Lobachevskii Journal of Mathematics

http://ljm.ksu.ru
ISSN 1818-9962

Vol. 22, 2006, 47–57 \begin{matrix}c \\ ◯ \end{matrix} M. Z. Sarikaya, H. Yildirim

Mehmet Zeki Sarikaya and Hüseyin Yildirim

ON HARDY TYPE INEQUALITY WITH NON-ISOTROPIC KERNELS

(submitted by F. G. Avkhadiev)

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

The λ-distance between points x = (x_{1}, ..., x_{n}) and y = (y_{1}, ..., y_{n}) is defined by the following formula given in [1,7-9,11];

|x − y|_{λ} := (|x_{1} − y_{1}|^{\frac{1}{λ_{1}}} + |x_{2} − y_{2}|^{\frac{1}{λ_{2}}} + ... + |x_{n} − y_{n}|^{\frac{1}{λn}})^{\frac{|λ|}{n}} .

where λ = (λ_{1}, λ_{2}, ..., λ_{n}), λ_{k} ≥ \frac{1}{2}, k = 1, 2, ..., n, |λ| = λ_{1} +λ_{2} +...+λ_{n}. Note that this distance has the following properties of homogeneity for any positive t,

(|t^{λ_{1}}x_{1}|^{\frac{1}{λ_{1}}} + ... + |t^{λ_{n}}x_{n}|^{\frac{1}{λn}} )^{\frac{|λ|}{n}} = t^{\frac{|λ|}{n}} |x|_{λ} , t > 0.

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

2000 Mathematical Subject Classification. 31B10, 44A15, 46E35.

Key words and phrases. Riesz Potential, Non-Isotropic distance, Hardy inequality, maximal function.

\begin{matrix}1 & |x|_{λ} = 0 ⇔ x = θ, θ = (0, 0, ..., 0) \\ 2. & |t^{λ}x|_{λ} = |t|^{\frac{|λ|}{n}} |x|_{λ} \\ 3. & |x + y|_{λ} ≤ k(|x|_{λ} + |y|_{λ}) \end{matrix}

where k = 2^{^{(1+ \frac{1}{λ_{min}} )}\frac{|λ|}{n}} , λ_{min} = min\{λ_{1}, λ_{2}, ..., λ_{n}\}.

Here we consider λ-spherical coordinates by the following formulas :

x_{1} = (ρ cos φ_{1})^{2λ_{1}}, ..., x_{n} = (ρ sin φ_{1}sinφ_{2}... sin φ_{n−1})^{2λ_{n}}.

We obtained that |x|_{λ} = ρ^{\frac{2|λ|}{n}} . It can be seen that the Jacobian J_{λ}(ρ, φ) of this transformation is J_{λ}(ρ, φ) = ρ^{2|λ|−1}Ω_{λ}(φ), where Ω_{λ}(φ) is the bounded function, which only depend on angles φ_{1}, φ_{2}, ..., φ_{n−1}. It is clear that if λ_{i} = \frac{1}{2}, i = 1, ..., n, then the λ-distance is Euclidean distance.

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

∥x^{β−α} ∫_{0}^{x} \frac{f(y)dy}{y^{β}(x−y)^{1−α}} ∥_{L_{p}(0,b)} ≤ c ∥f∥_{L_{p}(0,b)} , 0 < α < 1

where α − \frac{1}{p} < β < \frac{1}{q}, \frac{1}{p} + \frac{1}{q} = 1 and 0 < b ≤ ∞. Its generalization

∫_{ℝ^{n}} |x|_{λ}^{μ} |I_{α,λ}f(x)|^{p} dx ≤ c ∫_{ℝ^{n}} |x|_{λ}^{γ} |f(x)|^{p} dx

for the following generalized Riesz potential with the non-isotropic kernel depending on λ-distance,

I_{α,λ}f(x) = ∫_{ℝ^{n}} |x − y|_{λ}^{α−n} f(y)dy, 0 < α < n.  (1)

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

In this paper we consider the case λ_{i} ≥ \frac{1}{2}, i = 1, ..., n.

