Sharp estimates for the gradient of the generalized Poisson integral for a half-space

TL;DR

This paper derives explicit sharp estimates for the gradient of the generalized Poisson integral in a half-space, providing exact coefficients for specific cases and advancing understanding of boundary behavior in harmonic analysis.

Contribution

It provides explicit formulas for the sharp coefficient in gradient estimates of the generalized Poisson integral, including special cases for p=1 and p=2.

Findings

Explicit sharp coefficient formulas for the gradient estimate.

Exact coefficient values for p=1 and p=2 cases.

Enhanced understanding of boundary behavior in harmonic analysis.

Abstract

A representation of the sharp coefficient in a pointwise estimate for the gradient of the generalized Poisson integral of a function $f$ on ${\mathbb R}^n$ is obtained under the assumption that $f$ belongs to $L^p$. The explicit value of the coefficient is found for the cases $p=1$ and $p=2$.

Equations

$$\left|\nabla u(x)\right|\leq{\mathcal{K}}_{p}(x)\big{|}\!\big{|}u\big{|}\!\big{|}_{p}$$

$${\mathcal{K}}_p(x)=\frac{K_p}{x_{n+1}^{(n+p)/p}}$$

$$K_1=\frac{2n}{{\omega }_{n+1}},K_2=\sqrt{\frac{n(n+1)}{2^{n+1}{\omega }_{n+1}}},$$

$$u_{{}_f}(x)=k_{n,\alpha }{\int }_{{\mathbb{R}}^n}\frac{x_{n+1}^{\alpha }}{|y-x{|}^{n+\alpha }}f(y^{\prime })d{y}^{\prime }$$

$$\left|\nabla u_{{}_{f}}(x)\right|\leq{\mathcal{C}}_{p}(x)\big{|}\!\big{|}f\big{|}\!\big{|}_{p}\;,$$

$${\mathcal{C}}_p(x)=\frac{C_p}{x_{n+1}^{(n+p)/p}}$$

$$C_1=k_{n,\alpha }n$$

$$C_2=\sqrt{{\omega }_{n-1}}{k}_{n,\alpha }{\left\{\frac{\sqrt{\pi }(n+\alpha )n(n+2)\Gamma \left(\frac{n}{2}-1\right)\Gamma \left(\frac{n}{2}+\alpha \right)}{8(n+1+\alpha )\Gamma (n+\alpha )}\right\}}^{1/2}$$

$$||f|{|}_p={\left\{{\int }_{{\mathbb{R}}^n}|f(x^{\prime }){|}^{p}d{x}^{\prime }\right\}}^{1/p},$$

$$u_{{}_f}(x)=k_{n,\alpha }{\int }_{{\mathbb{R}}^n}\frac{x_{n+1}^{\alpha }}{|y-x{|}^{n+\alpha }}f(y^{\prime })d{y}^{\prime }$$

$$k_{n,\alpha }={\left\{{\int }_{{\mathbb{R}}^n}\frac{x_{n+1}^{\alpha }}{|y-x{|}^{n+\alpha }}d{y}^{\prime }\right\}}^{-1}=\frac{\Gamma \left(\frac{n+\alpha }{2}\right)}{{\pi }^{n/2}\Gamma \left(\frac{\alpha }{2}\right)},$$

$$\alpha >-\frac{n}{p}.$$

$$|\left(\nabla u_{{}_{f}}(x),\boldsymbol{z}\right)|\leq{\mathcal{C}}_{p}(x;\boldsymbol{z})\big{|}\!\big{|}f\big{|}\!\big{|}_{p}$$

$${\mathcal{C}}_p(x;\mathbf{z})=\frac{C_p(\mathbf{z})}{x_{n+1}^{(n+p)/p}},$$

$$C_{1}(\boldsymbol{z})=k_{n,\alpha}\sup_{\sigma\in{\mathbb{S}}^{n}_{+}}\big{|}\big{(}\alpha\boldsymbol{e}_{n+1}-(n+\alpha)(\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1})\boldsymbol{e}_{\sigma},\;\boldsymbol{z}\big{)}\big{|}\big{(}\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1}\big{)}^{n+\alpha},$$

