right)\mid \le (\mid \alpha \mid +1)\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha }}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +2}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(14) ) should be compared with formula (11(11) $\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le (\alpha +1)\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha }}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +2}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(11) ) in the proof of Theorem 2.5. We now proceed as in the proof of Theorem 2.5 to arrive at the conclusion of the theorem.

We next calculate the angular derivative of $P_{\alpha }$.

Theorem 2.11

Let $\alpha >-1$ and let $P_{\alpha }$ be as in (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ). Then $(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)=\left(\frac{z}{1-z}-(\alpha +1)\frac{\overline{z}}{1-\overline{z}}\right)\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }$ for $z\in \mathbb{D}$.

Proof.

From Lemma 2.7 we have that $z\mathrm{\partial }{P}_{\alpha }\left(z\right)=\left(\frac{z}{(1-z{)}^2}-\alpha \frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }$ for $z\in \mathbb{D}$. Using also Lemma 2.2 we have that (15) $\begin{array}{rl}(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)& =\left(\frac{z}{(1-z{)}^2}-\alpha \frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}-(\alpha +1)\frac{\overline{z}}{(1-\overline{z}{)}^2}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\\ & =\left(\frac{z}{(1-z{)}^2}-\frac{\overline{z}}{(1-\overline{z}{)}^2}-\alpha \left(\frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}+\frac{\overline{z}}{(1-\overline{z}{)}^2}\right)\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\end{array}$(15) for $z\in \mathbb{D}$, where the last equality is straightforward to check.

We now turn our attention to the leftmost factor in (15(15) $\begin{array}{rl}(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)& =\left(\frac{z}{(1-z{)}^2}-\alpha \frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}-(\alpha +1)\frac{\overline{z}}{(1-\overline{z}{)}^2}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\\ & =\left(\frac{z}{(1-z{)}^2}-\frac{\overline{z}}{(1-\overline{z}{)}^2}-\alpha \left(\frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}+\frac{\overline{z}}{(1-\overline{z}{)}^2}\right)\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\end{array}$(15) ). By a partial fraction decomposition we have that $\begin{array}{rl}\frac{z}{(1-z{)}^2}-\frac{\overline{z}}{(1-\overline{z}{)}^2}& =\frac{1}{(1-z{)}^2}-\frac{1}{(1-\overline{z}{)}^2}+\frac{1}{1-\overline{z}}-\frac{1}{1-z}\\ & =\left(\frac{1}{1-z}-\frac{1}{1-\overline{z}}\right)\left(\frac{1}{1-z}+\frac{1}{1-\overline{z}}\right)+\frac{1}{1-\overline{z}}-\frac{1}{1-z}\\ & =\left(\frac{1}{1-z}-\frac{1}{1-\overline{z}}\right)\left(\frac{1}{1-z}+\frac{1}{1-\overline{z}}-1\right)\end{array}$ for $z\in \mathbb{D}$. In view of the partial fraction formula (12(12) $\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}=\frac{1}{1-z}+\frac{\overline{z}}{1-\overline{z}}$(12) ) for the classical Poisson kernel we thus have that (16) $\frac{z}{(1-z{)}^2}-\frac{\overline{z}}{(1-\overline{z}{)}^2}=\left(\frac{1}{1-z}-\frac{1}{1-\overline{z}}\right)\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}$(16) for $z\in \mathbb{D}$.

We return to the leftmost factor in (15(15) $\begin{array}{rl}(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)& =\left(\frac{z}{(1-z{)}^2}-\alpha \frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}-(\alpha +1)\frac{\overline{z}}{(1-\overline{z}{)}^2}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\\ & =\left(\frac{z}{(1-z{)}^2}-\frac{\overline{z}}{(1-\overline{z}{)}^2}-\alpha \left(\frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}+\frac{\overline{z}}{(1-\overline{z}{)}^2}\right)\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\end{array}$(15) ). We have that (17) $\frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}+\frac{\overline{z}}{(1-\overline{z}{)}^2}=\frac{\overline{z}}{1-\overline{z}}\left(\frac{z}{1-z}+\frac{1}{1-\overline{z}}\right)=\frac{\overline{z}}{1-\overline{z}}\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}$(17) for $z\in \mathbb{D}$, where the last equality again follows by (12(12) $\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}=\frac{1}{1-z}+\frac{\overline{z}}{1-\overline{z}}$(12) ).

Recall formula (15(15) $\begin{array}{rl}(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)& =\left(\frac{z}{(1-z{)}^2}-\alpha \frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}-(\alpha +1)\frac{\overline{z}}{(1-\overline{z}{)}^2}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\\ & =\left(\frac{z}{(1-z{)}^2}-\frac{\overline{z}}{(1-\overline{z}{)}^2}-\alpha \left(\frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}+\frac{\overline{z}}{(1-\overline{z}{)}^2}\right)\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\end{array}$(15) ). From (16(16) $\frac{z}{(1-z{)}^2}-\frac{\overline{z}}{(1-\overline{z}{)}^2}=\left(\frac{1}{1-z}-\frac{1}{1-\overline{z}}\right)\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}$(16) ) and (17(17) $\frac{z}{1-z}\frac{\overline{z}}{1-\overline{z}}+\frac{\overline{z}}{(1-\overline{z}{)}^2}=\frac{\overline{z}}{1-\overline{z}}\left(\frac{z}{1-z}+\frac{1}{1-\overline{z}}\right)=\frac{\overline{z}}{1-\overline{z}}\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}$(17) ) we have that $\begin{array}{rl}(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)& =\left(\left(\frac{1}{1-z}-\frac{1}{1-\overline{z}}\right)\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}-\alpha \frac{\overline{z}}{1-\overline{z}}\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}\right){\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\\ & =\left(\frac{1}{1-z}-\frac{1}{1-\overline{z}}-\alpha \frac{\overline{z}}{1-\overline{z}}\right)\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\end{array}$ for $z\in \mathbb{D}$. We next rewrite the leftmost factor so that $\begin{array}{rl}(z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)& =\left(1+\frac{z}{1-z}-\left(1+\frac{\overline{z}}{1-\overline{z}}\right)-\alpha \frac{\overline{z}}{1-\overline{z}}\right)\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\\ & =\left(\frac{z}{1-z}-(\alpha +1)\frac{\overline{z}}{1-\overline{z}}\right)\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2}{\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\end{array}$ for $z\in \mathbb{D}$. This completes the proof of the theorem.

Theorem 2.11 leads to an important inequality for the angular derivative of the Poisson kernel.

Corollary 2.12

Let $\alpha >-1$ and let $P_{\alpha }$ be as in (5(5) $P_{\alpha }\left(z\right)={\left(\frac{1-\mid z{\mid }^2}{1-\overline{z}}\right)}^{\alpha }\frac{1-\mid z{\mid }^2}{\mid 1-z{\mid }^2},z\in \mathbb{D},$(5) ). Then $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})P_{\alpha }\left(z\right)\mid \le (\alpha +2)\mid z\mid \frac{(1-\mid z{\mid }^2{)}^{\alpha +1}}{\mid 1-z{\mid }^{\alpha +3}}$ for $z\in \mathbb{D}$.

Proof.

Recall Theorem 2.11. The result follows by the triangle inequality.

The point of Corollary 2.12 is that the angular derivative of $P_{\alpha }$ behaves somewhat better than $\overline{\mathrm{\partial }}{P}_{\alpha }$ or $\mathrm{\partial }{P}_{\alpha }$ (see formula (10(10) $\mid \overline{\mathrm{\partial }}{P}_{\alpha }\left(z\right)\mid =(\alpha +1)\frac{(1-\mid z{\mid }^2{)}^{\alpha }}{\mid 1-z{\mid }^{\alpha +2}}$(10) ) or Proposition 2.9).

Theorem 2.13

Let $\alpha >-1$ and $0<\text{β}\le 1$. Let $f\in C\left(\mathbb{T}\right)$ be such that $\mid f\left({\mathrm{e}}^{\mathrm{i}{\theta }_1}\right)-f\left({\mathrm{e}}^{\mathrm{i}{\theta }_2}\right)\mid \le C\mid {\mathrm{e}}^{\mathrm{i}{\theta }_1}-{\mathrm{e}}^{\mathrm{i}{\theta }_2}{\mid }^{\text{β}},{\mathrm{e}}^{\mathrm{i}{\theta }_1},{\mathrm{e}}^{\mathrm{i}{\theta }_2}\in \mathbb{T},$ where C is a finite positive constant. Set $u=P_{\alpha }\left[f\right]$. Then $\mid (z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }})u\left(z\right)\mid \le 2C(\alpha +2)M(\alpha -\text{β}+1)\frac{\mid z\mid }{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$ where $M(\alpha -\text{β}+1)$ is as in Theorem 2.4.

Proof.