For a positive r and any x ∈ ℝ^{n} we denote the open λ−ball B_{λ}(x, r) with radius r and a center x as

B_{λ}(x, r) = {y ∈ ℝ^{n} : |y − x|_{λ} < r }.

Let Ω be an open bounded set in ℝ^{n}, n ≥ 1 and p(x) a function on \overline{Ω} satisfying the conditions

1 < p_{0} ≤ p(x) ≤ P < ∞, x ∈ \overline{Ω}  (2)

and

|p(x) − p(y)| ≤ \frac{A}{ln \frac{1}{|x−y|_{λ}}} , |x − y|_{λ} ≤ \frac{1}{2}, x, y ∈ \overline{Ω}.  (3)

Let the weighted maximal function

M^{β,λ}f(x) = |x − x^{0}|{β \atop λ} sup_{r>0}^{\frac{1}{|B_{λ}(x,r)|}} ∫_{B_{λ}(x,r)∩Ω} \frac{|f(y)|}{|y−x_{0}|_{λ}^{β}} dy  (4)

where x_{0} ∈ \overline{Ω}. We write M = M_{0,λ} in the case where β = 0.

By L_{p(.)}(Ω) we denote the space of measurable functions f(x) on Ω such that

I_{p}(f) := ∫_{Ω} |f(x)|^{p(x)} dx < ∞.

This is a Banach space with respect to the norm

∥f∥_{p(.)} = inf\{τ > 0 : I_{p}(\frac{f}{τ}) ≤ 1\}.

The Hölder inequality holds in the form

∫_{Ω} |f(x)g(x)| dx ≤ K ∥f∥_{p(.)} ∥g∥_{q(.)}

with K = \frac{1}{p_{0}} + \frac{1}{q_{0}} . The functional I_{p}(f) and the norm ∥f∥_{p(.)} are simultaneously greater than one and simultaneously less than 1 :

∥f∥_{p(.)}^{P} ≤ I_{p}(f) ≤ ∥f∥_{p(.)}^{p_{0}} if ∥f∥_{p(.)} ≤ 1

and

∥f∥_{p(.)}^{p_{0}} ≤ I_{p}(f) ≤ ∥f∥_{p(.)}^{P} if ∥f∥_{p(.)} ≥ 1.

The imbedding

L_{p(x)} ⊆ L_{r(x)}, 1 ≤ r(x) < p(x) ≤ P < ∞

is valid if |Ω| < ∞. In that case

∥f∥_{r(.)} ≤ m ∥f∥_{p(.)} , m = a_{2} + (1 − a_{1}) |Ω|  (5)

where a_{1} = inf_{x∈Ω} \frac{r(x)}{p(x)} and a_{2} = sup_{x∈Ω} \frac{r(x)}{p(x)}.

Lemma 1: Let 0 < α < n. Then there is the following inequality.

||x − z|_{λ}^{α−n} − |z − y|_{λ}^{α−n}| ≤ M |x − y|_{λ} |x − z|_{λ}^{α−n−1} , for x, y, z ∈ ℝ^{n}

where |x − z|_{λ} ≥ 2 |x − y|_{λ} , and M is a constant which does not depend on x, y and z.

Lemma 1 is proved in [7].

Lemma 2: Let 0 < α < n. There is the following inequality

sup_{r>0} r^{−2|λ|} ∫_{|y|_{λ}<r} \frac{dy}{|y − x|_{λ}^{n−α}} ≤ C |x|_{λ}^{α−n}

where x, y ∈ ℝ^{n} ,λ = (λ_{1}, λ_{2}, ..., λ_{n}), λ_{k} ≥ \frac{1}{2}, k = 1, 2, ..., n, |λ| = λ_{1} + λ_{2} + ... + λ_{n} and the constant C is independent of x, y and r.

İspat: Passing to the λ-spherical coordinates we obtain

∫_{|y−x|_{λ}< \frac{|x|_{λ}}{2}} |y − x|_{λ}^{α−n} dy = Ω_{λ}(φ) ∫_{0}^{\frac{|x|_{λ}}{2}}ρ^{α−n+2|λ|−1}dρ = C |x|_{λ}^{α−n+2|λ|} .