$$C_{p}(\boldsymbol{z})\!=\!k_{n,\alpha}\!\left\{\int_{{\mathbb{S}}^{n}_{+}}\!\!\big{|}\!\big{(}\alpha\boldsymbol{e}_{n+1}\!-\!(n\!+\!\alpha)(\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1})\boldsymbol{e}_{\sigma},\boldsymbol{z}\big{)}\!\big{|}^{\frac{p}{p\!-\!1}}\big{(}\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1}\big{)}^{\frac{(\alpha\!-\!1)p\!+\!n\!+\!1}{p-1}}d\sigma\!\right\}^{\frac{p\!-\!1}{p}}$$

$$C_{\infty}(\boldsymbol{z})=k_{n,\alpha}\int_{{\mathbb{S}}^{n}_{+}}\big{|}\big{(}\alpha\boldsymbol{e}_{n+1}-(n+\alpha)(\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1})\boldsymbol{e}_{\sigma},\;\boldsymbol{z}\big{)}\big{|}\big{(}\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1}\big{)}^{\alpha-1}\;d\sigma.$$

$$\left|\nabla u_{{}_{f}}(x)\right|\leq{\mathcal{C}}_{p}(x)\big{|}\!\big{|}f\big{|}\!\big{|}_{p}$$

$${\mathcal{C}}_p(x)=\frac{C_p}{x_{n+1}^{(n+p)/p}},$$

$$C_p=\underset{|\mathbf{z}|=1}{\sup }{C}_p(\mathbf{z}).$$

$$\frac{\partial u_{{}_f}}{\partial x_i}=k_{n,\alpha }{\int }_{{\mathbb{R}}^n}\left[\frac{{\delta }_{ni}\alpha x_{n+1}^{\alpha -1}}{|y-x{|}^{n+\alpha }}+\frac{(n+\alpha )x_{n+1}^{\alpha }(y_i-x_i)}{|y-x{|}^{n+2+\alpha }}\right]f(y^{\prime })d{y}^{\prime },$$

$$\nabla u_{{}_f}(x)$$

$$(\nabla u_{{}_f}(x),\mathbf{z})=k_{n,\alpha }{x}_{n+1}^{\alpha -1}{\int }_{{\mathbb{R}}^n}\frac{(\alpha {\mathbf{e}}_{n+1}-(n+\alpha ){\mathbf{e}}_{xy},{\mathbf{e}}_{n+1}){\mathbf{e}}_{xy},\mathbf{z})}{|y-x{|}^{n+\alpha }}f(y^{\prime })d{y}^{\prime }.$$

$${\mathcal{C}}_1(x;\mathbf{z})=k_{n,\alpha }{x}_{n+1}^{\alpha -1}\underset{y\in {\partial \mathbb{R}}^n}{\sup }\frac{|(\alpha {\mathbf{e}}_{n+1}-(n+\alpha )({\mathbf{e}}_{xy},{\mathbf{e}}_{n+1}){\mathbf{e}}_{xy},\mathbf{z})|}{|y-x{|}^{n+\alpha }},$$

$${\mathcal{C}}_{p}(x;\boldsymbol{z})=k_{n,\alpha}x_{n+1}^{\alpha-1}\left\{\int_{{\mathbb{R}}^{n}}\frac{\big{|}\big{(}\alpha\boldsymbol{e}_{n+1}-(n+\alpha)(\boldsymbol{e}_{xy},\boldsymbol{e}_{n+1})\boldsymbol{e}_{xy},\boldsymbol{z}\big{)}\big{|}^{q}}{|y-x|^{(n+\alpha)q}}\;dy^{\prime}\right\}^{1/q}$$

