The integral representations of the products for the harmonic polynomials and the Poisson kernel and their growth properties in the half space

Abstract

In this paper, we study the integrals and growth properties of the products multiplying types of polynomials and the Poisson kernel in the half space. By gradually weakening the convergence conditions and redefining the measure, two special cases, namely the growth properties of the integrals multiplying two types of harmonic polynomials and the Poisson kernel, are obtained to solve the Dirichlet boundary value problem of the polyharmonic equation. Moreover, we generalize the results concerning the modified Poisson integral in the half space.

Keywords

Harmonic polynomial / growth property / modified Poisson kernel

1 Introduction and main results

In recent years, more and more attention has been paid on the Poisson integral and its growth properties on the half space H=\{x=\left(x^{\prime },x_n\right)\in R^{n-1}\times R:x_n>0\} of {\text{R}}^n[2, 5, 7, 8]. Scholars at home and abroad use Gegenbauer polynomials [1] to modify the Poisson kernel and study the modified Poisson integral and its growth properties [1, 3, 4].

For the Dirichlet boundary value problem of polyharmonic equation

(1){\mathrm{\Delta }}^{m}U(x)=0,U\in W^{m,p}(H),

with boundary conditions

{\left(\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_n}\right)}^{k}U\left(x^{\prime },0\right)=f_k\left(x^{\prime }\right),0\le k\le m-1,

when , and

It is proved in reference [6] that the K-potential

U_{k}f(x)={\int }_{R^n}\frac{(x-y{)}^{\lambda }}{|x-y{|}^n}f(y)\mathrm{d}y

is an integral representation of the polyharmonic Dirichlet problem (1), where \lambda is a multi-index of length m, l>0.

Reference [9] studied the simplest case of the \mathit{K}-potential, where the modified Poisson kernel is

P_1\left(x\right)=\frac{2x_n}{n{\omega }_n}\frac{x_1}{|x{|}^n},x\in H,

and under certain convergence conditions, the generalized Poisson integral is as follows:

U\left[P_1\right]f(x)=\frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{\left(x_1-y_1\right)f\left(y^{\prime }\right)}{{\left|x-y^{\prime }\right|}^n}\mathrm{d}{y}^{\prime }.

U\left[P_1\right]f(x) satisfies the growth property o\left(x_1\left|x\right|\right),\left|x\right|\to \mathrm{\infty } when removing an exceptional set. Reference [10] modified the Poisson kernel and, under corresponding convergence conditions, the integral of the product multiplying the harmonic polynomial {\left(x_2-y_2\right)}^2-{\left(x_1-y_1\right)}^2 and the Poisson kernel is as follows:

U\left[P_2\right]f(x)=\frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{{\left(x_2-y_2\right)}^2-{\left(x_1-y_1\right)}^2}{{\left|x-\left(y^{\prime },0\right)\right|}^n}f\left(y^{\prime }\right)\mathrm{d}{y}^{\prime }.

U\left[P_2\right]f(x) satisfies the growth property

\underset{|x|\to \mathrm{\infty },x\in H\mathrm{\setminus }G}{\text{lim}}|x{|}^{3}U\left[P_2\right]f(x)=0,

when removing an exceptional set G.

To complete the growth properties of the K-potential, this paper studies special multi-indices. First, modify the Poisson kernel as

P_3\left(x,y^{\prime }\right)=\frac{2x_n}{n{\omega }_n}\frac{x_1{x}_2-y_1{y}_2}{{\left|x-\left(y^{\prime },0\right)\right|}^n},x_1,y_1,x_2,y_2\in \mathrm{R}.

When the continuous function f on {\text{R}}^{n-1} satisfies

(2)

the integral of the product multiplying the harmonic polynomial x_1{x}_2-y_1{y}_2 and the Poisson kernel is

(3)U\left[P_3\right]f\left(x\right)=\frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{\left(x_1{x}_2-y_1{y}_2\right)f\left(y^{\prime }\right)}{{\left|x-\left(y^{\prime },0\right)\right|}^n}\mathrm{d}{y}^{\prime }.

Note that (2) is equivalent to the condition U\left[P_3\right]f(x)\not\equiv \mathrm{\infty }. If x_1{x}_2-y_1{y}_2=1, then P_3\left(x,y^{\prime }\right) is the classical Poisson kernel in the upper half space.

Theorem 1　 Let f be a measurable function on {\text{R}}^{n-1} satisfying (2) and U\left[P_3\right]f(x), defined by (3), be the integral of the product multiplying the harmonic polynomial x_1{x}_2-y_1{y}_2 and the Poisson kernel. The function U\left[P_3\right]f\left(x\right) satisfies

(4)U\left[P_3\right]f\left(x\right)=o\left(x_1{x}_2\left|x\right|\right),\left|x\right|\to \mathrm{\infty },x\in H\mathrm{\setminus }G,

where G={\bigcup }_{j=1}^{\mathrm{\infty }}B\left(x_j,{\rho }_j\right),B\left(x_j,{\rho }_j\right) denotes the ball with center x_j and radius {\rho }_j.