In case \frac{|x|_{λ}}{2} ≥ r, from Lemma 1 and the λ−spherical coordinates we have

\begin{aligned}∫_{|y|_{λ}<r} |y − x|_{λ}^{α−n} dy & ≤ & C_{1} ∫_{|y|_{λ}<r} ||x|_{λ} − |y|_{λ}|^{α−n} dy \\ & ≤ & C_{1}2^{n−α} |x|_{λ}^{α−n} \frac{r^{2|λ|}}{2|λ|} Ω_{λ}(φ) \\ & = & C_{2}r^{2|λ|} |x|_{λ}^{α−n} . \end{aligned}  (6)

In case \frac{|x|_{λ}}{2} < r, we can write the following inequality

∫_{|y|_{λ}<r} |y − x|_{λ}^{α−n} dy

≤ ∫_{|y|_{λ}<r} \frac{dy}{(\frac{|x|_{λ}}{2} )n−α} + ∫_{|y|_{λ}< \frac{|x|_{λ}}{2}} \frac{dy}{|y − x|_{λ}^{n−α}}

≤ 2^{n−α} |x|_{λ}^{α−n} ∫_{|y|_{λ}<r} dy + ∫_{|y|_{λ}< \frac{|x|_{λ}}{2}} 2^{n−α} |x|_{λ}^{α−n} dy

= C_{3}r^{2|λ|} |x|_{λ}^{α−n} + C_{4} |x|_{λ}^{α−n+2|λ|} . (7)

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

\begin{aligned}r^{−2|λ|} ∫_{|y|_{λ}<r} |y − x|_{λ}^{α−n} dy & ≤ & \{ \begin{matrix}C_{2} |x|_{λ}^{α−n} ,  \frac{|x|_{λ}}{2} ≥ r \\ (C_{3} |x|_{λ}^{α−n} + C_{4} \frac{|x|_{λ}^{α−n+2|λ|}}{r^{2|λ|}} ) , \frac{|x|_{λ}}{2} < r \end{matrix} \\ & ≤ & \{ \begin{matrix}C_{2} |x|_{λ}^{α−n} , \frac{|x|_{λ}}{2} ≥ r \\ C_{5} |x|_{λ}^{α−n} ,  \frac{|x|_{λ}}{2} < r \end{matrix} \end{aligned}

Now, for C = max{C_{2}, C_{5}} we obtain

sup_{r>0} r^{−2|λ|} ∫_{|y|_{λ}<r} |y − x|_{λ}^{α−n} dy ≤ C |x|_{λ}^{α−n} .

Theorem 1: Let p(x) satisfy conditions (2), (3). If

0 < β < \frac{n}{q(x_{0})},  (8)

then there is a following inequality

[M_{β,λ}f]^{p(x)} ≤ C (1 + \frac{1}{|B_{λ}(x,r)|} ∫_{B_{λ}(x,r)} |f(y)| dy)  (9)

for all f ∈ L_{p(.)}(Ω) such that ∥f∥_{p(.)} ≤ 1, where C = C(p, β, λ) 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(x) we conclude that there exists a d > 0 such that

βq(x) < n for all |x − x_{0}|_{λ} ≤ d  (10)

without loss of generality we assume that d ≤ 1. Let

p_{r}(x) = min_{|x−y|_{λ}≤r}p(y)

and \frac{1}{q_{r}(x)} = 1 − \frac{1}{p_{r}(x)}. From (8) it is easily seen that

βq_{r}(x) < n if |x − x_{0}|_{λ} ≤ \frac{d}{2} and 0 < r ≤ \frac{d}{4}.

