# RETRACTED ARTICLE: A note on the boundary behavior for a modified Green function in the upper-half space

• The Retraction Note to this article has been published in Boundary Value Problems 2020 2020:160

## Abstract

Motivated by (Xu et al. in Bound. Value Probl. 2013:262, 2013) and (Yang and Ren in Proc. Indian Acad. Sci. Math. Sci. 124(2):175-178, 2014), in this paper we aim to construct a modified Green function in the upper-half space of the n-dimensional Euclidean space, which generalizes the boundary property of general Green potential.

## Introduction and main results

Let $${\mathbf{R}}^{n}$$ ($$n\geq2$$) denote the n-dimensional Euclidean space. The upper half-space H is the set $$H=\{x=(x_{1},x_{2},\ldots,x_{n})\in{\mathbf{R}}^{n}: x_{n}>0\}$$, whose boundary and closure are ∂H and $$\overline{H}$$ respectively.

For $$x\in{\mathbf{R}}^{n}$$ and $$r>0$$, let $$B(x,r)$$ denote the open ball with center at x and radius r.

Set

$$E_{\alpha}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} -\log|x| & \mbox{if } \alpha=n=2, \\ |x|^{\alpha-n} & \mbox{if } 0< \alpha< n. \end{array}\displaystyle \displaystyle \displaystyle \right .$$

Let $$G_{\alpha}$$ be the Green function of order α for H, that is,

$$G_{\alpha}(x,y)=E_{\alpha}(x-y)-E_{\alpha}\bigl(x-y^{\ast}\bigr),\quad x,y\in\overline {H} , x\neq y, 0< \alpha\leq n,$$

where denotes reflection in the boundary plane ∂H just as $$y^{\ast}=(y_{1},y_{2},\ldots,-y_{n})$$.

In case $$\alpha=n=2$$, we consider the modified kernel function, which is defined by

$$E_{n,m}(x-y)=\left \{ \textstyle\begin{array}{l@{\quad}l} E_{n}(x-y) & \mbox{if } |y|< 1, \\ E_{n}(x-y) +\Re (\log y- \sum_{k=1}^{m-1}(\frac{x^{k} }{k y^{k} }) ) & \mbox{if } |y|\geq1. \end{array}\displaystyle \displaystyle \right .$$

In case $$0<\alpha<n$$, we define

$$E_{\alpha,m}(x-y)=\left \{ \textstyle\begin{array}{l@{\quad}l} E_{\alpha}(x-y) & \mbox{if } |y|< 1, \\ E_{\alpha}(x-y) -\sum_{k=0}^{m-1}\frac{|x|^{k}}{|y|^{n-\alpha+k}}C^{\frac {n-\alpha}{2}}_{k} ( \frac{x\cdot y}{|x||y|} ) & \mbox{if } |y|\geq1, \end{array}\displaystyle \displaystyle \right .$$

where m is a non-negative integer, $$C^{\omega}_{k}(t)$$ ($$\omega=\frac{n-\alpha}{2}$$) is the ultraspherical (or Gegenbauer) polynomial (see ). The expression arises from the generating function for Gegenbauer polynomials

$$\bigl(1-2tr+r^{2}\bigr)^{-\omega} = \sum _{k=0}^{\infty} C_{k}^{\omega}(t)r^{k},$$
(1.1)

where $$|r|<1$$, $$|t|\leq1$$ and $$\omega>0$$. The coefficient $$C^{\omega}_{k}(t)$$ is called the ultraspherical (or Gegenbauer) polynomial of degree k associated with ω, the function $$C^{\omega}_{k}(t)$$ is a polynomial of degree k in t.

Then we define the modified Green function $$G_{\alpha,m}(x,y)$$ by

$$G_{\alpha,m}(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} E_{n,m+1}(x-y)-E_{n,m+1}(x-y^{\ast}) & \mbox{if } \alpha=n=2, \\ E_{\alpha,m+1}(x-y)-E_{\alpha,m+1}(x-y^{\ast}) & \mbox{if } 0< \alpha< n, \end{array}\displaystyle \displaystyle \displaystyle \right .$$

where $$x, y\in\overline{H}$$ and $$x\neq y$$. We remark that this modified Green function is also used to give unique solutions of the Neumann and Dirichlet problem in the upper-half space .