Note 1　When x_1=x_2=1, Theorem 1 reduces to the classical growth result for the Poisson integral in [8].

Considering the complexity of the multi-index \lambda, the Poisson kernel is further modified as

P_4\left(x,y^{\prime }\right)=\frac{2x_n}{n{\omega }_n}\frac{x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}-y_1^{{\lambda }_1}{y}_2^{{\lambda }_2}}{{\left|x-y^{\prime }\right|}^n},x_1,y_1,x_2,y_2\in R,

where {\lambda }_1,{\lambda }_2 are positive real numbers. When the measurable function f on {\text{R}}^{n-1} satisfies

(5)

the integral of the product multiplying the polynomial \left(x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}-y_1^{{\lambda }_1}{y}_2^{{\lambda }_2}\right) and the Poisson kernel is

(6)U\left[P_4\right]f\left(x\right)=\frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{\left(x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}-y_1^{{\lambda }_1}{y}_2^{{\lambda }_2}\right)f\left(y^{\prime }\right)}{{\left|x-y^{\prime }\right|}^n}\mathrm{d}{y}^{\prime }.

Note that condition (5) is equivalent to U\left[P_4\right]f(x)\not\equiv \mathrm{\infty }.

When {\lambda }_1={\lambda }_2=1, P_4 is the modified Poisson kernel P_3 in the upper half space H; when x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}-y_1^{{\lambda }_1}{y}_2^{{\lambda }_2}=1, P_4 is the classical Poisson kernel in the upper half space. Theorem 2 is obtained by generalizing Theorem 1.

Theorem 2　 Let f be a measurable function on {\text{R}}^{n-1} satisfying (5) and U\left[P_4\right]f(x)\text{,} defined by (6), be the integral of the product multiplying the polynomial x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}-y_1^{{\lambda }_1}{y}_2^{{\lambda }_2} and the Poisson kernel U\left[P_4\right]f(x) satisfies

(7)U\left[P_4\right]f\left(x\right)=o\left(x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}\left|x\right|\right),\left|x\right|\to \mathrm{\infty },x\in H\mathrm{\setminus }G,

where G is defined as in Theorem 1.

Note 2　Theorems 1 and 2 establish growth properties while removing exceptional sets. The sizes of these exceptional sets can be controlled, and the results cannot be further improved. As a result, the growth properties of the K-potential [6] and other corresponding results are partly generalized.

2 Proof of theorems

Let \mu be a positive Borel measure on {\text{R}}^{n-1}, \alpha \ge 0. The measurable maximal function M(\mathrm{d}\mu )(x) on order \alpha is defined in [9, 10].

Lemma 1　 If f is a measurable function on {\mathrm{R}}^{n-1} satisfying (1), and

U_1\left[P_3\right]f(x)\triangleq \frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{x_1{x}_{2}f\left(y^{\prime }\right)}{{\left|x-\left(y^{\prime },0\right)\right|}^n}\mathrm{d}{y}^{\prime },

then

(8)U_1\left[P_3\right]f(x)=o\left(x_1{x}_2|x|\right),|x|\to \mathrm{\infty },x\in H\mathrm{\setminus }G.

Note 3　The result can be directly obtained with the method in [8].

Lemma 2　 If f is a measurable function on {\text{R}}^{n-1} satisfying (1), and

U_2\left[P_3\right]f\left(x\right)\triangleq \frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{y_1{y}_{2}f\left(y^{\prime }\right)}{{\left|x-\left(y^{\prime },0\right)\right|}^n}\text{d}{y}^{\prime },

then

(9)U_2\left[P_3\right]f\left(x\right)=o\left(\left|x\right|\right),\left|x\right|\to \mathrm{\infty },x\in H\mathrm{\setminus }G.

Proof　Let

\begin{array}{rl}K\left(x,y^{\prime }\right)& =\frac{2x_n}{n{\omega }_n}\frac{1+{\left|\left(y^{\prime },0\right)\right|}^n}{{\left|x-\left(y^{\prime },0\right)\right|}^n};\\ \text{d}m\left(y^{\prime }\right)& =\frac{y_1{y}_2\left|f\left(y^{\prime }\right)\right|d{y}^{\prime }}{1+\mid ({\left(y^{\prime },0\right)|}^n}.\end{array}

Then the function U_2\left[P_3\right]f(x) can be rewritten as

U_2\left[P_3\right]f(x)={\int }_{R^{n-1}}K\left(x,y^{\prime }\right)\mathrm{d}m\left(y^{\prime }\right).