In case |x − x_{0}|_{λ} ≤ \frac{d}{2} and 0 < r ≤ \frac{d}{4}, applying the Hölder inequality with the exponents p_{r}(x) and q_{r}(x) to the integral on the right-hand side of the equality

|M_{β,λ} (\frac{f(y)}{|y − x_{0}|_{λ}^{β}} )|^{p(x)} ≤ \frac{C}{r^{2|λ|p(x)}} ( ∫_{B_{λ}(x,r)} \frac{f(y)}{|y − x_{0}|_{λ}^{β}} dy)^{p(x)}

and taking into account (10), we get

\begin{matrix}|M_{β,λ} (\frac{f(y)}{|y − x_{0}|_{λ}^{β}} )| & p(x) \\ ≤ \frac{C}{r^{2|λ|p(x)}} ( ∫_{B_{λ}(x,r)} |f(y)|^{p_{r}(x)} dy)^{\frac{p(x)}{pr(x)}} ( ∫_{B_{λ}(x,r)} \frac{dy}{|y − x_{0}|_{λ}^{βq_{r}(x)}} )^{\frac{p(x)}{qr(x)}} \end{matrix} (11)

From Lemma 2, we obtain

|M_{β,λ} (\frac{f(y)}{|y−x_{0}|_{λ}^{β}} )|^{p(x)} ≤ _{\frac{C|x−x_{0}|_{λ}^{−βp(x)}}{r \frac{2|λ|p(x)}{p_{r}(x)}}} ( ∫_{B_{λ}(x,r)} |f(y)|^{p_{r}(x)} dy)^{\frac{p(x)}{pr(x)}} .

Hence

∫_{B_{λ}(x,r)} |f(y)|^{p_{r}(x)} dy ≤ ∫_{B_{λ}(x,r)} dy + ∫_{B_{λ}(x,r)_{{y: |f(y)|≥1}}} |f(y)|^{p(y)} dy

since p_{r}(x) ≤ p(y) for y ∈ B_{λ}(x, r). Since p(x) is bounded, we see that

|M_{β,λ} (\frac{f(y)}{|y−x_{0}|_{λ}^{β}} )|^{p(x)} ≤ _{\frac{C_{1}|x−x_{0}|_{λ}^{−βp(x)}}{r \frac{2|λ|p(x)}{p_{r}(x)}}} (r^{2|λ|} + \frac{1}{2} ∫_{B_{λ}(x,r)} |f(y)|^{p(y)} dy)^{\frac{p(x)}{pr(x)}} .

Since r ≤ \frac{d}{2} ≤ \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

\begin{aligned}|M_{β,λ}f|^{p(x)} & ≤ & _{\frac{C}{r \frac{2|λ|p(x)}{p_{r}(x)}}} (r^{2|λ|} + ∫_{B_{λ}(x,r)} |f(y)|^{p(y)} dy) \\ & ≤ & Cr^{2|λ| \frac{p_{r}(x)−p(x)}{p_{r}(x)}} (1 + \frac{1}{r^{2|λ|}} ∫_{B_{λ}(x,r)} |f(y)|^{p(y)} dy) . \end{aligned}

From here (10) follows, since r^{2|λ| \frac{p_{r}(x)−p(x)}{p_{r}(x)}} ≤ C.

In case |x − x_{0}|_{λ} ≥ \frac{d}{2} and 0 < r ≤ \frac{d}{4}. Then we have

|y − x_{0}|_{λ} ≥ K^{−1} |x − x_{0}|_{λ} − |x − y|_{λ} ≥ K^{−1} \frac{d}{2} − \frac{d}{4} = \frac{d}{2}(2K^{−1} − 1). (12)

Thus |y − x_{0}|_{λ}^{β} ≥ (\frac{d}{2}(2K^{−1} − 1))^{β} . Since |x − x_{0}|_{λ}^{β} ≤ (diam Ω)^{β} , it follows that M_{β,λ}f(x) ≤ CM_{λ}f, and one may proceed as above for the case β = 0 (the condition |x − x_{0}|_{λ} ≤ \frac{d}{2} is not need in this case).

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

M_{β,λ}f(x) ≤ \frac{C(diam Ω)}{( \frac{d}{4} )^{2|λ|}} ( ∫_{|y−x_{0}|_{λ}≤ \frac{d}{8}} \frac{f(y)dy}{|y−x_{0}|_{λ}^{β}} + ∫_{|y−x_{0}|_{λ}≥ \frac{d}{8}} \frac{f(y)dy}{|y−x_{0}|_{λ}^{β}} ) .

