RETRACTED ARTICLE: A remark on the Dirichlet problem in a half-plane

Abstract

In this paper, we prove that if the positive part $u^{+}(z)$ of a harmonic function $u(z)$ in a half-plane satisfies a slowly growing condition, then its negative part $u^{-}(z)$ can also be dominated by a similarly growing condition. Further, a solution of the Dirichlet problem in a half-plane for a fast growing continuous boundary function is constructed by the generalized Dirichlet integral with this boundary function.

1 Introduction and main theorem

Let R be the set of all real numbers and let C denote the complex plane with points $z=x+iy$, where $x,y\in \mathbf{R}$. The boundary and closure of an open set Ω are denoted by ∂ Ω and $\overline{\mathrm{\Omega }}$, respectively. The upper half-plane is the set ${\mathbf{C}}_{+}:=\{z=x+iy\in \mathbf{C}:y>0\}$, whose boundary is $\partial {\mathbf{C}}_{+}=\mathbf{R}$.

We use the standard notations $u^{+}=max\{u,0\}$, $u^{-}=-min\{u,0\}$, and $[d]$ is the integer part of the positive real number d. For positive functions $h_1$ and $h_2$, we say that $h_1\lesssim h_2$ if $h_1\le M{h}_2$ for some positive constant M.

Given a continuous function f in $\partial {\mathbf{C}}_{+}$, we say that h is a solution of the (classical) Dirichlet problem in ${\mathbf{C}}_{+}$ with f, if $\mathrm{\Delta }h=0$ in ${\mathbf{C}}_{+}$ and ${lim}_{z\in {\mathbf{C}}_{+},z\to t}h(z)=f(t)$ for every $t\in \partial {\mathbf{C}}_{+}$.

The classical Poisson kernel in ${\mathbf{C}}_{+}$ is defined by

$$P(z,t)=\frac{y}{\pi {|z-t|}^2},$$

where $z=x+iy\in {\mathbf{C}}_{+}$ and $t\in \mathbf{R}$.

It is well known (see [1]) that the Poisson kernel $P(z,t)$ is harmonic for $z\in \mathbf{C}-\{t\}$ and has the expansion

$$P(z,t)=\frac{1}{\pi }Im\sum _{k=0}^{\mathrm{\infty }}\frac{z^k}{t^{k+1}},$$

which converges for $|z|<|t|$. We define a modified Cauchy kernel of $z\in {\mathbf{C}}_{+}$ by

$$C_m(z,t)=\{\begin{array}{ll}\frac{1}{\pi }\frac{1}{t-z}& \text{when }|t|\le 1,\\ \frac{1}{\pi }\frac{1}{t-z}-\frac{1}{\pi }{\sum }_{k=0}^m\frac{z^k}{t^{k+1}}& \text{when }|t|>1,\end{array}$$

where m is a nonnegative integer.

To solve the Dirichlet problem in ${\mathbf{C}}_{+}$, as in [2], we use the modified Poisson kernel defined by

$$P_m(z,t)=Im{C}_m(z,t)=\{\begin{array}{ll}P(z,t)& \text{when }|t|\le 1,\\ P(z,t)-\frac{1}{\pi }Im{\sum }_{k=0}^m\frac{z^k}{t^{k+1}}& \text{when }|t|>1.\end{array}$$

We remark that the modified Poisson kernel $P_m(z,t)$ is harmonic in ${\mathbf{C}}_{+}$. About modified Poisson kernel in a cone, we refer readers to papers by I Miyamoto, H Yoshida, L Qiao and GT Deng (e.g. see [3–11]).

Put

$$U(f)(z)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}P(z,t)f(t)dt\text{and}{U}_m(f)(z)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{P}_m(z,t)f(t)dt,$$

where $f(t)$ is a continuous function in $\partial {\mathbf{C}}_{+}$.