The proof parallels that of Theorem 2.5. Let $z=r{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{D}$ with $r\ge 0$ and ${\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T}$. Let $A=z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }}$ be the angular derivative. By Lemmas 2.1 and 2.6 we have that $Au\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}A{P}_{\alpha }\left(z{\mathrm{e}}^{-\mathrm{i}\tau }\right)\left(f\right({\mathrm{e}}^{\mathrm{i}\tau })-f({\mathrm{e}}^{\mathrm{i}\theta }\left)\right)\mathrm{d}\tau .$ By the triangle inequality we have that $\mid Au\left(z\right)\mid \le \frac{1}{2\pi }{\int }_{\mathbb{T}}\mid A{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}(\theta -\tau )}\right)\mid \mid f\left({\mathrm{e}}^{\mathrm{i}\tau }\right)-f\left({\mathrm{e}}^{\mathrm{i}\theta }\right)\mid \mathrm{d}\tau .$ By the assumption on f we have that $\mid Au\left(z\right)\mid \le \frac{C}{2\pi }{\int }_{\mathbb{T}}\mid A{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}(\theta -\tau )}\right)\mid \mid {\mathrm{e}}^{\mathrm{i}\tau }-{\mathrm{e}}^{\mathrm{i}\theta }{\mid }^{\text{β}}\mathrm{d}\tau =\frac{C}{2\pi }{\int }_{\mathbb{T}}\mid A{P}_{\alpha }\left(r{\mathrm{e}}^{\mathrm{i}\tau }\right)\mid \mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau ,$ where the last equality follows by a change of variables. We next invoke Corollary 2.12 to the extent that (18) $\mid Au\left(z\right)\mid \le (\alpha +2)r\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha +1}}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +3}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(18) for $z\in \mathbb{D}$ with $r=\mid z\mid$.

As in the proof of Theorem 2.5 we have that $\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\le 2\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}$ for 0<r<1 and ${\mathrm{e}}^{\mathrm{i}\tau }\in \mathbb{T}$. From (18(18) $\mid Au\left(z\right)\mid \le (\alpha +2)r\frac{C}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha +1}}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha +3}}\mid 1-{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\text{β}}\mathrm{d}\tau$(18) ) we have that $\mid Au\left(z\right)\mid \le 2C(\alpha +2)r\frac{1}{2\pi }{\int }_{\mathbb{T}}\frac{(1-r^2{)}^{\alpha +1}}{\mid 1-r{\mathrm{e}}^{\mathrm{i}\tau }{\mid }^{\alpha -\text{β}+3}}\mathrm{d}\tau$ An application of Theorem 2.4 now yields the conclusion of the theorem.

It is important to notice that Theorem 2.13 carries no additional assumption on the Lipschitz parameter $0<\text{β}\le 1$ (compare Theorems 2.5 and 2.10). The case $\text{β}=1$ of Theorem 2.13 is closely related to Chen [10, Lemma 3.2].

3. Lipschitz continuity from gradient growth

It is well-known that Lipschitz continuity $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ of a function $u\in C^1\left(\mathbb{D}\right)$ is inferred from a gradient growth estimate of the form (7(7) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le C\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D}$(7) ). This section pertains to a discussion of this aspect of the theory.

Let $0\le \text{β}\le 1$. We shall consider the quantity $d_{\text{β}}(z_1,z_2)=\underset{\gamma }{inf}{\int }_{\gamma }\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}}\mid \mathrm{d}z\mid ,z_1,z_2\in \mathbb{D},$ where the infimum is taken over all piecewice continuously differentiable curves γ from $z_1$ to $z_2$ in $\mathbb{D}$ and the integration is with respect to arclength. It is straightforward to check that the function $d_{\text{β}}:\mathbb{D}\times \mathbb{D}\to \mathbb{R}$ is a metric on $\mathbb{D}$.

It is evident that $d_1(z_1,z_2)=\mid z_1-z_2\mid$ is usual Euclidean distance. Furthermore, conformal invariance leads to the formula $d_0(z_1,z_2)=\frac{1}{2}\log \left(\frac{1+{\rho }_{\mathbb{D}}(z_1,z_2)}{1-{\rho }_{\mathbb{D}}(z_1,z_2)}\right)={\tanh }^{-1}\left({\rho }_{\mathbb{D}}\right(z_1,z_2\left)\right)$ for $z_1,z_2\in \mathbb{D}$, where ${\rho }_{\mathbb{D}}(z_1,z_2)=∣\frac{z_1-z_2}{1-{\overline{z}}_1{z}_2}∣,z_1,z_2\in \mathbb{D}$ (see for instance [19, Theorem 2.2]). The function $d_0$ is known as the hyperbolic metric for $\mathbb{D}$, whereas the function ${\rho }_{\mathbb{D}}$ is known as the pseudo hyperbolic metric for $\mathbb{D}$.

We shall make use of the result that (19) $d_{\text{β}}(z_1,z_2)\le C\left(\text{β}\right)\mid z_1-z_2{\mid }^{\text{β}},z_1,z_2\in \mathbb{D},$(19) for $0<\text{β}\le 1$, where $C\left(\text{β}\right)$ is a finite positive constant. A careful analysis shows that the constant in (19(19) $d_{\text{β}}(z_1,z_2)\le C\left(\text{β}\right)\mid z_1-z_2{\mid }^{\text{β}},z_1,z_2\in \mathbb{D},$(19) ) can be chosen of the form $C\left(\text{β}\right)=C/\text{β}$, where C is an absolute constant. For proofs of (19(19) $d_{\text{β}}(z_1,z_2)\le C\left(\text{β}\right)\mid z_1-z_2{\mid }^{\text{β}},z_1,z_2\in \mathbb{D},$(19) ) we refer to the original papers Gehring and Martio [11, Section 2] or Hardy and Littlewood [6, Section 5].

Theorem 3.1

Let $u\in C^1\left(\mathbb{D}\right)$ be such that (20) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le \frac{C}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(20) for some $0<\text{β}\le 1$ and C>0. Then $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ and $\mid u\left(z_1\right)-u\left(z_2\right)\mid \le CC\left(\text{β}\right)\mid z_1-z_2{\mid }^{\text{β}}$ for $z_1,z_2\in \mathbb{D},$ where $C\left(\text{β}\right)$ is as in (19(19) $d_{\text{β}}(z_1,z_2)\le C\left(\text{β}\right)\mid z_1-z_2{\mid }^{\text{β}},z_1,z_2\in \mathbb{D},$(19) ).

Proof.

Let $z_1,z_2\in \mathbb{D}$. By the fundamental theorem of calculus we have that $u\left(z_2\right)-u\left(z_1\right)={\int }_{\gamma }\mathrm{\partial }u\left(z\right)\mathrm{d}z+\overline{\mathrm{\partial }}u\left(z\right)\mathrm{d}\overline{z},$ where γ is a piecewise $C^1$-curve from $z_1$ to $z_2$ (see for instance Gamelin [20, Section III.2]). By the triangle inequality we have that $\mid u\left(z_2\right)-u\left(z_1\right)\mid \le {\int }_{\gamma }(\mid \mathrm{\partial }u(z)\mid +\mid \overline{\mathrm{\partial }}u(z)\mid )\mid dz\mid \le C{\int }_{\gamma }\frac{1}{(1-\mid z{\mid }^2{)}^{1-\text{β}}}\mid \mathrm{d}z\mid ,$ where the last inequality follows by (20(20) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le \frac{C}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(20) ). Passing to an infimum over curves γ we have that $\mid u\left(z_2\right)-u\left(z_1\right)\mid \le C{d}_{\text{β}}(z_1,z_2)\le CC\left(\text{β}\right)\mid z_1-z_2{\mid }^{\text{β}},$ where the last equality follows by (19(19) $d_{\text{β}}(z_1,z_2)\le C\left(\text{β}\right)\mid z_1-z_2{\mid }^{\text{β}},z_1,z_2\in \mathbb{D},$(19) ). This completes the proof of the theorem.

The interest in Theorem 3.1 is when the function u has some interior rigidity.

Theorem 3.2

Let $\alpha >-1$ and $0<\text{β}\le 1$ be such that $\text{β}<\alpha +1$. Let $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ and set $u=P_{\alpha }\left[f\right]$. Then $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$.

Proof.

Recall Theorems 2.5 and 2.10. The result now follows from Theorem 3.1.

The case of analytic functions merits particular mention.

Theorem 3.3

Let $\alpha >-1$ and $0<\text{β}\le 1$. Let $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ be such that $\hat{f}\left(n\right)=0$ for n<0. Then $u=P_0\left[f\right]$ belongs to ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$.

Proof.

Recall Theorem 2.13. The spectral assumption means that $\overline{\mathrm{\partial }}u=0$ in $\mathbb{D}$. The result now follows from Theorem 3.1.

Remark 3.4

We point out that Theorem 3.3 can be regarded as a special case of Theorem 3.2. Indeed, if u is as in Theorem 3.3, then $u=P_{\alpha }\left[f\right]$ for all $\alpha >-1$ and an application of Theorem 3.2 with $\alpha >0$ yields that $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$.

Theorem 3.3 goes back to Hardy and Littlewood [6, Section 5].

Let us revisit a classic function (21) $f\left(x\right)=\log (x+\sqrt{x^2+1}),x\in \mathbb{R},$(21) from calculus. Notice that $f={\sinh }^{-1}$. Repeated differentiation shows that $f^{\left(n\right)}\left(x\right)=\frac{R_n\left(x\right)}{\sqrt{x^2+1}},x\in \mathbb{R},$ for $n\in {\mathbb{Z}}^{+}$, where $R_1\left(x\right)=1$ and $R_{n+1}\left(x\right)=R_n^{\prime }\left(x\right)-\frac{x}{x^2+1}{R}_n\left(x\right)$ for $x\in \mathbb{R}$. The function $R_n$ is a rational function. From the vanishing of $R_n$ at infinity we see that there is an estimate of the form (22) $\mid f^{\left(n\right)}\left(x\right)\mid \le C_n(x^2+1{)}^{-n/2},x\in \mathbb{R},$(22) for $n\in {\mathbb{Z}}^{+}$, where $C_n$ is a finite positive constant.

We next construct an example where Theorem 3.1 is of less interest.