Write

$$G_{\alpha,m}(x,\mu)=\int_{H} G_{\alpha,m}(x,y) \, d \mu(y),$$

where μ is a non-negative measure on H. Here note that $$G_{2,0}(x,\mu)$$ is nothing but the general Green potential.

Let k be a non-negative Borel measurable function on $${\mathbf{R}}^{n}\times{\mathbf{R}}^{n}$$, and set

$$k(y,\mu)=\int_{E}k(y,x) \, d\mu(x)\quad \mbox{and}\quad k( \mu,x)=\int_{E}k(y,x) \, d\mu(y)$$

for a non-negative measure μ on a Borel set $$E\subset{\mathbf{R}}^{n}$$. We define a capacity $$C_{k}$$ by

$$C_{k}(E)=\sup\mu\bigl({\mathbf{R}}^{n}\bigr), \quad E \subset H,$$

where the supremum is taken over all non-negative measures μ such that $$S_{\mu}$$ (the support of μ) is contained in E and $$k(y,\mu) \leq1$$ for every $$y\in H$$.

For $$\beta\leq0$$, $$\delta\leq0$$ and $$\beta\leq\delta$$, we consider the kernel function

$$k_{\alpha,\beta,\delta}(y,x)=x_{n}^{-\beta}y_{n}^{-\delta}G_{\alpha}(x,y).$$

Now we prove the following result. For related results in a smooth cone and tube, we refer the reader to the papers by Qiao (see [5, 6]) and Liao-Su (see ), respectively. The readers may also find some related interesting results with respect to the Schrödinger operator in the papers by Su (see ), by Polidoro and Ragusa (see ) and the references therein.

### Theorem

Let $$n+m-\alpha+\delta+2\geq0$$. Ifμis a non-negative measure onHsatisfying

$$\int_{H} \frac{y_{n}^{\delta+1}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y)< \infty,$$
(1.2)

then there exists a Borel set $$E\subset H$$with properties:

\begin{aligned} (1)&\quad \lim_{x_{n} \rightarrow0, x \in H-E} \frac{x_{n}^{n-\alpha-\beta+\delta+1}}{(1+|x|)^{n+m-\alpha+\delta+2}} G_{\alpha,m}(x, \mu)=0; \\ (2)&\quad \sum_{i=1}^{\infty}2^{i(n-\alpha+\beta+\delta)}C_{k_{\alpha,\beta,\delta }}(E_{i})< \infty, \end{aligned}

where $$E_{i}=\{x\in E: 2^{-i}\leq x_{n}<2^{-i+1}\}$$.

### Remark

By using Lemma 4 below, condition (2) in Theorem with $$\alpha=2$$, $$\beta=0$$, $$\delta=0$$ means that E is 2-thin at ∂H in the sense of .

## Some lemmas

Throughout this paper, let M denote various constants independent of the variables in questions, which may be different from line to line.

### Lemma 1

There exists a positive constantMsuch that $$G_{\alpha}(x,y)\leq M\frac{x_{n}y_{n}}{|x-y|^{n-\alpha+2}}$$, where $$0<\alpha\leq n$$, $$x=(x_{1},x_{2},\ldots,x_{n})$$and $$y=(y_{1},y_{2},\ldots, y_{n})$$inH.

This can be proved by a simple calculation.

### Lemma 2

Gegenbauer polynomials have the following properties:

1. (1)

$$|C_{k}^{\omega}(t)|\leq C_{k}^{\omega}(1)=\frac{\Gamma(2\omega +k)}{\Gamma(2\omega)\Gamma(k+1)}$$, $$|t|\leq1$$;

2. (2)

$$\frac{d}{dt}C_{k}^{\omega}(t)=2\omega C_{k-1}^{\omega+1}(t)$$, $$k \geq1$$;

3. (3)

$$\sum_{k=0}^{\infty} C_{k}^{\omega}(1)r^{k}=(1-r)^{-2\omega}$$;

4. (4)

$$|C^{\frac{n-\alpha}{2}}_{k} (t)-C^{\frac{n-\alpha}{2}}_{k} ( t^{\ast})| \leq(n-\alpha)C^{\frac{n-\alpha+2}{2}}_{k-1} (1)|t-t^{\ast}|$$, $$|t|\leq1$$, $$|t^{\ast}|\leq1$$.