$$\frac{x_{n+1}}{|y-x|}=({\mathbf{e}}_{xy},-{\mathbf{e}}_{n+1}),$$

$${\mathcal{C}}_1(x;\mathbf{z})$$

$$\frac{1}{|y-x{|}^{(n+\alpha )q}}=\frac{1}{x_{n+1}^{(n+\alpha )q-n}}{\left(\frac{x_{n+1}}{|y-x|}\right)}^{(n+\alpha )q-n-1}\frac{x_{n+1}}{|y-x{|}^{n+1}},$$

$$C_{p}(\boldsymbol{z})=k_{n,\alpha}\left\{\int_{{\mathbb{S}}^{n}_{-}}\big{|}\big{(}\alpha\boldsymbol{e}_{n+1}-(n+\alpha)(\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1})\boldsymbol{e}_{\sigma},\;\boldsymbol{z}\big{)}\big{|}^{\frac{p}{p-1}}\big{(}\boldsymbol{e}_{\sigma},-\boldsymbol{e}_{n+1}\big{)}^{\frac{(\alpha-1)p+n+1}{p-1}}\;d\sigma\right\}^{\frac{p-1}{p}},$$

$$\big{|}\nabla u_{{}_{f}}(x)\big{|}=k_{n,\alpha}x_{n+1}^{\alpha-1}\sup_{|\boldsymbol{z}|=1}\int_{{\mathbb{R}}^{n}}\frac{(\alpha\boldsymbol{e}_{n+1}-(n+\alpha)(\boldsymbol{e}_{xy},\boldsymbol{e}_{n+1})\boldsymbol{e}_{xy},\;\boldsymbol{z})}{|y-x|^{n+\alpha}}\;f(y^{\prime})dy^{\prime}.$$

Full text

Sharp estimates for the gradient of the generalized Poisson integral for a half-space

Gershon Kresin${}^{a}\!\!$    and Vladimir Maz’yab   

*a**Department of Mathematics, Ariel University, Ariel 40700, Israel

bDepartment of Mathematical Sciences, University of Liverpool, M$\&$O Building, Liverpool,

L69 3BX, UK; Department of Mathematics, Linköping University,SE-58183 Linköping,

Sweden* e-mail: kresin@ariel.ac.ile-mail: vladimir.mazya@liu.se

( )

Dedicated to Vakhtang Kokilashvili on the occasion of his 80th birthday

**Abstract. A representation of the sharp coefficient in a pointwise estimate for the gradient of the generalized Poisson integral of a function $f$ on ${\mathbb{R}}^{n}$ is obtained under the assumption that $f$ belongs to $L^{p}$. The explicit value of the coefficient is found for the cases $p=1$ and $p=2$.

Keywords: generalized Poisson integral, sharp estimate for the gradient, half-space

AMS Subject Classification: Primary: 31B10; Secondary: 31C99

**

1 Introduction

In the paper [3] (see also [6]) a representation for the sharp coefficient ${\mathcal{K}}_{p}(x)$ in the inequality

[TABLE]

was found, where $u$ is harmonic function in the half-space {\mathbb{R}}^{n+1}_{+}=\big{\{}x=(x^{\prime},x_{n+1}):x^{\prime}\in{\mathbb{R}}^{n},x_{n+1}>0\big{\}}, represented by the Poisson integral with boundary values in $L^{p}({\mathbb{R}}^{n})$, $||\cdot||_{p}$ is the norm in $L^{p}({\mathbb{R}}^{n})$, $1\leq p\leq\infty$, $x\in{\mathbb{R}}^{n+1}_{+}$. It was shown that

[TABLE]

and explicit formulas for $K_{1}$ and $K_{2}$ were given. Namely,

[TABLE]

where $\omega_{n}=2\pi^{n/2}/\Gamma(n/2)$ is the area of the unit sphere in ${\mathbb{R}}^{n}$.

In [3] it was shown that the sharp coefficients in pointwise estimates for the absolute value of the normal derivative and the modulus of the gradient of a harmonic function in the half-space coincide for the case $p=1$ as well as for the case $p=2$.

Similar results for the gradient and the radial derivative of a harmonic function in the multidimensional ball with boundary values from $L^{p}$ for $p=1,2$ in [4] were obtained.