Proposition 3.5

Let $0<\text{β}<1$ and let w be a positive function on $(-1,1)$. Then there exists a function $u\in C^{\mathrm{\infty }}(-1,1)$ such that $u\in {\mathrm{\Lambda }}_{\text{β}}(-1,1)$ and $\underset{x\to -1}{lim sup}{u}^{\prime }\left(x\right)/w\left(x\right)=\underset{x\to 1}{lim sup}{u}^{\prime }\left(x\right)/w\left(x\right)=+\mathrm{\infty }.$

Proof.

We shall make use of the functions $h_{\epsilon }\left(x\right)=(x+\sqrt{x^2+\epsilon }{)}^{\text{β}}={\epsilon }^{\text{β}/2}\exp (\text{β}f\left(x/\sqrt{\epsilon }\right)),x\in \mathbb{R},$ for $\epsilon >0$, where f is as in (21(21) $f\left(x\right)=\log (x+\sqrt{x^2+1}),x\in \mathbb{R},$(21) ). Notice that $h_{\epsilon }\left(x\right)\le 2^{\text{β}}(x^2+\epsilon {)}^{\text{β}/2}$ for $x\in \mathbb{R}$. An application of Lemma 2.3 shows that (23) $\mid h_{\epsilon }\left(x_2\right)-h_{\epsilon }\left(x_1\right)\mid \le 2^{\text{β}}\mid x_2-x_1{\mid }^{\text{β}},x_1,x_2\in \mathbb{R},$(23) for $\epsilon >0$. By differentiation we have that (24) $h_{\epsilon }^{\prime }\left(x\right)=\frac{\text{β}}{\sqrt{\epsilon }}{f}^{\prime }\left(x/\sqrt{\epsilon }\right)h_{\epsilon }\left(x\right)=\text{β}\frac{1}{\sqrt{x^2+\epsilon }}{h}_{\epsilon }\left(x\right),x\in \mathbb{R}.$(24) In particular, we have that $0<h_{\epsilon }^{\prime }\left(x\right)\le \text{β}{2}^{\text{β}}(x^2+\epsilon {)}^{(\text{β}-1)/2}$ for $x\in \mathbb{R}$ and $h_{\epsilon }^{\prime }\left(0\right)=\text{β}{\epsilon }^{(\text{β}-1)/2}$. Moreover, an induction using (22(22) $\mid f^{\left(n\right)}\left(x\right)\mid \le C_n(x^2+1{)}^{-n/2},x\in \mathbb{R},$(22) ) and (24(24) $h_{\epsilon }^{\prime }\left(x\right)=\frac{\text{β}}{\sqrt{\epsilon }}{f}^{\prime }\left(x/\sqrt{\epsilon }\right)h_{\epsilon }\left(x\right)=\text{β}\frac{1}{\sqrt{x^2+\epsilon }}{h}_{\epsilon }\left(x\right),x\in \mathbb{R}.$(24) ) shows that (25) $\mid h_{\epsilon }^{\left(n\right)}\left(x\right)\mid \le C_n(x^2+\epsilon {)}^{(\text{β}-n)/2},x\in \mathbb{R},$(25) for $\epsilon >0$ and $n\ge 0$, where $C_n$ is a finite positive constant.

Let us now construct a function u with the desired properties. Let $\{x_k{\}}_{k=1}^{\mathrm{\infty }}$ be a sequence of points in the interval $(-1,1)$ with accumulation points $-1$ and 1 only. Let $\{{\epsilon }_k{\}}_{k=1}^{\mathrm{\infty }}$ be a sequence of positive numbers such that ${\epsilon }_k\to 0$ as $k\to \mathrm{\infty }$. We consider a function u of the form (26) $u\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}{h}_{{\epsilon }_k}(x-x_k),-1<x<1,$(26) where the $h_{{\epsilon }_k}$'s are as above. From the bound (23(23) $\mid h_{\epsilon }\left(x_2\right)-h_{\epsilon }\left(x_1\right)\mid \le 2^{\text{β}}\mid x_2-x_1{\mid }^{\text{β}},x_1,x_2\in \mathbb{R},$(23) ) we see that the Lipschitz norms $\parallel h_{{\epsilon }_k}(\cdot -x_k){\parallel }_{{\mathrm{\Lambda }}_{\text{β}}(-1,1)}$ for $k=1,2,\dots$ are uniformly bounded. By completeness of the space ${\mathrm{\Lambda }}_{\text{β}}(-1,1)$ this yields that $u\in {\mathrm{\Lambda }}_{\text{β}}(-1,1)$.

We next show that $u\in C^{\mathrm{\infty }}(-1,1)$. Let $K\subset (-1,1)$ be a compact set and $n\ge 0$ an integer. Let 0<r<1 be such that $K\subset (-r,r)$. Since the sequence $\{x_k{\}}_{k=1}^{\mathrm{\infty }}$ accumulates only at the points $-1$ and 1 there exists $N\ge 1$ such that $\mid x_k\mid \ge r$ for k>N. From the bound (25(25) $\mid h_{\epsilon }^{\left(n\right)}\left(x\right)\mid \le C_n(x^2+\epsilon {)}^{(\text{β}-n)/2},x\in \mathbb{R},$(25) ) we have that $\underset{x\in K}{max}\mid h_{{\epsilon }_k}^{\left(n\right)}(x-x_k)\mid \le C(n,K,r)<+\mathrm{\infty }$ for k>N. This proves that $\sum _{k\ge 1}\left(1/k^2\right)\underset{x\in K}{max}\mid h_{{\epsilon }_k}^{\left(n\right)}(x-x_k)\mid <+\mathrm{\infty }$. Varying the compact K and the integer $n\ge 0$, we see that the series (26(26) $u\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}{h}_{{\epsilon }_k}(x-x_k),-1<x<1,$(26) ) defining u is convergent in $C^{\mathrm{\infty }}(-1,1)$.

By differentiation, we see that $u^{\prime }\left(x_k\right)\ge \text{β}{{\epsilon }_k}^{(\text{β}-1)/2}/k^2$. The final conclusion of the proposition is satisfied provided ${\epsilon }_k\to 0$ rapidly enough.

We mention that growth estimates of the form (25(25) $\mid h_{\epsilon }^{\left(n\right)}\left(x\right)\mid \le C_n(x^2+\epsilon {)}^{(\text{β}-n)/2},x\in \mathbb{R},$(25) ) appear in the definition of symbol classes for pseudo differential operators (see Hörmander [21, Chapter 18]).

4. The homogeneous expansion

In this section, we discuss some generalities related to the homogeneous expansion of α-harmonic functions. Applications of this material appear in later sections.

Let us recall the Beta function defined by $B(a,b)={\int }_0^1{t}^{a-1}(1-t{)}^{b-1}\mathrm{d}t$ for a,b>0. It is well-known that $B(a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)},$ where Γ is the Gamma function (see [22, Section 1.1]).

Let $\alpha >-1$ and set $f_{\alpha ,n}\left(x\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-xt{)}^{\alpha }\mathrm{d}t,0\le x<1,$ for $n\in {\mathbb{Z}}^{+}$, where B is the Beta function. Clearly, $f_{\alpha ,n}\left(1\right)=\underset{x\to 1}{lim}{f}_{\alpha ,n}\left(x\right)=1$ as follows by a passage to the limit.

We now introduce the functions (27) $e_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D},$(27) for $n\in {\mathbb{Z}}^{+}$. From [2, Section 1] we know that a complex-valued function u in $\mathbb{D}$ is α-harmonic if and only if it has a series expansion of the form (28) $u\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{e}_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n,z\in \mathbb{D},$(28) for some sequence of complex numbers $\{a_n{\}}_{n=-\mathrm{\infty }}^{\mathrm{\infty }}$ such that (29) $\underset{\mid n\mid \to \mathrm{\infty }}{lim sup}\mid a_n{\mid }^{1/\mid n\mid }\le 1.$(29) Furthermore, the series expansion (28(28) $u\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{e}_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n,z\in \mathbb{D},$(28) ) is convergent in $C^{\mathrm{\infty }}\left(\mathbb{D}\right)$ provided (29(29) $\underset{\mid n\mid \to \mathrm{\infty }}{lim sup}\mid a_n{\mid }^{1/\mid n\mid }\le 1.$(29) ) holds.

We proceed to study the functions $f_{\alpha ,n}$ in some more detail. The integral formula (30) $f_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(30) extends the function $f_{\alpha ,n}$ to an analytic function in the slit plane $\mathbb{C}\setminus [1,\mathrm{\infty })$. Here the power is defined in the usual way using a logarithm which is real on the positive real axis.

Lemma 4.1

Let $\alpha >-1$. Let $f_{\alpha ,n}$ be as in (30(30) $f_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(30) ) for some $n\in {\mathbb{Z}}^{+}$. Then $z{f}_{\alpha ,n}^{\prime }\left(z\right)+n{f}_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}(1-z{)}^{\alpha }$ for $z\in \mathbb{C}\setminus [1,\mathrm{\infty })$.

Proof.

Differentiating under the integral sign in (30(30) $f_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(30) ) we have that $f_{\alpha ,n}^{\prime }\left(z\right)=-\frac{\alpha }{B(n,\alpha +1)}{\int }_0^1{t}^n(1-tz{)}^{\alpha -1}\mathrm{d}t$ for $z\in \mathbb{C}\setminus [1,\mathrm{\infty })$. An integration by parts now gives that $\begin{array}{rl}z{f}_{\alpha ,n}^{\prime }\left(z\right)& =\frac{1}{B(n,\alpha +1)}(1-z{)}^{\alpha }-\frac{n}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t\\ & =\frac{1}{B(n,\alpha +1)}(1-z{)}^{\alpha }-n{f}_{\alpha ,n}(z)\end{array}$ for $z\in \mathbb{C}\setminus [1,\mathrm{\infty })$, where the last equality again follows by (30(30) $f_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(30) ). This yields the conclusion of the lemma.