### Proof

(1) and (2) can be derived from , p.232. Equality (3) follows from expression (1.1) by taking $$t=1$$; property (4) is an easy consequence of the mean value theorem, (1) and also (2). □

### Lemma 3

For $$x, y\in{\mathbf{R}}^{n}$$ ($$\alpha=n=2$$), we have the following properties:

1. (1)

$$|\Im \sum_{k=0}^{m}\frac{x^{k}}{y^{k+1}}|\leq\sum_{k=0}^{m-1} \frac{2^{k} x_{n} |x|^{k}}{|y|^{k+2}}$$;

2. (2)

$$|\Im\sum_{k=0}^{\infty}\frac{x^{k+m+1}}{y^{k}}|\leq 2^{m+1}x_{n} |x|^{m}$$;

3. (3)

$$|G_{n,m}(x,y)-G_{n}(x,y)|\leq M \sum_{k=1}^{m} \frac{k x_{n} y_{n} |x|^{k-1}}{|y|^{k+1}}$$;

4. (4)

$$|G_{n,m}(x,y)|\leq M \sum_{k=m+1}^{\infty} \frac{k x_{n} y_{n} |x|^{k-1}}{|y|^{k+1}}$$.

The following lemma can be proved by using Fuglede (see , Théorèm 7.8).

### Lemma 4

For any Borel setEinH, we have $$C_{k_{\alpha}}(E)=\hat{C}_{k_{\alpha}}(E)$$, where $$\hat{C}_{k_{\alpha}}(E)=\inf\lambda(H)$$, $$k_{\alpha}=k_{\alpha,0,0}$$, the infimum being taken over all non-negative measuresλonHsuch that $$k_{\alpha}(\lambda,x)\geq1$$for every $$x \in E$$.

Following , we say that a set $$E\subset H$$ is α-thin at the boundary ∂H if

$$\sum_{i=1}^{\infty}2^{i(n-\alpha)}C_{k_{\alpha}}(E_{i})< \infty,$$

where $$E_{i}=\{x\in E: 2^{-i}\leq x_{n} <2^{-i+1}\}$$.

## Proof of Theorem

We write

\begin{aligned} G_{\alpha,m}(x,\mu) =&\int_{G_{1}}G_{\alpha}(x,y) \,d \mu(y)+\int_{G_{2}}G_{\alpha}(x,y) \,d\mu(y)+\int _{G_{3}}\bigl[G_{\alpha,m}(x,y)-G_{\alpha}(x,y)\bigr] \,d\mu(y) \\ &{}+\int_{G_{4}}G_{\alpha,m}(x,y) \,d\mu(y)+\int _{G_{5}}G_{\alpha,m}(x,y) \,d\mu(y) \\ =&U_{1}(x)+U_{2}(x)+U_{3}(x)+U_{4}(x)+U_{5}(x), \end{aligned}

where

\begin{aligned}& G_{1}=\biggl\{ y\in H:|x-y|\leq\frac{x_{n}}{2}\biggr\} , \qquad G_{2}=\biggl\{ y\in H:|y|\geq1, \frac{x_{n}}{2}< |x-y|\leq3|x|\biggr\} , \\& G_{3}=\bigl\{ y\in H:|y|\geq1,\vert x-y\vert \leq3|x|\bigr\} , \qquad G_{4}=\bigl\{ y\in H:|y|\geq1,\vert x-y\vert > 3|x|\bigr\} , \\& G_{5}=\biggl\{ y\in H:|y|< 1,|x-y|> \frac{x_{n}}{2}\biggr\} . \end{aligned}

We distinguish the following two cases.

Case 1. $$0<\alpha<n$$.