Thus, the $L^{1},L^{2}$-analogues of Khavinson’s problem [1] were solved in [3, 4] for harmonic functions in the multidimensional half-space and the ball.

We note that explicit sharp coefficients in the inequality for the first derivative of analytic function in the half-plane and the disk with boundary values of the real-part from $L^{p}$ in [2, 5, 7] were found.

In this paper we treat a generalization of the problem considered in our work [3]. Here we consider the generalized Poisson integral

[TABLE]

with $f\in L^{p}({\mathbb{R}}^{n})$, $\alpha>-(n/p)$, $1\leq p\leq\infty$, where $x\in{\mathbb{R}}^{n+1}_{+}$, $y=(y^{\prime},0)$, $y^{\prime}\in{\mathbb{R}}^{n}$, and $k_{n,\alpha}$ is a normalization constant. In the case $\alpha=1$ the last integral coincides with the Poisson integral for a half-space.

In Section 2 we obtain a representation for the sharp coefficient ${\mathcal{C}}_{p}(x)$ in the inequality

[TABLE]

where

[TABLE]

and the constant $C_{p}$ is characterized in terms of an extremal problem on the unit sphere ${\mathbb{S}}^{n}$ in ${\mathbb{R}}^{n+1}$.

In Section 3 we reduce this extremal problem to that of finding of the supremum of a certain double integral, depending on a scalar parameter and show that

[TABLE]

if $-n<\alpha\leq n$, and

[TABLE]

if $-(n/2)<\alpha\leq n(n+1)/2$.

It is shown that the sharp coefficients in pointwise estimates for the absolute value of the normal derivative and the modulus of the gradient of the generalized Poisson integral for a half-space coincide in the case $p=1$ as well as in the case $p=2$.

2 Representation for the sharp constant in inequality for the gradient in terms of an extremal problem on the unit sphere

We introduce some notation used henceforth. Let {\mathbb{R}}^{n+1}_{+}=\big{\{}x=(x^{\prime},x_{n+1}):x^{\prime}=(x_{1},\dots,x_{n})\in{\mathbb{R}}^{n},x_{n+1}>0\big{\}}, ${\mathbb{S}}^{n}=\{x\in{\mathbb{R}}^{n+1}:|x|=1\}$, ${\mathbb{S}}^{n}_{+}=\{x\in{\mathbb{R}}^{n+1}:|x|=1,\;x_{n+1}>0\}$ and ${\mathbb{S}}^{n}_{-}=\{x\in{\mathbb{R}}^{n+1}:|x|=1,\;x_{n+1}<0\}$. Let $\boldsymbol{e}_{\sigma}$ stand for the $n+1$-dimensional unit vector joining the origin to a point $\sigma$ on the sphere ${\mathbb{S}}^{n}$.

By $||\cdot||_{p}$ we denote the norm in the space $L^{p}({\mathbb{R}}^{n})$, that is

[TABLE]

if $1\leq p<\infty$, and $||f||_{\infty}=\mbox{ess}\;\sup\{|f(x^{\prime})|:x^{\prime}\in{\mathbb{R}}^{n}\}$.

Let the function in ${\mathbb{R}}^{n+1}_{+}$ be represented as the generalized Poisson integral

[TABLE]

with $f\in L^{p}({\mathbb{R}}^{n})$, $1\leq p\leq\infty$, where $y=(y^{\prime},0)$, $y^{\prime}\in{\mathbb{R}}^{n}$,

[TABLE]

and

[TABLE]

Now, we find a representation for the best coefficient ${\mathcal{C}}_{p}(x;\boldsymbol{z})$ in the inequality for the absolute value of derivative of $u_{{}_{f}}(x)$ in an arbitrary direction $\boldsymbol{z}\in{\mathbb{S}}^{n}$, $x\in{\mathbb{R}}^{n+1}_{+}$. In particular, we obtain a formula for the constant in a similar inequality for the modulus of the gradient.