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

p_{\frac{d}{8}} = min_{|y−x_{0}|_{λ}≤ \frac{d}{8}}p(y) and q_{\frac{d}{8}} = p′_{\frac{d}{8}}

as in (11), which is possible since αq_{\frac{d}{8}} < n. The estimate of the second integral is same as (12) since |y − x_{0}|_{λ} ≥ \frac{d}{8}.

Corollary: Let 0 < β < \frac{n}{q(x_{0})}. If conditions (2), (3) are satisfied, then

|M_{β,λ}f|^{p(x)} ≤ C (1 + M [|f(.)|^{p(.)}] (x))  (13)

for all f ∈ L_{p(.)}(Ω) such that ∥f∥_{p(.)} ≤ 1.

Theorem 2: Let p(x) satisfy conditions (2), (3). The operator M_{β,λ} with x_{0} ∈ Ω is bounded in L_{p(x)}(Ω) if

−\frac{n}{p(x_{0})} < β < \frac{n}{q(x_{0})}.

Proof. We have to show that

∥M_{β,λ}f∥_{p(.)} ≤ c

in some ball ∥f∥_{p(.)} ≤ R, which is equivalent to the inequality

I_{p}(M_{β,λ}f) ≤ c for ∥f∥_{p(.)} ≤ R.

We observe that

|x − x_{0}|_{λ}^{βp(x)} ∼ |x − x_{0}|_{λ}^{βp(x_{0})}  (14)

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

\begin{aligned}I_{p}(M_{β,λ}f) & ≤ & c ∫_{Ω} |x − x_{0}|_{λ}^{βp(x)} |M(\frac{f(y)}{|y−x_{0}|_{λ}^{β}})|^{p(x)} dx \\ & ≤ & c ∫_{Ω} |x − x_{0}|_{λ}^{βp(x_{0})} |M(\frac{f(y)}{|y−x_{0}|_{λ}^{β}})|^{p(x)} dx. \end{aligned}

For r(x) = \frac{p(x)}{p_{0}}, we have the following inequality

I_{p}(M_{β,λ}f) ≤ c ∫_{Ω} (|x − x_{0}|_{λ}^{βr(x_{0})} |M(\frac{f(y)}{|y−x_{0}|_{λ}^{β}})|^{r(x)})^{p_{0}} dx.

We will proof the theorem breaks up into two case β ≤ 0 and β ≥ 0. Case 1. Let −\frac{n}{p(x_{0})} < β ≤ 0. Estimate (13) with β = 0 says that

|M_{λ}ϕ(x)|^{r(x)} ≤ C (1 + M [ϕ^{r(.)}] (x))  (15)

for all ϕ ∈ L_{r(.)}(Ω) with ∥ϕ∥_{r(.)} ≤ 1. For ϕ(x) = \frac{|f(x)|}{|x−x_{0}|_{λ}^{β}}, we have

∥ϕ∥_{r(.)} ≤ a_{0} ∥f∥_{r(.)} , a_{0} = (diam Ω)^{|β|},

where we took into account that β ≤ 0. From imbedding (5) we obtain

∥ϕ∥_{r(.)} ≤ a_{0}.k ∥f∥_{p(.)} ≤ a_{0}kR.

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

I_{p}(M_{β,λ}f) ≤ C ∫_{Ω} (|x − x_{0}|_{λ}^{βr(x_{0})} [1 + M (|\frac{f(y)}{|y−x_{0}|_{λ}^{β}} |^{r(y)})])^{p_{0}} dx.