For any positive real number α, We denote by ${\mathcal{A}}_{\alpha }$ the space of all measurable functions $f(x+iy)$ in ${\mathbf{C}}_{+}$ satisfying

$${\iint }_{{\mathbf{C}}_{+}}\frac{y|f(x+iy)|}{1+{|x+iy|}^{\alpha +2}}dxdy<\mathrm{\infty }$$

(1.1)

and by ${\mathcal{B}}_{\alpha }$ the set of all measurable functions $g(x)$ in $\partial {\mathbf{C}}_{+}$ such that

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{|g(x)|}{1+{|x|}^{\alpha }}dx<\mathrm{\infty }.$$

(1.2)

We also denote by ${\mathcal{D}}_{\alpha }$ the set of all continuous functions $u(x+iy)$ in ${\overline{\mathbf{C}}}_{+}$, harmonic in ${\mathbf{C}}_{+}$ with $u^{+}(x+iy)\in A_{\alpha }$ and $u^{+}(x)\in {\mathcal{B}}_{\alpha }$.

About the solution of the Dirichlet problem with continuous data in ${\mathbf{C}}_{+}$, we refer readers to the following result (see [12, 13]).

Theorem A Let u be a real-valued function harmonic in ${\mathbf{C}}_{+}$ and continuous in ${\overline{\mathbf{C}}}_{+}$. If $u(z)\in {\mathcal{B}}_2$, then there exists a constant $d_1$ such that $u(z)=d_{1}y+U(u)(z)$ for all $z=x+iy\in {\mathbf{C}}_{+}$.

Inspired by Theorem A, we first prove the following.

Theorem 1 If $\alpha \ge 2$ and $u\in {\mathcal{D}}_{\alpha }$, then $u\in {\mathcal{B}}_{\alpha }$.

Then we are concerned with the growth property of $U_m(f)(z)$ at infinity in ${\mathbf{C}}_{+}$.

Theorem 2 If $\alpha -2\le m<\alpha -1$ and $f\in {\mathcal{D}}_{\alpha }$, then

$$\underset{|z|\to \mathrm{\infty },z\in {\mathbf{C}}_{+}}{lim}y{|z|}^{-\alpha }{U}_m(f)(z)=0.$$

(1.3)

We say that u is of order λ if

$$\lambda =\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{log({sup}_{H\cap B(r)}|u|)}{logr}.$$

If $\lambda <\mathrm{\infty }$, then u is said to be of finite order. See Hayman-Kennedy [[14], Definition 4.1].

Our next aim is to give solutions of the Dirichlet problem for harmonic functions of infinite order in ${\mathbf{C}}_{+}$. For this purpose, we define a nondecreasing and continuously differentiable function $\rho (R)\ge 1$ on the interval $[0,+\mathrm{\infty })$. We assume further that

$${ϵ}_0=\underset{R\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\rho }^{\prime }(R)RlogR}{\rho (R)}<1.$$

(1.4)

Remark For any ϵ ($0<ϵ<1-{ϵ}_0$), there exists a sufficiently large positive number R such that $r>R$, by (1.4) we have

$$\rho (r)<\rho (e){(lnr)}^{{ϵ}_0+ϵ}.$$

Let $\mathcal{E}(\rho ,\beta )$ be the set of continuous functions f in $\partial {\mathbf{C}}_{+}$ such that

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{|f(t)|}{1+{|t|}^{\rho (|t|)+\beta +1}}dt<\mathrm{\infty },$$

(1.5)

where β is a positive real number.

Theorem 3 If $f\in \mathcal{E}(\rho ,\beta )$, then the integral $U_{[\rho (|t|)+\beta ]}(f)(x)$ is a solution of the Dirichlet problem in ${\mathbf{C}}_{+}$ with f.

The following result immediately follows from Theorem 2 (the case $\alpha =m+2$) and Theorem 3 (the case $[\rho (|t|)+\beta ]=m$).

Corollary 1 If f is a continuous function in ${\mathbf{C}}_{+}$ satisfying

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{|f(t)|}{1+{|t|}^{m+2}}dt<\mathrm{\infty },$$

then $U_m(f)(z)$ is a solution of the Dirichlet problem in ${\mathbf{C}}_{+}$ with f satisfying

$$\underset{|z|\to \mathrm{\infty },z\in {\mathbf{C}}_{+}}{lim}{|z|}^{-m-1}{U}_m(f)(z)=0.$$

For harmonic functions of finite order in ${\mathbf{C}}_{+}$, we have the following integral representations.