Proposition 1**.**

Let $x$ be an arbitrary point in ${\mathbb{R}}^{n+1}_{+}$ and let $\boldsymbol{z}\in{\mathbb{S}}^{n}$. The sharp coefficient ${\mathcal{C}}_{p}(x;\boldsymbol{z})$ in the inequality

[TABLE]

is given by

[TABLE]

where

[TABLE]

[TABLE]

for $1<p<\infty$, and

[TABLE]

In particular, the sharp coefficient ${\mathcal{C}}_{p}(x)$ in the inequality

[TABLE]

is given by

[TABLE]

where

[TABLE]

Proof.

Let $x=(x^{\prime},x_{n+1})$ be a fixed point in ${\mathbb{R}}^{n+1}_{+}$. The representation (2.1) implies

[TABLE]

that is

[TABLE]

where $\boldsymbol{e}_{xy}=(y-x)|y-x|^{-1}$. For any $\boldsymbol{z}\in{\mathbb{S}}^{n}$,

[TABLE]

Hence,

[TABLE]

and

[TABLE]

for $1<p\leq\infty$, where $p^{-1}+q^{-1}=1$.

Taking into account the equality

[TABLE]

by (2.11) we obtain

[TABLE]

Replacing here $\boldsymbol{e}_{\sigma}$ by $-\boldsymbol{e}_{\sigma}$, we arrive at (2.4) for $p=1$ with the sharp constant (2.5).

Let $1<p\leq\infty$. Using (2.13) and the equality

[TABLE]

and replacing $q$ by $p/(p-1)$ in (2.12), we conclude that (2.4) holds with the sharp constant

[TABLE]

where ${\mathbb{S}}^{n}_{-}=\{\sigma\in{\mathbb{S}}^{n}:(\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1})<0\}$. Replacing here $\boldsymbol{e}_{\sigma}$ by $-\boldsymbol{e}_{\sigma}$, we arrive at (2.6) for $1<p<\infty$ and at (2.7) for $p=\infty$.

By (2.10) we have

[TABLE]

Hence, by the permutation of suprema, (2.12), (2.11) and (2.4),

[TABLE]

for $1<p\leq\infty$, and

[TABLE]

Using the notation (2.9) in (2.14) and (2.15), we arrive at (2.8). ∎

Remark. Formula (2.6) for the coefficient $C_{p}(\boldsymbol{z}),1<p<\infty,$ can be written with the integral over the whole sphere ${\mathbb{S}}^{n}$ in ${\mathbb{R}}^{n+1}$,

[TABLE]

A similar remark relates (2.7):

[TABLE]

as well as formula (2.5):

[TABLE]

3 Reduction of the extremal problem to finding of the supremum by parameter of a double integral.

The cases $p=1$ and $p=2$

The next assertion is based on the representation for $C_{p}$, obtained in Proposition 1.

Proposition 2**.**

Let $f\in L^{p}({\mathbb{R}}^{n})$, and let $x$ be an arbitrary point in ${\mathbb{R}}^{n+1}_{+}$. The sharp coefficient ${\mathcal{C}}_{p}(x)$ in the inequality

[TABLE]

is given by

[TABLE]

where

[TABLE]

if $1<p<\infty$. Here

[TABLE]

with

[TABLE]

In addition,

[TABLE]

if $-n<\alpha\leq n$.

In particular,

[TABLE]

for $-(n/2)<\alpha\leq n(n+1)/2$.

For $p=1$ and $p=2$ the coefficient ${\mathcal{C}}_{p}(x)$ is sharp in conditions of the Proposition also in the weaker inequality obtained from (\ref{EH_2_3}) by replacing $\nabla u_{{}_{f}}$ by $\partial u_{{}_{f}}/\partial x_{n+1}$.

Proof.

The equality (3.2) was proved in Proposition 1.

(i) Let $p=1$. Using (2.5), (2.9) and the permutability of two suprema, we find

[TABLE]

Taking into account the equality

[TABLE]

and using (2.3), (3.7), we arrive at the sharp constant (3.6) for $-n<\alpha\leq n$.