By assumption (1.2) we can find a sequence $$\{a_{i}\}$$ of positive numbers such that $$\lim_{i\rightarrow\infty} a_{i}=\infty$$ and $$\sum_{i=1}^{\infty}a_{i}b_{i}<\infty$$, where

$$b_{i}=\int_{\{y \in H:2^{-i-1}< y_{n}< 2^{-i+2}\}}\frac{y_{n}^{\delta+1}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y).$$

Consider the sets

$$E_{i}=\biggl\{ x\in H: 2^{-i}\leq x_{n}< 2^{-i+1}, \frac{x_{n}^{n-\alpha-\beta+\delta+1}}{(1+|x|)^{n+m-\alpha+\delta +2}}U_{1}(x)\geq a_{i}^{-1}2^{(i-1)\beta} \biggr\}$$

for $$i=1,2,\ldots$$ . Set

$$G=\bigcup_{x\in E_{i}}B\biggl(x,\frac{x_{n}}{2}\biggr).$$

Then $$G\subset\{y \in H:2^{-i-1}< y_{n}< 2^{-i+2}\}$$. Let ν be a non-negative measure on H such that $$S_{\nu}\subset E_{i}$$, where $$S_{\nu}$$ is the support of ν. Then we have $$k_{\alpha,\beta,\delta}(y,\nu)\leq1$$ for $$y\in H$$ and

\begin{aligned} \int_{H} d\nu \leq& a_{i}2^{(-i+1)\beta} \int _{H} \frac{x_{n}^{n-\alpha-\beta+\delta+1}}{(1+|x|)^{n+m-\alpha+\delta +2}}U_{1}(x)\,d\nu(x) \\ \leq& M a_{i}2^{(-i+1)\beta} 2^{(-i+1)(n-\alpha+\delta+1)}\int _{G}k_{\alpha,\beta,\delta}(y,\nu )\frac{y_{n}^{\delta}}{(1+|y|)^{n+m-\alpha+\delta+2}} \,d\mu(y) \\ \leq& M a_{i}2^{(-i+1)\beta} 2^{(-i+1)(n-\alpha+\delta+1)} 2^{i+1} \int _{\{y\in H: 2^{-i-1}< y_{n}< 2^{-i+2}\}}\frac{y_{n}^{\delta+1}}{(1+|y|)^{n+m-\alpha+\delta+2}}\,d\mu(y) \\ \leq& M 2^{n-\alpha+\beta+\delta+2}2^{-i{(n-\alpha+\beta+\delta)}}a_{i} b_{i}. \end{aligned}

So that

$$C_{k_{\alpha,\beta,\delta}}(E_{i})\leq M 2^{-i{(n-\alpha+\beta+\delta )}}a_{i} b_{i},$$

which yields

$$\sum_{i=1}^{\infty}2^{i(n-\alpha+\beta+\delta)}C_{k_{\alpha,\beta,\delta }}(E_{i})< \infty.$$

Setting $$E=\bigcup_{i=1}^{\infty}E_{i}$$, we see that (2) in Theorem is satisfied and

$$\lim_{x_{n} \rightarrow0, x\in H-E} \frac{x_{n}^{n-\alpha-\beta+\delta+1}}{(1+|x|)^{n+m-\alpha+\delta+2}}U_{1}(x)=0.$$
(3.1)

For $$U_{2}(x)$$, by Lemma 1 we have

\begin{aligned} \bigl\vert U_{2}(x)\bigr\vert \leq& M x_{n} \int_{G_{2}}\frac{y_{n}}{|x-y|^{n-\alpha+2}} \, d\mu(y) \\ \leq& M x_{n}^{\alpha-n-1}|x|^{n+m-\alpha+\delta+2}\int _{G_{2}} \frac{1}{y_{n}^{\delta}}\frac{y_{n}^{\delta+1}}{(1+|y|)^{n+m-\alpha+\delta +2}}\, d\mu(y) \\ \leq& M x_{n}^{\alpha-n-1}|x|^{n+m-\alpha+2}\int _{G_{2}} \frac{y_{n}^{\delta+1}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y). \end{aligned}
(3.2)

Note that $$C^{\omega}_{0} ( t )\equiv1$$. By (3) and (4) in Lemma 2, we take $$t=\frac{x\cdot y}{|x||y|}$$, $$t^{\ast}=\frac{x\cdot y^{\ast}}{|x||y^{\ast}|}$$ in Lemma 2(4) and obtain

\begin{aligned} \bigl\vert U_{3}(x)\bigr\vert \leq& \int _{G_{3}}\sum_{k=1}^{m} \frac{|x|^{k}}{|y|^{n-\alpha+k}} 2(n-\alpha)C^{\frac{n-\alpha+2}{2}}_{k-1} (1) \frac{x_{n}y_{n}}{|x||y|} \frac{2|y|^{n+m-\alpha+\delta+2}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m} \sum_{k=1}^{m} \frac{1}{4^{k-1}}C^{\frac{n-\alpha+2}{2}}_{k-1} (1) \int_{G_{3}} \frac{y_{n}^{\delta+1}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m}. \end{aligned}
(3.3)