Corollary 2 Let $u\in {\mathcal{D}}_{\alpha }$ ($\alpha \ge 2$) and let m be an integer such that $m+2<\alpha \le m+3$.

• (I) If $\alpha =2$, then $U(u)(z)$ is a harmonic function in ${\mathbf{C}}_{+}$ and can be continuously extended to ${\overline{\mathbf{C}}}_{+}$ such that $u(z^{\prime })=U(u)(z^{\prime })$ for $z^{\prime }\in \partial {\mathbf{C}}_{+}$. There exists a constant $d_2$ such that $u(z)=d_{2}y+U(u)(z)$ for all $z\in {\mathbf{C}}_{+}$.

• (II) If $\alpha >2$, then $U_m(u)(z)$ is a harmonic function in ${\mathbf{C}}_{+}$ and can be continuously extended to ${\overline{\mathbf{C}}}_{+}$ such that $u(z^{\prime })=U_m(u)(z^{\prime })$ for $z^{\prime }\in \partial {\mathbf{C}}_{+}$. There exists a harmonic polynomial $Q_m(u)(z)$ of degree at most $m-1$ which vanishes in $\partial {\mathbf{C}}_{+}$ such that $u(z)=U_m(u)(z)+Q_m(u)(z)$ for all $z\in {\mathbf{C}}_{+}$.

Finally, we prove the following.

Theorem 4 Let u be a real-valued function harmonic in ${\mathbf{C}}_{+}$ and continuous in ${\overline{\mathbf{C}}}_{+}$. If $u\in \mathcal{E}(\rho ,\beta )$, then we have

$$u(z)=U_{[\rho (|t|)+\beta ]}(u)(z)+Im\mathrm{\Pi }(z)$$

for all $z\in {\overline{\mathbf{C}}}_{+}$, where $\mathrm{\Pi }(z)$ is an entire function in ${\mathbf{C}}_{+}$ and vanishes continuously in $\partial {\mathbf{C}}_{+}$.

2 Main lemmas

The Carleman formula refers to holomorphic functions in ${\mathbf{C}}_{+}$ (see [15, 16]).

Lemma 1 If $R>1$ and $u(z)$ ($z=x+iy$) is a harmonic function in ${\mathbf{C}}_{+}$ with continuous boundary in $\partial {\mathbf{C}}_{+}$, then we have

$$\begin{array}{c}{m}_{-}(R)+\frac{1}{2\pi }{\int }_1^R(\frac{1}{x^2}-\frac{1}{R^2})g_{-}(x)dx\hfill \\ =m_{+}(R)+\frac{1}{2\pi }{\int }_1^R(\frac{1}{x^2}-\frac{1}{R^2})g_{+}(x)dx-d_3-\frac{d_4}{R^2},\hfill \end{array}$$

where

$$\begin{array}{c}{m}_{\pm }(R)=\frac{1}{\pi R}{\int }_0^{\pi }{u}^{\pm }\left(R{e}^{i\theta }\right)sin\theta d\theta ,g_{\pm }(x)=u^{\pm }(x)+u^{\pm }(-x),\hfill \\ d_3=\frac{1}{2\pi }{\int }_0^{\pi }(u\left(R{e}^{i\theta }\right)+\frac{\partial u(R{e}^{i\theta })}{\partial n})sin\theta d\theta \hfill \end{array}$$

and

$$d_4=\frac{1}{2\pi }{\int }_0^{\pi }(u\left(R{e}^{i\theta }\right)-\frac{\partial u(R{e}^{i\theta })}{\partial n})sin\theta d\theta .$$

Lemma 2 For any $z=x+iy\in {\mathbf{C}}_{+}$, $|z|>1$, and $t\in \mathbf{R}$, we have

$$|C_m(z,t)|\lesssim y^{-1}{|z|}^{m+1}{|t|}^{-m-1},$$

(2.1)

where $1<|t|\le 2|z|$,

$$|C_m(z,t)|\lesssim {|z|}^{m+1}{|t|}^{-m-2},$$

(2.2)

where $|t|>max\{1,2|z|\}$,

$$|C_m(z,t)|\lesssim y^{-1},$$

(2.3)

where $|t|\le 1$.

