Lobachevskii Journal of Mathematics http://ljm.ksu.ru ISSN 1818-9962 Vol. 22, 2006, 47–57

© M. Z. Sarikaya, H. Yildirim

Abstract. In the present paper we establish a Stein-Weiss type generalization of the Hardy type inequality with non-isotropic kernels depending on $\lambda$-distance for the spaces $L_{p\left(.\right)}\left(\Omega \right)$ with variable exponent $p\left(x\right)$ in the case of bounded domains $\Omega$ in $R^n.$

The $\lambda$-distance between points $x=\left(x_1,...,x_n\right)$ and $y=\left(y_1,...,y_n\right)$ is defined by the following formula given in [1,7-9,11];

where $\lambda =\left({\lambda }_1,{\lambda }_2,...,{\lambda }_n\right),{\lambda }_k\ge \frac{1}{2},$ $k=1,2,...,n$, $\left|\lambda \right|={\lambda }_1+{\lambda }_2+...+{\lambda }_n$. Note that this distance has the following properties of homogeneity for any positive $t,$

From this relation it follows that the $\lambda$-distance is the $a$-homogeneous function [1,7-11] where $a=\frac{\left|\lambda \right|}{n}$. So the non-isotropic $\lambda$-distance has the following properties:

where $k=2^{{}^{\left(1+\frac{1}{{\lambda }_{min}}\right)\frac{\left|\lambda \right|}{n}}},{\lambda }_{min}=min\left\{{\lambda }_1,{\lambda }_2,...,{\lambda }_n\right\}.$

Here we consider $\lambda$-spherical coordinates by the following formulas :

We obtained that $\mid x{\mid }_{\lambda }={\rho }^{\frac{2\left|\lambda \right|}{n}}.$ It can be seen that the Jacobian $J_{\lambda }\left(\rho ,\varphi \right)$of this transformation is $J_{\lambda }\left(\rho ,\varphi \right)={\rho }^{2\left|\lambda \right|-1}{\Omega }_{\lambda }\left(\varphi \right),$where ${\Omega }_{\lambda }\left(\varphi \right)$is the bounded function, which only depend on angles${\varphi }_1,{\varphi }_2,...,{\varphi }_{n-1}.$ It is clear that if ${\lambda }_i=\frac{1}{2},i=1,...,n,$ then the $\lambda$-distance is Euclidean distance.

In [3], the classical Hardy inequality for fractional integrals states that

where $\alpha -\frac{1}{p}<\beta <\frac{1}{q},\frac{1}{p}+\frac{1}{q}=1$ and $0<b\le \infty .$ Its generalization

for the following generalized Riesz potential with the non-isotropic kernel depending on $\lambda$-distance,

where $x\in {\mathbb{ℝ}}^n$. (1) equality is well-known the classical Riesz potential for ${\lambda }_i=\frac{1}{2},i=1,...,n.$ For classical Riesz potentials the Hardy type inequality was investigated by [6]. Here particular importance of the non-isotropic kernel is that it doesn’t have the classical triangle inequality.

In this paper we consider the case ${\lambda }_i\ge \frac{1}{2},i=1,...,n.$

For a positive $r$ and any $x\in {\mathbb{ℝ}}^n$ we denote the open $\lambda -$ball $B_{\lambda }\left(x,r\right)$ with radius $r$ and a center $x$ as

Let $\Omega$ be an open bounded set in ${\mathbb{ℝ}}^n$, $n\ge 1$ and $p\left(x\right)$ a function on $\overline{\Omega }$ satisfying the conditions

and

Let the weighted maximal function

where $x_0\in \overline{\Omega }.$ We write $M=M_{0,\lambda }$ in the case where $\beta =0.$

By $L_{p\left(.\right)}\left(\Omega \right)$ we denote the space of measurable functions $f\left(x\right)$ on $\Omega$ such that