Similarly, we have by (3) and (4) in Lemma 2

\begin{aligned} \bigl\vert U_{4}(x)\bigr\vert \leq& \int _{G_{4}}\sum_{k=m+1}^{\infty} \frac{|x|^{k}}{|y|^{n-\alpha+k}} 2(n-\alpha)C^{\frac{n-\alpha+2}{2}}_{k-1}(1)\frac{x_{n}y_{n}}{|x||y|} \frac{2|y|^{n+m-\alpha+\delta+2}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m} \sum_{k=m+1}^{\infty} \frac{1}{2^{k-1}}C^{\frac{n-\alpha+2}{2}}_{k-1} (1) \int_{G_{4}} \frac{y_{n}^{\delta+1}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m}. \end{aligned}
(3.4)

Finally, by Lemma 1, we have

$$\bigl\vert U_{5}(x)\bigr\vert \leq M x_{n}^{\alpha-n-1} \int_{G_{5}}\frac{y_{n}^{\delta +1}}{(1+|y|)^{n+m-\alpha+\delta+2}}\, d\mu(y).$$
(3.5)

Combining (3.1), (3.2), (3.3), (3.4) and (3.5), by Lebesgue’s dominated convergence theorem, we prove Case 1.

Case 2. $$\alpha=n=2$$.

In this case, $$U_{1}(x)$$, $$U_{2}(x)$$ and $$U_{5}(x)$$ can be proved similarly as in Case 1. Here we omit the details and state the following facts:

$$\lim_{x_{n} \rightarrow0, x_{n}\in H-E} \frac{x_{n}^{\delta-\beta+1}}{(1+|x|)^{m+\delta+2}}U_{1}(x)=0,$$
(3.6)

where $$E=\bigcup_{i=1}^{\infty}E_{i}$$ and $$\sum_{i=1}^{\infty}2^{i(\beta+\delta)}C_{k_{\alpha,\beta,\delta }}(E_{i})<\infty$$,

$$\lim_{x_{n} \rightarrow0, x_{n}\in H} \frac{x_{n}^{\delta-\beta+1}}{(1+|x|)^{m+\delta+2}}\bigl[U_{2}(x)+U_{5}(x) \bigr]=0.$$
(3.7)

By Lemma 3(3), we obtain

\begin{aligned} \bigl\vert U_{3}(x)\bigr\vert \leq& \int _{G_{3}}\sum_{k=1}^{m} \frac{k x_{n} y_{n} |x|^{k-1}}{|y|^{k+1}} \frac{2|y|^{m+\delta+2}}{y_{n}^{\delta+1}} \frac{y_{n}^{\delta+1}}{(1+|y|)^{m+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m} \sum_{k=1}^{m} \frac{k}{4^{k-1}}\int_{G_{3}}\frac{y_{n}^{\delta +1}}{(1+|y|)^{m+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m}. \end{aligned}
(3.8)

By Lemma 3(4), we have

\begin{aligned} \bigl\vert U_{4}(x)\bigr\vert \leq& \int _{G_{4}}\sum_{k=m+1}^{\infty} \frac{k x_{n} y_{n} |x|^{k-1}}{|y|^{k+1}} \frac{2|y|^{m+\delta+2}}{y_{n}^{\delta+1}} \frac{y_{n}^{\delta+1}}{(1+|y|)^{m+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m} \sum_{k=m+1}^{\infty} \frac{k}{2^{k-1}}\int_{G_{4}}\frac {y_{n}^{\delta+1}}{(1+|y|)^{m+\delta+2}}\, d\mu(y) \\ \leq& M x_{n}|x|^{m}. \end{aligned}
(3.9)

Combining (3.6), (3.7), (3.8) and (3.9), we prove Case 2.

Hence the proof of the theorem is completed.

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## Acknowledgements

The authors are highly grateful for the referees’ careful reading and comments on this paper. This work was completed while the authors were visiting the Department of Mathematical Sciences at the University of Wollongong, and they are grateful for the kind hospitality of the Department.

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Correspondence to Valery Piskarev.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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