Let E be an arbitrary Lebesgue measurable set in {\text{R}}^n. Define m^{(\epsilon )}(E) and E(\epsilon ) as in [10], namely

\begin{array}{rl}{m}^{(\epsilon )}(E)& =m\left(E\cap \left\{x^{\prime }\in R^{n-1}:\left|x^{\prime }\right|>R_{\epsilon }\right\}\right),\\ E\left(\epsilon \right)& =\{x^{\prime }\in R^n:\left|x^{\prime }\right|>R_{\epsilon },\text{there exists}t>0,\\ & \text{such that}{m}^{(\epsilon )}\left(B(x,t)\cap R^{n-1}\right)>\epsilon \frac{t^{n-\alpha }}{|x{|}^{n-\alpha }}\},\end{array}

where R_{\epsilon } is a sufficiently large positive number related to \epsilon. Let

U_2^{(\epsilon )}\left[P_3\right]f(x)={\int }_{R^{n-1}}K\left(x,y^{\prime }\right)\text{d}{m}^{(\epsilon )}\left(y^{\prime }\right).

Define

Then

U_2^{(\epsilon )}\left[P_3\right]f(x)=U_{21}\left[P_3\right]f(x)+U_{22}\left[P_3\right]f(x)+U_{23}\left[P_3\right]f(x).

When |x|\ge 2R_{\epsilon } and x\notin E(\epsilon ), where R_{\epsilon } is sufficiently large, there is

Similarly,

Likewise,

\left|U_{23}\left[P_3\right]f(x)\right|\le {\int }_{\left\{y^{\prime }\in R^{n-1}:\left|x-y^{\prime }\right|\ge 2|x|\right\}}\frac{2x_n\left(1+{\left|y^{\prime }\right|}^n\right)}{n{\omega }_n{\left|x-y^{\prime }\right|}^n}\text{d}{m}^{(\epsilon )}\left(y^{\prime }\right)\ge M\epsilon |x|.

Here M is a positive constant related to \epsilon, which may represent different constants in different places. According to the proof method of Theorem 1 in [8], it is known that (9) holds when removing the exceptional set G. The proof of Lemma 2 is completed.

From Lemma 1 and Lemma 2, Theorem 1 holds.

The proof of Theorem 2 requires Lemma 3 and Lemma 4.

Lemma 3　 Let f be a measurable function on {\text{R}}^{n-1} satisfying (5). Then

U_1\left[P_4\right]f\left(x\right)\triangleq \frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}f\left(y^{\prime }\right)\text{d}{y}^{\prime }}{{\left|x-\left(y^{\prime },0\right)\right|}^n}

satisfies

(10)U_1\left[P_4\right]f\left(x\right)=o\left(x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}\left|x\right|\right),\left|x\right|\to \mathrm{\infty },x\in H\mathrm{\setminus }G.

Proof　Define

\begin{array}{rl}& L\left(x,y^{\prime }\right)=\frac{2x_n}{n{\omega }_n}\frac{1+{\left|y^{\prime }\right|}^n}{{\left|x-y^{\prime }\right|}^n};\\ & \text{d}{m}_1\left(y^{\prime }\right)=\frac{y_1{y}_2\left|f\left(y^{\prime }\right)\right|\text{d}{y}^{\prime }}{1+{\left|y^{\prime }\right|}^n}.\end{array}

Let E and E(\epsilon ) be respectively defined as in Lemma 2, and

m_1^{(\epsilon )}(E)\triangleq m_1\left(E\cap \left\{x^{\prime }\in R^{n-1}:\left|x^{\prime }\right|\ge R_{\epsilon }\right\}\right).

Then the function U_1\left[P_4\right]f(x) can be rewritten as

U_1\left[P_4\right]f(x)={\int }_{R^{n-1}}L\left(x,y^{\prime }\right)\text{d}{m}_1\left(y^{\prime }\right).

Let

U_1^{(\epsilon )}\left[P_4\right]f(x)={\int }_{R^{n-1}}L\left(x,y^{\prime }\right)\text{d}{m}_1^{(\epsilon )}\left(y^{\prime }\right)\text{.}

Define

Then

U_1^{(\epsilon )}\left[P_4\right]f(x)=U_{11}\left[P_4\right]f(x)+U_{12}\left[P_4\right]f(x)+U_{13}\left[P_4\right]f(x).

(10) is proved by the proof method similar to Lemma 2 in this paper.

Lemma 4　 Under the conditions of Lemma 3, the integral of the product multiplying the polynomial y_1^{{\lambda }_1}{y}_2^{{\lambda }_2} and the Poisson kernel

U_2\left[P_4\right]f(x)\triangleq \frac{2x_n}{n{\omega }_n}{\int }_{R^{n-1}}\frac{y_1^{{\lambda }_1}{y}_2^{{\lambda }_2}f\left(y^{\prime }\right)\text{d}{y}^{\prime }}{{\left|x-\left(y^{\prime },0\right)\right|}^n}

satisfies the growth property

(11)U_2\left[P_4\right]f\left(x\right)=o\left(\left|x\right|\right),\left|x\right|\to \mathrm{\infty },x\in H\mathrm{\setminus }G.

Note 4　Although the results of Lemma 2 and Lemma 4 are the same, their respective convergence conditions are different, so the removed exceptional sets are different.

According to Lemma 3 and Lemma 4, Theorem 2 holds.

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