This is a Banach space with respect to the norm

The Hölder inequality holds in the form

with $K=\frac{1}{p_0}+\frac{1}{q_0}.$ The functional $I_p\left(f\right)$ and the norm ${∥f∥}_{p\left(.\right)}$ are simultaneously greater than one and simultaneously less than $1:$

and

The imbedding

is valid if $\left|\Omega \right|<\infty .$ In that case

where $a_1=\underset{x\in \Omega }{inf}\frac{r\left(x\right)}{p\left(x\right)}$ and $a_2=\underset{x\in \Omega }{sup}\frac{r\left(x\right)}{p\left(x\right)}.$

Lemma 1: Let $0<\alpha <n.$ Then there is the following inequality.

where ${\left|x-z\right|}_{\lambda }\ge 2{\left|x-y\right|}_{\lambda },$and $M$ is a constant which does not depend on $x,y$ and $z.$

Lemma 1 is proved in [7].

Lemma 2: Let $0<\alpha <n$. There is the following inequality

where $x,y\in {\mathbb{ℝ}}^n$ ,$\lambda =\left({\lambda }_1,{\lambda }_2,...,{\lambda }_n\right),{\lambda }_k\ge \frac{1}{2},$ $k=1,2,...,n$, $\left|\lambda \right|={\lambda }_1+{\lambda }_2+...+{\lambda }_n$ and the constant $C$ is independent of $x,y$ and $r.$

İspat: Passing to the $\lambda$-spherical coordinates we obtain

In case $\frac{{\left|x\right|}_{\lambda }}{2}\ge r,$ from Lemma 1 and the $\lambda -$spherical coordinates we have

In case $\frac{{\left|x\right|}_{\lambda }}{2}<r,$ we can write the following inequality

Thus, by (6), (7) we get

Now, for $C=max\left\{C_2,C_5\right\}$ we obtain

Theorem 1: Let $p\left(x\right)$ satisfy conditions (2), (3). If

then there is a following inequality

for all $f\in L_{p\left(.\right)}\left(\Omega \right)$ such that ${∥f∥}_{p\left(.\right)}\le 1,$ where $C=C\left(p,\beta ,\lambda \right)$ is a constant not depending on $x,r$ and $x_0.$

Proof. We will adapt to our paper the proof given by Kokilashvili and Samko [4] for classical Maximal operator. From (8) and the continuity of $p\left(x\right)$ we conclude that there exists a $d>0$ such that

without loss of generality we assume that $d\le 1.$ Let

and $\frac{1}{q_r\left(x\right)}=1-\frac{1}{p_r\left(x\right)}.$ From (8) it is easily seen that

In case ${\left|x-x_0\right|}_{\lambda }\le \frac{d}{2}$  and  $0<r\le \frac{d}{4},$ applying the Hölder inequality with the exponents $p_r\left(x\right)$ and $q_r\left(x\right)$ to the integral on the right-hand side of the equality

and taking into account (10), we get

From Lemma 2, we obtain

Hence

since $p_r\left(x\right)\le p\left(y\right)$ for $y\in B_{\lambda }\left(x,r\right).$ Since $p\left(x\right)$ is bounded, we see that

Since $r\le \frac{d}{2}\le \frac{1}{2}$ and the second term in the brackets is also less than or equal to $\frac{1}{2}$, we arrive at the estimate

From here (10) follows, since $r^{2\left|\lambda \right|\frac{p_r\left(x\right)-p\left(x\right)}{p_r\left(x\right)}}\le C.$

In case ${\left|x-x_0\right|}_{\lambda }\ge \frac{d}{2}$  and  $0<r\le \frac{d}{4}.$ Then we have

Thus ${\left|y-x_0\right|}_{\lambda }^{\beta }\ge {\left(\frac{d}{2}\left(2K^{-1}-1\right)\right)}^{\beta }.$ Since ${\left|x-x_0\right|}_{\lambda }^{\beta }\le {\left(diam\Omega \right)}^{\beta },$ it follows that $M_{\beta ,\lambda }f\left(x\right)\le C{M}_{\lambda }f,$ and one may proceed as above for the case $\beta =0$ (the condition ${\left|x-x_0\right|}_{\lambda }\le \frac{d}{2}$ is not need in this case).