Proof If $t\in \mathbf{R}$ and $1<|t|\le 2|z|$, we have $|t-z|\ge y$, which gives

$$|C_m(z,t)|=\frac{1}{\pi }|\frac{1}{t-z}-\frac{1-{(\frac{z}{t})}^{m+1}}{t-z}|=\frac{1}{\pi }\frac{{|\frac{z}{t}|}^{m+1}}{|t-z|}\lesssim \frac{{|z|}^{m+1}}{y{|t|}^{m+1}}.$$

If $|t|>max\{1,2|z|\}$, we obtain

$$|C_m(z,t)|=\frac{1}{\pi }\left|\sum _{k=m+1}^{\mathrm{\infty }}\frac{z^k}{t^{k+1}}\right|\lesssim \sum _{k=m+1}^{\mathrm{\infty }}\frac{{|z|}^k}{{|t|}^{k+1}}\lesssim \frac{{|z|}^{m+1}}{{|t|}^{m+2}}.$$

If $t\in \mathbf{R}$ and $|t|\le 1$, then we also have $|t-z|\ge y$, which yields

$$|C_m(z,t)|\lesssim y^{-1}.$$

Thus this lemma is proved. □

Lemma 3 (see [[17], Theorem 10])

Let $h(z)$ be a harmonic function in ${\mathbf{C}}_{+}$ such that $h(z)$ vanishes continuously in $\partial {\mathbf{C}}_{+}$. If

$$\underset{|z|\to \mathrm{\infty },z\in {\mathbf{C}}_{+}}{lim}{|z|}^{-m-1}{h}^{+}(z)=0,$$

then $h(z)=Q_m(h)(z)$ in ${\mathbf{C}}_{+}$, where $Q_m(h)$ is a polynomial of $(x,y)\in {\mathbf{C}}_{+}$ of degree less than m and even with respect to the variable y.

3 Proof of Theorem 1

We distinguish the following two cases.

Case 1. $\alpha =2$.

If $R>2$, Lemma 1 gives

$$\begin{array}{c}{m}_{-}(R)+\frac{3}{4}{\int }_{1<x<R/2}\frac{g^{-}(x)}{x^2}dx\hfill \\ \lesssim m_{-}(R)+{\int }_{1<x<R}{g}^{-}(x)(\frac{1}{x^2}-\frac{1}{R^2})dx\hfill \\ \lesssim m_{+}(R)+{\int }_{1<x<R}\frac{g^{+}(x)}{x^2}dx+|d_3|+|d_4|.\hfill \end{array}$$

(3.1)

Since $u\in {\mathcal{C}}_2$, we obtain

$$\begin{array}{rcl}{\int }_1^{\mathrm{\infty }}\frac{m_{+}(R)}{R}dR& \lesssim & {\iint }_{\{z\in {\mathbf{C}}_{+}:|z|>1\}}\frac{y|f(x+iy)|}{{|x+iy|}^4}dxdy\\ \lesssim & {\iint }_{z\in {\mathbf{C}}_{+}}\frac{y|f(x+iy)|}{1+{|x+iy|}^4}dxdy\\ <& \mathrm{\infty }\end{array}$$

from (1.1) and hence

$$\underset{R\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{m}_{+}(R)=0.$$

(3.2)

Then from (1.2), (3.1), and (3.2) we have

$$\underset{R\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_{1<x<R/2}\frac{g^{-}(x)}{x^2}dx<\mathrm{\infty },$$

which gives

$${\int }_1^{\mathrm{\infty }}\frac{g^{-}(x)}{1+x^2}dx<\mathrm{\infty }.$$

Thus $u\in {\mathcal{B}}_2$ from $|u|=u^{+}+u^{-}$.

Case 2. $\alpha >2$.