Lemma 4.1 leads to differential equations for the $e_{\alpha ,-n}$'s.

Proposition 4.2

Let $\alpha >-1$. Let $e_{\alpha ,-n}$ be as in (27(27) $e_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D},$(27) ) for some $n\in {\mathbb{Z}}^{+}$. Then $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}{e}_{\alpha ,-n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }{e}_{\alpha ,-n}\left(z\right)+n{e}_{\alpha ,-n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$ for $z\in \mathbb{D}$.

Proof.

Recall formula (27(27) $e_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D},$(27) ). Differentiating we have that $\overline{\mathrm{\partial }}{e}_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}^{\prime }(\mid z{\mid }^2)z{\overline{z}}^n+f_{\alpha ,n}(\mid z{\mid }^2)n{\overline{z}}^{n-1}=(\mid z{\mid }^2{f}_{\alpha ,n}^{\prime }(\mid z{\mid }^2)+n{f}_{\alpha ,n}(\mid z{\mid }^2\left)\right){\overline{z}}^{n-1}$ for $z\in \mathbb{D}$. We now use Lemma 4.1 to conclude that $\overline{\mathrm{\partial }}{e}_{\alpha ,-n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^{n-1}(1-\mid z{\mid }^2{)}^{\alpha }$ for $z\in \mathbb{D}$. This yields the first equation in the proposition.

Recall formula (27(27) $e_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D},$(27) ). Differentiating we have that $z\mathrm{\partial }{e}_{\alpha ,-n}\left(z\right)=z{f}_{\alpha ,n}^{\prime }(\mid z{\mid }^2)\overline{z}{\overline{z}}^n=\mid z{\mid }^2{f}_{\alpha ,n}^{\prime }(\mid z{\mid }^2){\overline{z}}^n$ for $z\in \mathbb{D}$. We now use Lemma 4.1 to conclude that $\begin{array}{rl}\overline{\mathrm{\partial }}{e}_{\alpha ,-n}\left(z\right)& =\left(\frac{1}{B(n,\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }-n{f}_{\alpha ,n}(\mid z{\mid }^2)\right){\overline{z}}^n\\ & =\frac{1}{B(n,\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }{\overline{z}}^n-n{e}_{\alpha ,-n}(z)\end{array}$ for $z\in \mathbb{D}$, where the last equality again follows by (27(27) $e_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D},$(27) ). This yields the second equation in the proposition.

Let us digress on the differential equations from Proposition 4.2.

Theorem 4.3

Let $\alpha >-1$ and $n\in {\mathbb{Z}}^{+}$. Then a function $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ satisfies the differential equations (31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) for $z\in \mathbb{D}\setminus \left\{0\right\}$ if and only if it has the form (32) $u\left(z\right)=e_{\alpha ,-n}\left(z\right)+c/z^n,z\in \mathbb{D}\setminus \left\{0\right\},$(32) for some $c\in \mathbb{C}$.

Proof.

Assume first that $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ satisfies (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ). In view of Proposition 4.2, the first equation in (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ) implies that $u\left(z\right)=e_{\alpha ,-n}\left(z\right)+f\left(z\right),z\in \mathbb{D}\setminus \left\{0\right\},$ where f is analytic in $\mathbb{D}\setminus \left\{0\right\}$. Now the second equation in (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ) and Proposition 4.2 implies that $z{f}^{\prime }\left(z\right)+nf\left(z\right)=0,z\in \mathbb{D}\setminus \left\{0\right\}.$ Solving for f we see that $f\left(z\right)=c/z^n$ for some constant $c\in \mathbb{C}$.

Conversely, using Proposition 4.2 it is straightforward to check that every function u of the form (32(32) $u\left(z\right)=e_{\alpha ,-n}\left(z\right)+c/z^n,z\in \mathbb{D}\setminus \left\{0\right\},$(32) ) satisfies (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ).

Following earlier practice from [2–4], a function u in $\mathbb{D}\setminus \left\{0\right\}$ is called homogeneous of order $n\in \mathbb{Z}$ with respect to rotations if it has the property that $u\left({\mathrm{e}}^{\mathrm{i}\tau }z\right)={\mathrm{e}}^{\mathrm{i}n\tau }u\left(z\right),z\in \mathbb{D}\setminus \left\{0\right\},$ for ${\mathrm{e}}^{\mathrm{i}\tau }\in \mathbb{T}$. Notice that the function $e_{\alpha ,-n}$ is homogeneous of order $-n$.

Lemma 4.4

Let $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ be homogeneous of order $n\in \mathbb{Z}$. Then $z\mathrm{\partial }u\left(z\right)-\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=nu\left(z\right)$ for $z\in \mathbb{D}\setminus \left\{0\right\}$.

Proof.

Assume first that $n\ge 0$. Since $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ is homogeneous of order $n\ge 0$ it has the form $u\left(z\right)=f(\mid z{\mid }^2)z^n,z\in \mathbb{D}\setminus \left\{0\right\},$ for some $f\in C^1(0,1)$. Differentiating we see that $z\mathrm{\partial }u\left(z\right)=(\mid z{\mid }^2{f}^{\prime }(\mid z{\mid }^2)+nf(\mid z{\mid }^2\left)\right)z^n\mathrm{a}\mathrm{n}\mathrm{d}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\mid z{\mid }^2{f}^{\prime }(\mid z{\mid }^2)z^n$ for $z\in \mathbb{D}\setminus \left\{0\right\}$. By these two differentiation formulas we see that $z\mathrm{\partial }u\left(z\right)-\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=nf(\mid z{\mid }^2)z^n=nu\left(z\right)$ for $z\in \mathbb{D}\setminus \left\{0\right\}$. This yields the conclusion of the lemma for $n\ge 0$.

If $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ is homogeneous of order $n\in \mathbb{Z}$, then its complex conjugate $\overline{u}$ has homogeneity $-n$. The case n<0 thus follows by complex conjugation.

Let $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ and consider its homogeneous parts defined by (33) $u_n\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}u\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}n\theta }\mathrm{d}\theta ,z\in \mathbb{D}\setminus \left\{0\right\},$(33) for $n\in \mathbb{Z}$. Notice that the function $u_n$ is homogeneous of order n with respect to rotations and that $u_n\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ for $n\in \mathbb{Z}$. From a classical result of Fejér we have that (34) $u\left(z\right)=\underset{N\to \mathrm{\infty }}{lim}\sum _{n=-N}^N\left(1-\frac{\mid n\mid }{N+1}\right)u_n\left(z\right),z\in \mathbb{D}\setminus \left\{0\right\},$(34) with convergence in the space $C^1(\mathbb{D}\setminus \{0\left\}\right)$, where the $u_n$'s are as in (33(33) $u_n\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}u\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}n\theta }\mathrm{d}\theta ,z\in \mathbb{D}\setminus \left\{0\right\},$(33) ) (see Katznelson [7, Section I.2]). We shall refer to the expansion (33(33) $u_n\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}u\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}n\theta }\mathrm{d}\theta ,z\in \mathbb{D}\setminus \left\{0\right\},$(33) )–(34(34) $u\left(z\right)=\underset{N\to \mathrm{\infty }}{lim}\sum _{n=-N}^N\left(1-\frac{\mid n\mid }{N+1}\right)u_n\left(z\right),z\in \mathbb{D}\setminus \left\{0\right\},$(34) ) as the homogeneous expansion of u.

The differential equations (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ) in Theorem 4.3 exhibit certain structure owing to homogeneity.

Theorem 4.5

Let $n\in \mathbb{Z}$. Then a function $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ satisfies the differential equation (35) $z\mathrm{\partial }u\left(z\right)-\overline{z}\mathrm{\partial }u\left(z\right)=nu\left(z\right)$(35) for $z\in \mathbb{D}\setminus \left\{0\right\}$ if and only if it is homogeneous of order n with respect to rotations.

Proof.

By Lemma 4.4 we know that $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ satisfies (35(35) $z\mathrm{\partial }u\left(z\right)-\overline{z}\mathrm{\partial }u\left(z\right)=nu\left(z\right)$(35) ) if u is homogeneous of order n with respect to rotations. Assume next that $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ is a general solution of (35(35) $z\mathrm{\partial }u\left(z\right)-\overline{z}\mathrm{\partial }u\left(z\right)=nu\left(z\right)$(35) ). We consider the k-th homogeneous part $u_k$ of u defined as in (33(33) $u_n\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}u\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}n\theta }\mathrm{d}\theta ,z\in \mathbb{D}\setminus \left\{0\right\},$(33) ). Let $A=z\mathrm{\partial }-\overline{z}\overline{\mathrm{\partial }}$ be the angular derivative. Differentiating under the integral in (33(33) $u_n\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}u\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}n\theta }\mathrm{d}\theta ,z\in \mathbb{D}\setminus \left\{0\right\},$(33) ) we have that $A{u}_k\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}Au\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}k\theta }\mathrm{d}\theta$ for $z\in \mathbb{D}\setminus \left\{0\right\}$. Since u satisfies (35(35) $z\mathrm{\partial }u\left(z\right)-\overline{z}\mathrm{\partial }u\left(z\right)=nu\left(z\right)$(35) ) we conclude that $A{u}_k\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}nu\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}k\theta }\mathrm{d}\theta =n{u}_k\left(z\right)$ for $z\in \mathbb{D}\setminus \left\{0\right\}$, where the last equality follows by (33(33) $u_n\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}u\left({\mathrm{e}}^{\mathrm{i}\theta }z\right){\mathrm{e}}^{-\mathrm{i}n\theta }\mathrm{d}\theta ,z\in \mathbb{D}\setminus \left\{0\right\},$(33) ). Also, since $u_k$ is homogeneous of order k, we have by Lemma 4.4 that $A{u}_k=k{u}_k$. We conclude that $u_k=0$ if $k\ne n$. By (34(34) $u\left(z\right)=\underset{N\to \mathrm{\infty }}{lim}\sum _{n=-N}^N\left(1-\frac{\mid n\mid }{N+1}\right)u_n\left(z\right),z\in \mathbb{D}\setminus \left\{0\right\},$(34) ) we have that $u=u_n$ is homogeneous of order n.