In case $r\ge \frac{d}{4}.$ It suffices to show that the left-hand side of (9) is bounded. We have have

Here the first integral is estimated via the Hölder inequality with exponents

as in (11), which is possible since $\alpha q_{\frac{d}{8}}<n.$ The estimate of the second integral is same as (12) since ${\left|y-x_0\right|}_{\lambda }\ge \frac{d}{8}.$

Corollary: Let $0<\beta <\frac{n}{q\left(x_0\right)}.$ If conditions (2), (3) are satisfied, then

for all $f\in L_{p\left(.\right)}\left(\Omega \right)$ such that ${∥f∥}_{p\left(.\right)}\le 1.$

Theorem 2: Let $p\left(x\right)$ satisfy conditions (2), (3). The operator $M_{\beta ,\lambda }$ with $x_0\in \Omega$ is bounded in $L_{p\left(x\right)}\left(\Omega \right)$ if

Proof. We have to show that

in some ball ${∥f∥}_{p\left(.\right)}\le R,$ which is equivalent to the inequality

We observe that

in case $p\left(x\right)$ satisfies the condition (3). Following the idea in [2] and so from (14) we have the following inequality

For $r\left(x\right)=\frac{p\left(x\right)}{p_0}$, we have the following inequality

We will proof the theorem breaks up into two case $\beta \le 0$ and $\beta \ge 0.$

Case 1. Let $-\frac{n}{p\left(x_0\right)}<\beta \le 0.$ Estimate (13) with $\beta =0$ says that

for all $\phi \in L_{r\left(.\right)}\left(\Omega \right)$ with ${∥\phi ∥}_{r\left(.\right)}\le 1.$ For $\phi \left(x\right)=\frac{\left|f\left(x\right)\right|}{{\left|x-x_0\right|}_{\lambda }^{\beta }}$, we have

where we took into account that $\beta \le 0.$ From imbedding (5) we obtain

Therefore we choose $R=\frac{1}{a_{0}k}.$ Then ${∥\phi ∥}_{r\left(.\right)}\le 1,$ so that (15) is applicable. From (15), we obtain

Thus we have

where $\gamma =\beta r\left(x_0\right)=\frac{\beta p\left(x_0\right)}{p_0}.$ As is know [5], the weighted maximal operator $M^{\gamma }$ is bounded in $L_{p_0}$ with a constant $p_0$ if $-\frac{n}{p_0}<\gamma <\frac{n}{p_0^{\prime }},$ which is satisfied since $-\frac{n}{p\left(x_0\right)}<\beta \le 0.$ Therefore, we obtain

Case 2. Let $0\le \beta \le \frac{n}{q\left(x_0\right)}.$ We represent the functional $I_p\left(M_{\beta ,\lambda }f\right)$ in the form

with $r\left(x\right)=\frac{p\left(x\right)}{\tau }>1,\tau >1,$ where $\tau$ will be chosen in the interval $1<\tau <p_0.$ From above similar estimate we have

if ${∥f∥}_{r\left(.\right)}\le c$ and

The condition ${∥f∥}_{r\left(.\right)}\le c$ is satisfied since $r\left(x\right)\le p\left(x\right).$ Condition (16) is fulfilled if $\tau <\frac{n-\beta }{n}p\left(x_0\right).$ Thus, under the choice

we have

by the boundedness of the maximal operator $M$ in $L_{\tau }\left(\Omega \right),\tau >1.$ Hence

This proves the theorem.

Theorem 3: Let $p\left(x\right)$ satisfy conditions (2), (3) and $\Omega$ be a bounded domain in ${\mathbb{ℝ}}^n.$ Then the Hardy-type inequality is valid.

for all $\beta$ in the interval

Proof. For simplicity we take $x_0=0\in \overline{\Omega }.$ We may consider non-negative functions $f$ and assume that $f$ is continued as zero outside the domain $\Omega .$

We take

Hence we can split $I_{\alpha ,\lambda }^{\beta }f$ as follow

Since $\alpha +2\left|\lambda \right|>n$ with ${\lambda }_i\ge \frac{1}{2}$we obtain