Since $u\in {\mathcal{C}}_{\alpha }$, we see from (1.1) that

$$\begin{array}{rcl}{\int }_1^{\mathrm{\infty }}\frac{m_{+}(R)}{R^{\alpha -1}}dR& \lesssim & {\iint }_{\{z\in {\mathbf{C}}_{+}:|z|>1\}}\frac{y|f(x+iy)|}{{|x+iy|}^{\alpha +2}}dxdy\\ \lesssim & {\iint }_{z\in {\mathbf{C}}_{+}}\frac{y|f(x+iy)|}{1+{|x+iy|}^{\alpha +2}}dxdy\\ <& \mathrm{\infty },\end{array}$$

(3.3)

and we see from (1.2) that

$$\begin{array}{c}{\int }_1^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}{\int }_1^R{g}_{+}(x)(\frac{1}{x^2}-\frac{1}{R^2})dxdR\hfill \\ ={\int }_1^{\mathrm{\infty }}{g}_{+}(x){\int }_x^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}(\frac{1}{x^2}-\frac{1}{R^2})dRdx\hfill \\ \lesssim {\int }_1^{\mathrm{\infty }}\frac{g_{+}(x)}{x^{\alpha }}dx\hfill \\ <\mathrm{\infty }.\hfill \end{array}$$

(3.4)

We have from (3.3), (3.4), and Lemma 1

$$\begin{array}{c}{\int }_1^{\mathrm{\infty }}{g}_{-}(x){\int }_x^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}(\frac{1}{x^2}-\frac{1}{R^2})dRdx\hfill \\ \le 2\pi {\int }_1^{\mathrm{\infty }}\frac{m_{+}(R)}{R^{\alpha -1}}dR-2\pi {\int }_1^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}(d_3+\frac{d_4}{R^2})dR\hfill \\ +{\int }_1^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}{\int }_1^R{g}_{+}(x)(\frac{1}{x^2}-\frac{1}{R^2})dxdR\hfill \\ <\mathrm{\infty }.\hfill \end{array}$$

Set

$$I(\alpha )=\underset{x\to \mathrm{\infty }}{lim}{x}^{\alpha }{\int }_x^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}(\frac{1}{x^2}-\frac{1}{R^2})dR.$$

We have

$$I(\alpha )=\frac{2}{\alpha (\alpha -2)}$$

from the L’Hospital’s rule and hence we have

$$x^{-\alpha }\lesssim {\int }_x^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}(\frac{1}{x^2}-\frac{1}{R^2})dR.$$

So

$$\begin{array}{rcl}{\int }_1^{\mathrm{\infty }}\frac{g_{-}(x)}{x^{\alpha }}dx& \lesssim & {\int }_1^{\mathrm{\infty }}{g}_{-}(x){\int }_x^{\mathrm{\infty }}\frac{1}{R^{\alpha -1}}(\frac{1}{x^2}-\frac{1}{R^2})dRdx\\ <& \mathrm{\infty }.\end{array}$$

Then $u\in {\mathcal{B}}_{\alpha }$ from $|u|=u^{+}+u^{-}$. We complete the proof of Theorem 1.

4 Proof of Theorem 2

For any $ϵ>0$, there exists $R_{ϵ}>2$ such that

$${\int }_{|t|\ge R_{ϵ}}\frac{|f(t)|}{1+{|t|}^{\alpha }}dt<ϵ$$

(4.1)

from Theorem 1. For any fixed $z\in {\mathbf{C}}_{+}$ and $2|z|>R_{ϵ}$, we write

$$U_m(f)(x)=\sum _{i=1}^4{V}_i(x),$$

where

$$\begin{array}{c}{V}_1(x)={\int }_{0\le |t|<1}{P}_m(z,t)f(t)dt,V_2(x)={\int }_{1<|t|\le R_{ϵ}}{P}_m(z,t)f(t)dt,\hfill \\ V_3(x)={\int }_{R_{ϵ}<|t|\le 2|z|}{P}_m(z,t)f(t)dt\text{and}{V}_4(x)={\int }_{|t|>2|z|}{P}_m(z,t)f(t)dt.\hfill \end{array}$$

By (2.1), (2.2), (2.3), and (4.1), we have the following estimates:

$$\begin{array}{c}|V_1(z)|\lesssim y^{-1}{\int }_{0\le |t|<1}|f(t)|dt\hfill \\ \phantom{|V_1(z)|}\lesssim y^{-1},\hfill \end{array}$$