Furthermore, by (2.5),

[TABLE]

Hence, by $C_{1}\geq C_{1}(\boldsymbol{e}_{n+1})$ and by (3.6) we obtain $C_{1}=C_{1}(\boldsymbol{e}_{n+1})$, which completes the proof in the case $p=1$.

(ii) Let $1<p<\infty$. Since the integrand in (2.6) does not change when $\boldsymbol{z}\in{\mathbb{S}}^{n}$ is replaced by $-\boldsymbol{z}$, we may assume that $z_{n+1}=(\boldsymbol{e}_{n+1},\boldsymbol{z})>0$ in (2.9).

Let $\boldsymbol{z}^{\prime}=\boldsymbol{z}-z_{n+1}\boldsymbol{e}_{n+1}$. Then $(\boldsymbol{z}^{\prime},\boldsymbol{e}_{n+1})=0$ and hence $z^{2}_{n+1}+|\boldsymbol{z}^{\prime}|^{2}=1$. Analogously, with $\sigma=(\sigma_{1},\dots,\sigma_{n},\sigma_{n+1})\in{\mathbb{S}}^{n}_{+}$, we associate the vector $\boldsymbol{\sigma}^{\prime}=\boldsymbol{e}_{\sigma}-\sigma_{n+1}\boldsymbol{e}_{n+1}$.

Using the equalities $(\boldsymbol{\sigma}^{\prime},\boldsymbol{e}_{n+1})=0$, $\sigma_{n+1}=\sqrt{1-|\boldsymbol{\sigma}^{\prime}|^{2}}$ and $(\boldsymbol{z}^{\prime},\boldsymbol{e}_{n+1})=0$, we find an expression for (\alpha\boldsymbol{e}_{n+1}-(n+\alpha)(\boldsymbol{e}_{\sigma},\boldsymbol{e}_{n+1})\boldsymbol{e}_{\sigma},\;\boldsymbol{z}\big{)} as a function of $\boldsymbol{\sigma}^{\prime}$:

[TABLE]

Let ${\mathbb{B}}^{n}=\{x^{\prime}=(x_{1},\dots,x_{n})\in{\mathbb{R}}^{n}:|x^{\prime}|<1\}$. By (2.6) and (3.8), taking into account that $d\sigma=d\sigma^{\prime}/\sqrt{1-|\boldsymbol{\sigma}^{\prime}|^{2}}$, we may write (2.9) as

[TABLE]

where

[TABLE]

Using the well known formula (see e.g. [8], 3.3.2(3)),

[TABLE]

we obtain

[TABLE]

Making the change of variable $r=\sin\vartheta$ in the right-hand side of the last equality, we find

[TABLE]

where, by (3.10),

[TABLE]

Introducing here the parameter $\gamma=|\boldsymbol{z}^{\prime}|/z_{n+1}$ and using the equality $|\boldsymbol{z}^{\prime}|^{2}+z^{2}_{n+1}=1$, we obtain

[TABLE]

where ${\mathcal{G}}_{n}(\varphi,\vartheta;\gamma)$ is given by (3.5).

By (3), taking into account (3.11) and (3.12), we arrive at (3.3).

(iii) Let $p=2$. By (3.3), (3.4) and (3.5),

[TABLE]

where

[TABLE]

The last equality and (3.13) imply

[TABLE]

where

[TABLE]

and

[TABLE]

By (3.14) we have

[TABLE]

Further, by (3) and (3.16),

[TABLE]

Therefore,

[TABLE]

Taking into account (3.17) and that $n+2\alpha>0$ for $p=2$ by (2.3), we see that inequality

[TABLE]

holds for $\alpha\leq n(n+1)/2$. So, we arrive at the representation for $C_{2}$ with $-(n/2)<\alpha\leq n(n+1)/2$ given in formulation of the Proposition.

Since $\boldsymbol{z}\in{\mathbb{S}}^{n}$ and the supremum in $\gamma=|\boldsymbol{z}^{\prime}|/z_{n+1}$ in (3.13) is attained for $\gamma=0$, we have $C_{2}=C_{2}(\boldsymbol{e}_{n+1})$ under requirements of the Proposition. ∎