It is straightforward to check that a function $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ is homogeneous of order m=−n<0 with respect to rotations if and only if it has the form (36) $u\left(z\right)=f(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D}\setminus \left\{0\right\},$(36) for some function $f\in C^1(0,1)$.

Let us return to Equation (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ).

Theorem 4.6

Let $\alpha >-1$ and $n\in {\mathbb{Z}}^{+}$. Then a function $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ satisfies Equation (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ) if and only if it has the form (36(36) $u\left(z\right)=f(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D}\setminus \left\{0\right\},$(36) ) for some function $f\in C^1(0,1)$ such that (37) $x{f}^{\prime }\left(x\right)+nf\left(x\right)=\frac{1}{B(n,\alpha +1)}(1-x{)}^{\alpha }$(37) for 0<x<1.

Proof.

A Gauss elimination shows that (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ) is equivalently stated saying that (38) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ \overline{z}\overline{\mathrm{\partial }}u\left(z\right)-z\mathrm{\partial }u\left(z\right)=nu\left(z\right)\end{array}\right.$(38) for $z\in \mathbb{D}\setminus \left\{0\right\}$. By Theorem 4.5, the latter equation in (38(38) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ \overline{z}\overline{\mathrm{\partial }}u\left(z\right)-z\mathrm{\partial }u\left(z\right)=nu\left(z\right)\end{array}\right.$(38) ) is equivalent to $u\in C^1(\mathbb{D}\setminus \{0\left\}\right)$ having the form (36(36) $u\left(z\right)=f(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D}\setminus \left\{0\right\},$(36) ) for some function $f\in C^1(0,1)$. For such u, a differentiation shows that $\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=(\mid z{\mid }^2{f}^{\prime }(\mid z{\mid }^2)+nf(\mid z{\mid }^2\left)\right){\overline{z}}^n$ for $z\in \mathbb{D}\setminus \left\{0\right\}$. In view of this latter differentiation formula, we see that the first equation in (38(38) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ \overline{z}\overline{\mathrm{\partial }}u\left(z\right)-z\mathrm{\partial }u\left(z\right)=nu\left(z\right)\end{array}\right.$(38) ) is equivalent to (37(37) $x{f}^{\prime }\left(x\right)+nf\left(x\right)=\frac{1}{B(n,\alpha +1)}(1-x{)}^{\alpha }$(37) ). This yields the conclusion of the theorem.

Notice that the choice $f=f_{\alpha ,n}$ in (37(37) $x{f}^{\prime }\left(x\right)+nf\left(x\right)=\frac{1}{B(n,\alpha +1)}(1-x{)}^{\alpha }$(37) ) gives by (36(36) $u\left(z\right)=f(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D}\setminus \left\{0\right\},$(36) ) rise to the α-harmonic function $u=e_{\alpha ,-n}$.

Theorem 4.7

Let $\alpha >-1$. Let $e_{\alpha ,-n}$ be as in (27(27) $e_{\alpha ,-n}\left(z\right)=f_{\alpha ,n}(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D},$(27) ) for some $n\in {\mathbb{Z}}^{+}$. Then $e_{\alpha ,-n}\left(z\right)=\frac{1}{z^n}-\frac{1}{B(n,\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha +1}{\int }_0^1(1-(1-\mid z{\mid }^2)t{)}^{n-1}{t}^{\alpha }\mathrm{d}t\frac{1}{z^n}$ for $z\in \mathbb{D}\setminus \left\{0\right\}$.

Proof.

We shall exhibit a solution of Equation (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ). An interesting solution f of (37(37) $x{f}^{\prime }\left(x\right)+nf\left(x\right)=\frac{1}{B(n,\alpha +1)}(1-x{)}^{\alpha }$(37) ) is the function $f\left(x\right)=-\frac{1}{B(n,\alpha +1)}\frac{1}{x^n}{\int }_x^1{t}^{n-1}(1-t{)}^{\alpha }\mathrm{d}t$ for 0<x<1. The change of variables $t=sx+(1-s)1$ leads to the formula (39) $f\left(x\right)=-\frac{1}{B(n,\alpha +1)}\frac{(1-x{)}^{\alpha +1}}{x^n}{\int }_0^1(1-(1-x)s{)}^{n-1}{s}^{\alpha }\mathrm{d}s$(39) for 0<x<1. We observe also that $\underset{x\to 0}{lim}{x}^{n}f\left(x\right)=-\frac{1}{B(n,\alpha +1)}{\int }_0^1(1-t{)}^{n-1}{t}^{\alpha }\mathrm{d}t=-1,$ where the last equality follows by a standard formula for the Beta function.

We now consider the function u given by (36(36) $u\left(z\right)=f(\mid z{\mid }^2){\overline{z}}^n,z\in \mathbb{D}\setminus \left\{0\right\},$(36) ) with f as in (39(39) $f\left(x\right)=-\frac{1}{B(n,\alpha +1)}\frac{(1-x{)}^{\alpha +1}}{x^n}{\int }_0^1(1-(1-x)s{)}^{n-1}{s}^{\alpha }\mathrm{d}s$(39) ). Since f satisfies (37(37) $x{f}^{\prime }\left(x\right)+nf\left(x\right)=\frac{1}{B(n,\alpha +1)}(1-x{)}^{\alpha }$(37) ), we have from Theorem 4.6 that u satisfies Equation (31(31) $\left\{\begin{array}{l}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ z\mathrm{\partial }u\left(z\right)+nu\left(z\right)=\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\end{array}\right.$(31) ). An application of Theorem 4.3 shows that the function u has the form $u\left(z\right)=e_{\alpha ,-n}\left(z\right)+c/z^n,z\in \mathbb{D}\setminus \left\{0\right\},$ for some $c\in \mathbb{C}$. Observe that $c=\underset{z\to 0}{lim}{z}^{n}u\left(z\right)=\underset{x\to 0}{lim}{x}^{n}f\left(x\right)=-1.$ Thus $u\left(z\right)=e_{\alpha ,-n}\left(z\right)-1/z^n$ for $z\in \mathbb{D}\setminus \left\{0\right\}$. Solving for $e_{\alpha ,-n}$, we arrive at the conclusion of the theorem.

Theorem 4.7 displays interesting structure in the function $e_{\alpha ,-n}$. Notice that the factor ${\int }_0^1(1-(1-\mid z{\mid }^2)t{)}^{n-1}{t}^{\alpha }\mathrm{d}t=\sum _{j=0}^{n-1}\frac{(-1{)}^j}{\alpha +j+1}\left(\genfrac{}{}{0em}{}{n-1}{j}\right)(1-\mid z{\mid }^2{)}^j$ is a polynomial.

We mention that an earlier version of Theorem 4.7 goes back to Olofsson and Wittsten [2, Theorem 1.9]. The case n=2 of Theorem 4.7 has appeared recently in Chen [10, Example 2.1].

5. The case of big smoothness

In this section, we discuss Lipschitz continuity $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ of an α-harmonic function u in the case when $-1<\alpha <0$ and $\text{β}>\alpha +1$. We refer to this parameter range as the case of big smoothness.

Recall that the space ${\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ is a commutative Banach algebra under pointwise multiplication of functions. We shall use that the algebra ${\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ is inverse closed in the sense that $1/f\in {\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ if $f\in {\mathrm{\Lambda }}_{\text{β}}\left(E\right)$ and $f\left(z\right)\ne 0$ for $z\in \overline{E}$ (see Sherbert [5, Proposition 1.7]).

Lemma 5.1

Let $\alpha >-1$ and $0<\text{β}\le 1$. Let $n\ge 1$ be an integer. Then $e_{\alpha ,-n}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ if and only if $\text{β}\le \alpha +1$.

Proof.

From Theorem 4.7 we have that (40) $z^n{e}_{\alpha ,-n}\left(z\right)=1-\frac{1}{B(n,\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha +1}{\int }_0^1(1-(1-\mid z{\mid }^2)t{)}^{n-1}{t}^{\alpha }\mathrm{d}t$(40) for $z\in \mathbb{D}$. We set $w_{\alpha +1}\left(z\right)=(1-\mid z{\mid }^2{)}^{\alpha +1}$ and (41) $p\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1(1-(1-\mid z{\mid }^2)t{)}^{n-1}{t}^{\alpha }\mathrm{d}t.$(41) Observe that $p\left(z\right)\ge \frac{1}{B(n,\alpha +1)}{\int }_0^1(1-t{)}^{n-1}{t}^{\alpha }\mathrm{d}t=1$ for $z\in \mathbb{D}$.

Assume now that $e_{\alpha ,-n}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. By the algebra property of ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ we have from (40(40) $z^n{e}_{\alpha ,-n}\left(z\right)=1-\frac{1}{B(n,\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha +1}{\int }_0^1(1-(1-\mid z{\mid }^2)t{)}^{n-1}{t}^{\alpha }\mathrm{d}t$(40) ) that $p{w}_{\alpha +1}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Since ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ is inverse closed we can divide by p to conclude that $w_{\alpha +1}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. This yields that $\text{β}\le \alpha +1$.