(4.2)

$$\begin{array}{c}|V_2(z)|\lesssim y^{-1}{|z|}^{m+1}{\int }_{1<|t|\le R_{ϵ}}{|t|}^{-m-1}|f(t)|dt\hfill \\ \phantom{|V_2(z)|}\lesssim R_{ϵ}^{\alpha -m-1}{y}^{-1}{|z|}^{m+1}{\int }_{1<|t|\le R_{ϵ}}{|t|}^{-\alpha }\left|f\left(y^{\prime }\right)\right|dx\hfill \\ \phantom{|V_2(z)|}\lesssim R_{ϵ}^{\alpha -m-1}{y}^{-1}{|z|}^{m+1},\hfill \end{array}$$

(4.3)

$$\begin{array}{c}|V_3(z)|\lesssim {|z|}^{m+1}{y}^{-1}{\int }_{R_{ϵ}<|t|\le 2|z|}{t}^{-m-1}|f(t)|dt\hfill \\ \phantom{|V_3(z)|}\lesssim ϵy^{-1}{|z|}^{\alpha },\hfill \end{array}$$

(4.4)

$$\begin{array}{c}|V_4(z)|\lesssim {|z|}^{m+1}{\int }_{|t|>2|z|}{|t|}^{-m-2}|f(t)|dt\hfill \\ \phantom{|V_4(z)|}\lesssim {|z|}^{\alpha -1}{\int }_{|t|>2|z|}{|t|}^{-\alpha }|f(t)|dt\hfill \\ \phantom{|V_4(z)|}\lesssim ϵ{|z|}^{\alpha -1}.\hfill \end{array}$$

(4.5)

Combining (4.2)-(4.5), (1.3) holds. Thus we complete the proof of Theorem 2.

5 Proof of Theorem 3

Take a number r satisfying $r>R_1$, where $R_1$ is a sufficiently large positive number. For any ϵ ($0<ϵ<1-{ϵ}_0$), we have

$$\rho (r)<\rho (e){(lnr)}^{({ϵ}_0+ϵ)}$$

from the remark, which shows that there exists a positive constant $M(r)$ dependent only on r such that

$$k^{-\beta /2}{r}^{\rho (k+1)+\beta +1}\le M(r)$$

(5.1)

for any $k>k_r=[2r]+1$.

For any $z\in {\mathbf{C}}_{+}$ and $|z|\le r$, we have $|t|\ge 2|z|$ and

$$\begin{array}{c}\sum _{k=k_r}^{\mathrm{\infty }}{\int }_{k\le |t|<k+1}\frac{{|z|}^{[\rho (|t|)+\beta ]+1}}{{|t|}^{[\rho (|t|)+\beta ]+2}}|f(t)|dt\hfill \\ \lesssim \sum _{k=k_r}^{\mathrm{\infty }}\frac{r^{\rho (k+1)+\beta +1}}{k^{\beta /2}}{\int }_{k\le |t|<k+1}\frac{2|f(t)|}{1+{|t|}^{\rho (|t|)+\beta /2+1}}dt\hfill \\ \lesssim M(r){\int }_{|t|\ge k_r}\frac{|f(t)|}{1+{|t|}^{\rho (|t|)+\beta /2+1}}dt\hfill \\ <\mathrm{\infty }\hfill \end{array}$$

from (1.5), (2.2), and (5.1). Thus $U_{[\rho (|t|)+\beta ]}(f)(z)$ is finite for any $z\in {\mathbf{C}}_{+}$. $P_{[\rho (|t|)+\beta ]}(z,t)$ is a harmonic function of $z\in {\mathbf{C}}_{+}$ for any fixed $t\in \partial {\mathbf{C}}_{+}$. $U_{[\rho (|t|)+\beta ]}(f)(z)$ is also a harmonic function of $z\in {\mathbf{C}}_{+}$.