Assume next that $\text{β}\le \alpha +1$. From Proposition 4.2 we have that the function $u=e_{\alpha ,-n}$ satisfies an estimate of the form (20(20) $\mid \mathrm{\partial }u\left(z\right)\mid +\mid \overline{\mathrm{\partial }}u\left(z\right)\mid \le \frac{C}{(1-\mid z{\mid }^2{)}^{1-\text{β}}},z\in \mathbb{D},$(20) ). An application of Theorem 3.1 yields that $e_{\alpha ,-n}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$.

Remark 5.2

Let p be as in (41(41) $p\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1(1-(1-\mid z{\mid }^2)t{)}^{n-1}{t}^{\alpha }\mathrm{d}t.$(41) ). The fact that p is a polynomial makes division by p a simpler operation not dependent on the full strength of commutative Banach algebras (see end of Section 4).

Remark 5.3

As a byproduct from the proof of Lemma 5.1 we have the norm bound that $\parallel e_{\alpha ,-n}{\parallel }_{{\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)}\le C{n}^{\alpha +1}$ for $n\in {\mathbb{Z}}^{+}$ if $-1<\alpha \le 0$ and $\text{β}=\alpha +1$.

We now return to the study of α-harmonic functions.

Theorem 5.4

Let $-1<\alpha <0$ and $0<\text{β}\le 1$ be such that $\text{β}>\alpha +1$. Let u be an α-harmonic function such that $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Then u is analytic in $\mathbb{D}$.

Proof.

We consider the series expansion (28(28) $u\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{e}_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n,z\in \mathbb{D},$(28) ) for u. We shall show that $a_n=0$ for n<0. Consider the n-th homogeneous part $u_n\left(z\right)=\frac{1}{2\pi }{\int }_{\mathbb{T}}u\left(z{\mathrm{e}}^{\mathrm{i}\tau }\right){\mathrm{e}}^{-\mathrm{i}n\tau }\mathrm{d}\tau ,z\in \mathbb{D},$ of u for $n\in \mathbb{Z}$. The assumption that $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ gives that $u_n\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ for $n\in \mathbb{Z}$. Indeed, since $\mid u\left(z_1\right)-u\left(z_2\right)\mid \le C\mid z_1-z_2{\mid }^{\text{β}}$ for $z_1,z_2\in \mathbb{D}$ by assumption, we have by the triangle inequality that $\mid u_n\left(z_1\right)-u_n\left(z_2\right)\mid \le C\mid z_1-z_2{\mid }^{\text{β}}$ for $z_1,z_2\in \mathbb{D}$.

We shall next evaluate the constraint $u_n\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Let $n\in {\mathbb{Z}}^{+}$. By (28(28) $u\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{e}_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n,z\in \mathbb{D},$(28) ) and homogeneity we have that $\begin{array}{rl}{u}_{-n}\left(z\right)& =\frac{1}{2\pi }{\int }_{\mathbb{T}}\left(\sum _{m=1}^{\mathrm{\infty }}{a}_{-m}{e}_{\alpha ,-m}\left(z\right){\mathrm{e}}^{-\mathrm{i}m\tau }+\sum _{m=0}^{\mathrm{\infty }}{a}_m{z}^m{\mathrm{e}}^{\mathrm{i}m\tau }\right){\mathrm{e}}^{\mathrm{i}n\tau }\mathrm{d}\tau \\ & =\sum _{m=1}^{\mathrm{\infty }}{a}_{-m}{e}_{\alpha ,-m}\left(z\right)\frac{1}{2\pi }{\int }_{\mathbb{T}}{\mathrm{e}}^{\mathrm{i}(n-m)\tau }\mathrm{d}\tau +\sum _{m=0}^{\mathrm{\infty }}{a}_m{z}^m\frac{1}{2\pi }{\int }_{\mathbb{T}}{\mathrm{e}}^{\mathrm{i}(m+n)\tau }\mathrm{d}\tau \\ & =a_{-n}{e}_{\alpha ,-n}\left(z\right)\end{array}$ for $z\in \mathbb{D}$, where the last equality follows by an orthogonality property of exponential monomials. Observe that the change of order of integration and summation is permitted by uniform convergence. Thus $u_{-n}=a_{-n}{e}_{\alpha ,-n}$ for $n\in {\mathbb{Z}}^{+}$.

Since $\text{β}>\alpha +1$ by assumption, we have from Lemma 5.1 that $e_{\alpha ,-n}\notin {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ for $n\in {\mathbb{Z}}^{+}$. Now, since $u_{-n}=a_{-n}{e}_{\alpha ,-n}$ belongs to ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ by the results of the previous two paragraphs, we conclude that $a_n=0$ for n<0. From (28(28) $u\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{e}_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n,z\in \mathbb{D},$(28) ) we then have that $u\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n,z\in \mathbb{D},$ is analytic in $\mathbb{D}$.

We point out that Theorem 5.4 generalizes an earlier result by Olofsson and Wittsten [2, Theorem 1.11] to the scale of spaces ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$.

A variant of Theorem 5.4 goes as follows.

Theorem 5.5

Let $-1<\alpha <0$ and $0<\text{β}\le 1$ be such that $\text{β}>\alpha +1$. Let $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$. Then $P_{\alpha }\left[f\right]\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ if and only if $\hat{f}\left(n\right)=0$ for n<0.

Proof.

Assume first that $u=P_{\alpha }\left[f\right]$ belongs to ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. By Theorem 5.4 we conclude that u is analytic in $\mathbb{D}$. By passage to boundary values we see that $\hat{f}\left(n\right)=0$ for n<0.

Conversely, assume that $\hat{f}\left(n\right)=0$ for n<0. Then $u=P_{\alpha }\left[f\right]$ is analytic in $\mathbb{D}$ and Theorem 3.3 yields that $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$.

6. The case of critical smoothness

In this section, we discuss Lipschitz continuity $u\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$ of an α-harmonic function u in the case when $-1<\alpha \le 0$ and $\text{β}=\alpha +1$. We refer to this parameter range as the case of critical smoothness.

We shall need some more properties of the functions $f_{\alpha ,n}$ defined in (30(30) $f_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(30) ) for $\alpha >-1$ and $n\in {\mathbb{Z}}^{+}$. By differentiation we see that (42) $f_{\alpha ,n}^{\left(j\right)}\left(z\right)=\frac{1}{B(n,\alpha +1)}\frac{\mathrm{\Gamma }(j-\alpha )}{\mathrm{\Gamma }(-\alpha )}{\int }_0^1{t}^{n+j-1}(1-tz{)}^{\alpha -j}\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(42) for $j\ge 0$.

Lemma 6.1

Let $-1<\alpha <0$. Let $f_{\alpha ,n}$ be as in (30(30) $f_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(30) ) for some $n\in {\mathbb{Z}}^{+}$. Then $f_{\alpha ,n}$ is totally monotone on the interval $(-\mathrm{\infty },1)$ in the sense that $f_{\alpha ,n}^{\left(j\right)}\left(x\right)\ge 0$ for x<1 and $j\ge 0$.

Proof.

The conclusion of the lemma is evident from formula (42(42) $f_{\alpha ,n}^{\left(j\right)}\left(z\right)=\frac{1}{B(n,\alpha +1)}\frac{\mathrm{\Gamma }(j-\alpha )}{\mathrm{\Gamma }(-\alpha )}{\int }_0^1{t}^{n+j-1}(1-tz{)}^{\alpha -j}\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(42) ).

The study of the functions $f_{\alpha ,n}$ belongs to the subject of hypergeometry. We observe that (43) $f_{\alpha ,n}^{\left(j\right)}\left(z\right)=\frac{1}{B(n,\alpha +1)}\frac{\mathrm{\Gamma }(j-\alpha )}{\mathrm{\Gamma }(-\alpha )}\frac{1}{n+j}F(j-\alpha ,n+j;n+j+1;z),$(43) where $F(a,b;c;z)=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(a{)}_n\right(b{)}_n}{(c{)}_n}\frac{z^n}{n!}=\frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+n)\mathrm{\Gamma }(b+n)}{\mathrm{\Gamma }(c+n)}\frac{z^n}{n!}$ is a standard hypergeometric function. In fact, formula (42(42) $f_{\alpha ,n}^{\left(j\right)}\left(z\right)=\frac{1}{B(n,\alpha +1)}\frac{\mathrm{\Gamma }(j-\alpha )}{\mathrm{\Gamma }(-\alpha )}{\int }_0^1{t}^{n+j-1}(1-tz{)}^{\alpha -j}\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(42) ) is a specialization of the Euler integral formula $F(a,b;c;z)=\frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(c-b)}{\int }_0^1{t}^{b-1}(1-t{)}^{c-b-1}(1-tz{)}^{-a}\mathrm{d}t$ valid when c>b>0 (see [22, Theorem 2.2.1]).

The study of asymptotic behaviour of hypergeometric functions is a classical topic which goes back to Gauss. We shall use the result that (44) $\underset{x\to 1}{lim}\frac{F(a,b;c;x)}{(1-x{)}^{c-a-b}}=\frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(a+b-c)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}$(44) when c<a+b (see [22, Theorem 2.1.3]). The asymptotic result (44(44) $\underset{x\to 1}{lim}\frac{F(a,b;c;x)}{(1-x{)}^{c-a-b}}=\frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(a+b-c)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}$(44) ) belongs to the realm of the famous Gauss summation formula.

Lemma 6.2

Let $-1<\alpha <0$. Let $f_{\alpha ,n}$ be as in (30(30) $f_{\alpha ,n}\left(z\right)=\frac{1}{B(n,\alpha +1)}{\int }_0^1{t}^{n-1}(1-tz{)}^{\alpha }\mathrm{d}t,z\in \mathbb{C}\setminus [1,\mathrm{\infty }),$(30) ) for some $n\in {\mathbb{Z}}^{+}$. Then $\underset{x\to 1}{lim}\frac{1-f_{\alpha ,n}\left(x\right)}{(1-x{)}^{\alpha +1}}=\frac{\mathrm{\Gamma }(n+\alpha +1)}{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }(\alpha +2)}.$

Proof.

We consider the derivative $f_{\alpha ,n}^{\prime }$. From formula (43(43) $f_{\alpha ,n}^{\left(j\right)}\left(z\right)=\frac{1}{B(n,\alpha +1)}\frac{\mathrm{\Gamma }(j-\alpha )}{\mathrm{\Gamma }(-\alpha )}\frac{1}{n+j}F(j-\alpha ,n+j;n+j+1;z),$(43) ) above we have that $f_{\alpha ,n}^{\prime }\left(z\right)=\frac{1}{B(n,\alpha +1)}\frac{\mathrm{\Gamma }(1-\alpha )}{\mathrm{\Gamma }(-\alpha )}\frac{1}{n+1}F(1-\alpha ,n+1;n+2;z).$ By (44(44) $\underset{x\to 1}{lim}\frac{F(a,b;c;x)}{(1-x{)}^{c-a-b}}=\frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(a+b-c)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}$(44) ) we have that $\underset{x\to 1}{lim}\frac{f_{\alpha ,n}^{\prime }\left(x\right)}{(1-x{)}^{\alpha }}=\frac{1}{B(n,\alpha +1)}.$ The conclusion of the lemma now follows from L'Hospital's rule.

We shall consider the Fourier multiplier $M_{\alpha }$ defined by $\left(M_{\alpha }f\right)\stackrel{ˆ}{}\left(n\right)=\left\{\begin{array}{ll}\frac{\mathrm{\Gamma }(\mid n\mid +\alpha +1)}{\mathrm{\Gamma }(\mid n\mid )}\hat{f}\left(n\right)& \mathrm{f}\mathrm{o}\mathrm{r}n<0,\\ 0& \mathrm{f}\mathrm{o}\mathrm{r}n\ge 0,\end{array}\right.$ for $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$. From Stirling's formula we have that (45) $\mathrm{\Gamma }(n+\alpha +1)/\mathrm{\Gamma }\left(n\right)\approx n^{\alpha +1}$(45) asymptotically in the sense that the quotient tends to 1 as $n\to +\mathrm{\infty }$ (see [22, Section 1.4]).

Recall that the space $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ is the dual of $L^1\left(\mathbb{T}\right)$ and has as such a natural weak${}^{\ast }$ topology. A sequence $\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ in $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ converges to a function f in $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ in the weak${}^{\ast }$ topology of $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ if and only if $\underset{k\ge 1}{sup}\parallel f_k{\parallel }_{\mathrm{\infty }}<+\mathrm{\infty }$ and $\underset{k\to \mathrm{\infty }}{lim}{\hat{f}}_k\left(n\right)=\hat{f}\left(n\right)$ for every $n\in \mathbb{Z}$. It follows by a well-known diagonal procedure that every bounded sequence in $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ has a weak${}^{\ast }$ convergent subsequence. An interesting discussion of Fourier coefficients for linear functionals is found in Katznelson [7, Section I.7]; see [2, Section 4] for information on a related notion of relative completeness.

Theorem 6.3

Let $-1<\alpha <0$ and set $\text{β}=\alpha +1$. Let $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ be such that $P_{\alpha }\left[f\right]\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Then $M_{\alpha }f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$.

Proof.

Since $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ and $0<\text{β}<1$ we have from Theorem 3.2 that $P_0\left[f\right]$ belongs to ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Set $v=P_0\left[f\right]-P_{\alpha }\left[f\right]$. By assumption we have that $v\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$. Since also v=0 on $\mathbb{T}$ by construction, there exists a finite positive constant C such that (46) $\mid v\left(z\right)\mid \le C(1-\mid z{\mid }^2{)}^{\text{β}}$(46) for $z\in \mathbb{D}$.

We consider now the functions $h_r\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=v\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)/(1-r^2{)}^{\text{β}},{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T},$ for 0<r<1. Clearly $h_r\in C^{\mathrm{\infty }}\left(\mathbb{T}\right)$ for 0<r<1. Furthermore, from (46(46) $\mid v\left(z\right)\mid \le C(1-\mid z{\mid }^2{)}^{\text{β}}$(46) ) we have that $\parallel h_r{\parallel }_{\mathrm{\infty }}\le C$ for 0<r<1. The functions $\left\{h_r\right\}$ thus form a bounded sequence in $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$.

We now turn our attention to Fourier coefficients. From (9(9) $P_{\alpha }\left[f\right]\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)e_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}\hat{f}\left(n\right)z^n,z\in \mathbb{D},$(9) ) we have that (47) $v\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)(1-f_{\alpha ,n}(\mid z{\mid }^2\left)\right){\overline{z}}^n$(47) for $z\in \mathbb{D}$. This latter series expansion (47(47) $v\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)(1-f_{\alpha ,n}(\mid z{\mid }^2\left)\right){\overline{z}}^n$(47) ) makes evident that $\left(h_r\right)\stackrel{ˆ}{}\left(n\right)=0$ for $n\ge 0$. Let $n\in {\mathbb{Z}}^{+}$. Again from (47(47) $v\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)(1-f_{\alpha ,n}(\mid z{\mid }^2\left)\right){\overline{z}}^n$(47) ) we have that $\left(h_r\right)\stackrel{ˆ}{}(-n)=\hat{f}(-n)\frac{1-f_{\alpha ,n}\left(r^2\right)}{(1-r^2{)}^{\alpha +1}}{r}^n$ for 0<r<1. By Lemma 6.2 we have that $\underset{r\to 1}{lim}\left(h_r\right)\stackrel{ˆ}{}(-n)=\frac{\mathrm{\Gamma }(n+\alpha +1)}{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }(\alpha +2)}\hat{f}(-n)$ for $n\in {\mathbb{Z}}^{+}$.

From the result of the previous paragraph we know that the limit $\underset{r\to 1}{lim}\left(h_r\right)\stackrel{ˆ}{}\left(n\right)$ exists for every $n\in \mathbb{Z}$. Since the sequence $\left\{h_r\right\}$ is bounded in $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$, it follows that the limit $h=\underset{r\to 1}{lim}{h}_r$ exists in the weak${}^{\ast }$ topology of $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$. We conclude that $M_{\alpha }f=\mathrm{\Gamma }(\alpha +2)h$ belongs to $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ and $\parallel M_{\alpha }f{\parallel }_{\mathrm{\infty }}\le \mathrm{\Gamma }(\alpha +2)C$, where C is as in (46(46) $\mid v\left(z\right)\mid \le C(1-\mid z{\mid }^2{)}^{\text{β}}$(46) ).

The Fourier multiplier $M_{\alpha }$ is naturally derived from the binomial series: (48) $\frac{1}{(1-z{)}^a}=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(n+a)}{n!\mathrm{\Gamma }\left(a\right)}{z}^n,z\in \mathbb{D}.$(48) Let $\alpha >-1$ and consider the anti-analytic function $h\left(z\right)=\frac{\overline{z}}{(1-\overline{z}{)}^{\alpha +2}},z\in \mathbb{D}.$ From the binomial series (48(48) $\frac{1}{(1-z{)}^a}=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(n+a)}{n!\mathrm{\Gamma }\left(a\right)}{z}^n,z\in \mathbb{D}.$(48) ) we have that $h\left(z\right)=\frac{1}{\mathrm{\Gamma }(\alpha +2)}\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(n+\alpha +1)}{\mathrm{\Gamma }\left(n\right)}{\overline{z}}^n$ for $z\in \mathbb{D}$. From here we see that $M_{\alpha }f=\mathrm{\Gamma }(\alpha +2)(m_{\alpha }\ast f),f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right),$ where $m_{\alpha }=\underset{r\to 1}{lim}{h}_r$ in ${\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ is the distributional boundary limit of the function h, $h_r$ is as in (4(4) $u_r\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=u\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right),{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T},$(4) ) and $\ast$ is convolution in ${\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$.

The Fourier coefficients of a Lipschitz function $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ decay as $\hat{f}\left(n\right)=O(1/\mid n{\mid }^{\text{β}})$ as $\mid n\mid \to \mathrm{\infty }$ (see Zygmund [8, Theorem II.4.7]). An interesting example is provided by the Weierstrass function (49) $f_{\text{β}}\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=\sum _{k=1}^{\mathrm{\infty }}{b}^{-k\text{β}}{\mathrm{e}}^{-\mathrm{i}{b}^k\theta },{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T},$(49) where b>1 is an integer. It is known that $f_{\text{β}}\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ for $0<\text{β}<1$ (see Zygmund [8, Theorem II.4.9]).