Now we shall prove the boundary behavior of $U_{[\rho (|t|)+\beta ]}(f)(z)$. For any fixed $z^{\prime }\in \partial {\mathbf{C}}_{+}$, we can choose a number $R_2$ such that $R_2>|z^{\prime }|+1$. We write

$$U_{[\rho (|t|)+\beta ]}(f)(z)=X(z)-Y(z)+Z(z),$$

where

$$\begin{array}{c}X(z)={\int }_{|t|\le R_2}P(z,t)f(t)dt,\hfill \\ Y(z)=Im\sum _{k=0}^{[\rho (|t|)+\beta ]}{\int }_{1<|t|\le R_2}\frac{z^k}{\pi t^{k+1}}f(t)dt,\hfill \\ Z(z)={\int }_{|t|>R_2}{P}_{[\rho (|t|+\beta )]}(z,t)f(t)dt.\hfill \end{array}$$

Since $X(z)$ is the Poisson integral of $f(t){\chi }_{[-R_2,R_2]}(t)$, it tends to $f(z^{\prime })$ as $z\to z^{\prime }$. Clearly, $Y(z)$ vanishes in $\partial {\mathbf{C}}_{+}$. Further, $Z(z)=O(y)$, which tends to zero as $z\to z^{\prime }$. Thus the function $U_{[\rho (|t|)+\beta ]}(f)(z)$ can be continuously extended to ${\overline{\mathbf{C}}}_{+}$ such that $U_{[\rho (|t|)+\beta ]}(f)(z^{\prime })=f(z^{\prime })$ for any $z^{\prime }\in \partial {\mathbf{C}}_{+}$. Then Theorem 3 is proved.

6 Proof of Corollary 2

We prove (II). Consider the function $u(z)-U_m(u)(z)$. Then it follows from Corollary 1 that this is harmonic in ${\mathbf{C}}_{+}$ and vanishes continuously in $\partial {\mathbf{C}}_{+}$. Since

$$0\le {(u(z)-U_m(u)(z))}^{+}\le u^{+}(z)+U_m{(u)}^{-}(z)$$

(6.1)

for any $z\in {\mathbf{C}}_{+}$ and

$$\underset{|z|\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{|z|}^{-m-1}{u}^{+}(z)=0$$

(6.2)

from (1.1), for every $z\in {\mathbf{C}}_{+}$ we have

$$u(z)=U_m(u)(z)+Q_m(u)(z)$$

from (6.1), (6.2), Corollary 1, and Lemma 3, where $Q_m(u)$ is a polynomial in ${\mathbf{C}}_{+}$ of degree at most $m-1$ and even with respect to the variable y. From this we evidently obtain (II).

If $u\in {\mathcal{C}}_2$, then $u\in {\mathcal{C}}_{\alpha }$ for $\alpha >2$. (II) shows that there exists a constant $d_5$ such that

$$u(z)=d_{5}y+U_1(u)(z).$$

Put

$$d_2=d_5-\frac{1}{\pi }{\int }_{t\ge 1}\frac{f(t)}{{|t|}^2}dt.$$

It immediately follows that $u(z)=d_{2}y+U(u)(z)$ for every $z=x+iy\in {\mathbf{C}}_{+}$, which is the conclusion of (I). Thus we complete the proof of Corollary 2.

7 Proof of Theorem 4

Consider the function $u(z)-U_{[\rho (|t|)+\beta ]}(u)(z)$, which is harmonic in ${\mathbf{C}}_{+}$, can be continuously extended to ${\overline{\mathbf{C}}}_{+}$ and vanishes in $\partial {\mathbf{C}}_{+}$.

The Schwarz reflection principle [[12], p.68] applied to $u(z)-U_{[\rho (|t|)+\beta ]}(u)(z)$ shows that there exists a harmonic function $\mathrm{\Pi }(z)$ in ${\mathbf{C}}_{+}$ satisfying $\mathrm{\Pi }(\overline{z})=\overline{\mathrm{\Pi }(z)}$ such that $Im\mathrm{\Pi }(z)=u(z)-U_{[\rho (|t|)+\beta ]}(u)(z)$ for $z\in {\overline{\mathbf{C}}}_{+}$. Thus $u(z)=U_{[\rho (|t|)+\beta ]}(u)(z)+Im\mathrm{\Pi }(z)$ for all $z\in {\overline{\mathbf{C}}}_{+}$, where $\mathrm{\Pi }(z)$ is an entire function in ${\mathbf{C}}_{+}$ and vanishes continuously in $\partial {\mathbf{C}}_{+}$. Thus we complete the proof of Theorem 4.