Let $-1<\alpha <0$ and set $\text{β}=\alpha +1$. Using (45(45) $\mathrm{\Gamma }(n+\alpha +1)/\mathrm{\Gamma }\left(n\right)\approx n^{\alpha +1}$(45) ) and (49(49) $f_{\text{β}}\left({\mathrm{e}}^{\mathrm{i}\theta }\right)=\sum _{k=1}^{\mathrm{\infty }}{b}^{-k\text{β}}{\mathrm{e}}^{-\mathrm{i}{b}^k\theta },{\mathrm{e}}^{\mathrm{i}\theta }\in \mathbb{T},$(49) ) it is straightforward to check that the Fourier coefficients of the distribution $M_{\alpha }{f}_{\text{β}}$ are not vanishing at infinity. By Riemann–Lebesgue lemma we conclude that $M_{\alpha }{f}_{\text{β}}\notin L^1\left(\mathbb{T}\right)$ for such parameters. The conclusion $M_{\alpha }f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$ in Theorem 6.3 is thus quite demanding of a Lipschitz function $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$.

Lemma 6.4

Let $\alpha >-1$. Let $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ and set $u=P_{\alpha }\left[f\right]$. Then $\overline{z}\overline{\mathrm{\partial }}u\left(z\right)=\frac{1}{\mathrm{\Gamma }(\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }{P}_0[M_{\alpha }f\left]\right(z)$ for $z\in \mathbb{D}$.

Proof.

Recall the homogeneous expansion of the Poisson integral (9(9) $P_{\alpha }\left[f\right]\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)e_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}\hat{f}\left(n\right)z^n,z\in \mathbb{D},$(9) ). Differentiating we have that $\overline{\mathrm{\partial }}u\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)\overline{\mathrm{\partial }}{e}_{\alpha ,-n}\left(z\right)$ for $z\in \mathbb{D}$. We now use the first of the differential equations in Proposition 4.2 to conclude that $\begin{array}{rl}\overline{z}\overline{\mathrm{\partial }}u\left(z\right)& =\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }\\ & =\frac{1}{\mathrm{\Gamma }(\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }\sum _{n=1}^{\mathrm{\infty }}(M_{\alpha }f\left)\stackrel{ˆ}{}\right(-n){\overline{z}}^n\end{array}$ for $z\in \mathbb{D}$, where the last equality is straightforward to check. Since $\left(M_{\alpha }f\right)\stackrel{ˆ}{}\left(n\right)=0$ for $n\ge 0$, this yields the conclusion of the lemma.

We now return to the study of Lipschitz classes.

Theorem 6.5

Let $-1<\alpha \le 0$ and set $\text{β}=\alpha +1$. Let $f\in {\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{T}\right)$ be such that $M_{\alpha }f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$. Then the function $u=P_{\alpha }\left[f\right]$ belongs to ${\mathrm{\Lambda }}_{\text{β}}\left(\mathbb{D}\right)$.

Proof.

We shall apply Theorem 3.1. Recall Lemma 6.4. A standard estimation of $P_0\left[M_{\alpha }f\right]$ gives that $\mid \overline{z}\overline{\mathrm{\partial }}u\left(z\right)\mid \le \frac{1}{\mathrm{\Gamma }(\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }\parallel M_{\alpha }f{\parallel }_{\mathrm{\infty }}$ for $z\in \mathbb{D}$. Recall Theorem 2.13. An application of Theorem 3.1 now yields the conclusion of the theorem.

We denote by $f^{\prime }$ the distributional derivative of $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$. Recall that $\left(f^{\prime }\right)\stackrel{ˆ}{}\left(n\right)=in\hat{f}\left(n\right)$ for $n\in \mathbb{Z}$.

Lemma 6.4 is accompanied with the following lemma.

Lemma 6.6

Let $\alpha >-1$. Let $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ and set $u=P_{\alpha }\left[f\right]$. Then $z\mathrm{\partial }u\left(z\right)=\frac{1}{\mathrm{\Gamma }(\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }{P}_0[M_{\alpha }f\left]\right(z)-i{P}_{\alpha }[f^{\prime }\left]\right(z)$ for $z\in \mathbb{D}$.

Proof.

Recall the homogeneous expansion of the Poisson integral (9(9) $P_{\alpha }\left[f\right]\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)e_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}\hat{f}\left(n\right)z^n,z\in \mathbb{D},$(9) ). Differentiating we have that $\mathrm{\partial }u\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)\mathrm{\partial }{e}_{\alpha ,-n}\left(z\right)+\sum _{n=1}^{\mathrm{\infty }}\hat{f}\left(n\right)n{z}^{n-1}$ for $z\in \mathbb{D}$. We now use the second of the differential equations in Proposition 4.2 to conclude that $\begin{array}{rl}z\mathrm{\partial }u\left(z\right)& =\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)\left(\frac{1}{B(n,\alpha +1)}{\overline{z}}^n(1-\mid z{\mid }^2{)}^{\alpha }-n{e}_{\alpha ,-n}(z)\right)+\sum _{n=1}^{\mathrm{\infty }}n\hat{f}\left(n\right)z^n\\ & =\frac{1}{\mathrm{\Gamma }(\alpha +1)}(1-\mid z{\mid }^2{)}^{\alpha }\sum _{n=1}^{\mathrm{\infty }}(M_{\alpha }f\left)\stackrel{ˆ}{}\right(-n){\overline{z}}^n-\sum _{n=1}^{\mathrm{\infty }}n\hat{f}(-n\left)e_{\alpha ,-n}\right(z)+\sum _{n=1}^{\mathrm{\infty }}n\hat{f}(n)z^n\end{array}$ for $z\in \mathbb{D}$, where the last equality is straightforward to check. In view of the series expansion (9(9) $P_{\alpha }\left[f\right]\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\hat{f}(-n)e_{\alpha ,-n}\left(z\right)+\sum _{n=0}^{\mathrm{\infty }}\hat{f}\left(n\right)z^n,z\in \mathbb{D},$(9) ), this yields the conclusion of the lemma.

Let $f\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$. The conjugate function of f is the distribution $\tilde{f}\in {\mathcal{D}}^{\prime }\left(\mathbb{T}\right)$ defined by the condition that (50) $\left(\tilde{f}\right)\stackrel{ˆ}{}\left(n\right)=-i\mathrm{s}\mathrm{g}\mathrm{n}\left(n\right)\hat{f}\left(n\right),n\in \mathbb{Z},$(50) where $\mathrm{s}\mathrm{g}\mathrm{n}\left(n\right)=n/\mid n\mid$ for $n\ne 0$ and $\mathrm{s}\mathrm{g}\mathrm{n}\left(0\right)=0$. From (50(50) $\left(\tilde{f}\right)\stackrel{ˆ}{}\left(n\right)=-i\mathrm{s}\mathrm{g}\mathrm{n}\left(n\right)\hat{f}\left(n\right),n\in \mathbb{Z},$(50) ) we have that the α-harmonic functions $u=P_{\alpha }\left[f\right]$ and $v=P_{\alpha }\left[\tilde{f}\right]$ are such that the function g=u+iv is analytic in $\mathbb{D}$ and $v\left(0\right)=0$.

The following result is well-known but included here for the sake of completeness.

Theorem 6.7

Let $f\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$ and set $u=P_0\left[f\right]$. Then $u\in {\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$ if and only if $\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$.

Proof.

Assume first that $u\in {\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$. We set $v=P_0\left[\tilde{f}\right]$. Since $u\in {\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$ we have that $\mid u\left(z_1\right)-u\left(z_2\right)\mid \le C\mid z_1-z_2\mid$ for $z_1,z_2\in \mathbb{D}$, where C is a finite positive constant. This latter Lipschitz condition gives that the partial derivatives of u are uniformly bounded in $\mathbb{D}$. Since v is a harmonic conjugate for u we conclude that v also has uniformly bounded partial derivatives in $\mathbb{D}$. By Theorem 3.1 we have that $v\in {\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$. Passing to the boundary limit of v we see that $\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$.

Assume next that $\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$. Now the function $u+iv=P_0[f+i\tilde{f}]$ is analytic in $\mathbb{D}$ and $f+i\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$. By Theorem 3.3 we have that $u+iv\in {\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$. Similarly, the function $u-iv=P_0[f-i\tilde{f}]$ is anti-analytic in $\mathbb{D}$ and $f-i\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$. An application of Theorem 3.3 gives that $u-iv\in {\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$. We now conclude that $u,v\in {\mathrm{\Lambda }}_1\left(\mathbb{D}\right)$.

The Fourier multiplier $M_0$ is naturally expressed in terms of the conjugate function. It is straightforward to check that (51) $M_{0}f=\frac{i}{2}(f-i\tilde{f}{)}^{\prime },f\in {\mathcal{D}}^{\prime }(\mathbb{T}),$(51) where the prime indicates distributional derivative.

A classical result is that the kernel of the distributional derivative ${\mathcal{D}}^{\prime }\left(\mathbb{T}\right)\ni f\mapsto f^{\prime }$ is the space of constant functions (see Hörmander [23, Theorem 3.1.4]). It is also well-known that the distributional derivative maps the Lipschitz space ${\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$ onto $L^{\mathrm{\infty }}\left(\mathbb{T}\right)$.

Proposition 6.8

Let $f\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$. Then $\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$ if and only if $M_{0}f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$.

Proof.

Recall formula (51(51) $M_{0}f=\frac{i}{2}(f-i\tilde{f}{)}^{\prime },f\in {\mathcal{D}}^{\prime }(\mathbb{T}),$(51) ). If $\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$, then $f-i\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$ and a differentiation gives that $M_{0}f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$. Conversely, if $M_{0}f\in L^{\mathrm{\infty }}\left(\mathbb{T}\right)$, then $f-i\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$ which yields that $\tilde{f}\in {\mathrm{\Lambda }}_1\left(\mathbb{T}\right)$.

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