Thus we have I_{p}(M_{β,λ}f)

≤ C ∫_{Ω} \{|x − x_{0}|_{λ}^{βp(x_{0})} + (|x − x_{0}|_{λ}^{βr(x_{0})} M (\begin{matrix}|f(y)| & r(y) \\ |y − x_{0}| & {βr(x_{0}) \atop λ} \end{matrix} ))^{p_{0}}\} dx

≤ C + C ∫_{Ω} M^{γ} (|f(.)|^{r(.)})^{p_{0}} dx

where γ = βr(x_{0}) = \frac{βp(x_{0})}{p_{0}}. As is know [5], the weighted maximal operator M^{γ} is bounded in L_{p_{0}} with a constant p_{0} if −\frac{n}{p_{0}} < γ < \frac{n}{p′_{0}} , which is satisfied since −\frac{n}{p(x_{0})} < β ≤ 0. Therefore, we obtain

\begin{aligned}I_{p}(M_{β,λ}f) & ≤ & C + C ∫_{Ω} |f(y)|^{r(y)p_{0}} dy \\ & ≤ & c + c ∫_{Ω} |f(y)|^{p(y)} dy < ∞. \end{aligned}

Case 2. Let 0 ≤ β ≤ \frac{n}{q(x_{0})}. We represent the functional I_{p}(M_{β,λ}f) in the form

I_{p}(M_{β,λ}f) = ∫_{Ω} (|M_{β,λ}f(x)|^{r(x)})^{τ} dx

with r(x) = \frac{p(x)}{τ} > 1, τ > 1, where τ will be chosen in the interval 1 < τ < p_{0}. From above similar estimate we have

|M_{β,λ}f(x)|^{r(x)} ≤ c (1 + M (f^{r(.)}) (x))

if ∥f∥_{r(.)} ≤ c and

β < \frac{n}{[r(x_{0})]′} .  (16)

The condition ∥f∥_{r(.)} ≤ c is satisfied since r(x) ≤ p(x). Condition (16) is fulfilled if τ < \frac{n−β}{n} p(x_{0}). Thus, under the choice

1 < τ < min (p_{0}, \frac{n − β}{n} p(x_{0}))

we have

\begin{aligned}I_{p}(M_{β,λ}f) & ≤ & c + c ∫_{Ω} |M (|f^{r(.)}|)|^{τ} dx \\ & ≤ & c + c ∫_{Ω} (|f(x)|^{r(x)})^{τ} dx \end{aligned}

by the boundedness of the maximal operator M in L_{τ}(Ω), τ > 1. Hence

I_{p}(M_{β,λ}f) ≤ c + c ∫_{Ω} |f(x)|^{p(x)} dx.

This proves the theorem.

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

∥|x − x^{0}|_{λ}^{β−α} ∫_{Ω} \frac{|f(y)|}{|y−x_{0}|_{λ}^{β}|x−y|_{λ}^{n−α}} dy∥_{L_{p(.)}} ≤ c ∥f∥_{L_{p(.)}} , 0 < α < n (17)

for all β in the interval

α − \frac{n}{p(x_{0})} < β < \frac{n}{q(x_{0})}.  (18)

Proof. For simplicity we take x_{0} = 0 ∈ \overline{Ω}. We may consider nonnegative functions f and assume that f is continued as zero outside the domain Ω.

We take

I_{α,λ}^{β}f(x) = |x|_{λ}^{β−α} ∫_{Ω} \frac{|f(y)|}{|y|_{λ}^{β}|x−y|_{λ}^{n−α}} dy.

Hence we can split I_{α,λ}^{β}f as follow

I_{α,λ}^{β}f(x)

= |x|_{λ}^{β−α} ∫_{|x−y|_{λ}<2k|x|_{λ}} \frac{|f(y)|}{|y|_{λ}^{β} |x − y|_{λ}^{n−α}} dy

+ |x|_{λ}^{β−α} ∫_{|x−y|_{λ}≥2k|x|_{λ}} \frac{|f(y)|}{|y|_{λ}^{β} |x − y|_{λ}^{n−α}} dy

= J^{1} + J^{2}.

Since α + 2 |λ| > n with λ_{i} ≥ \frac{1}{2}we obtain

J^{1} = |x|_{λ}^{β−α} ∑_{m=1}^{∞} ∫_{2^{−m}k|x|_{λ}<|x−y|_{λ}<2^{−m+1}k|x|_{λ}} \frac{|f(y)|}{|y|_{λ}^{β} |x − y|_{λ}^{n−α}} dy

≤ 2^{2|λ|} |x|_{λ}^{β+2|λ|−n} k^{α+2|λ|−n}

× ∑_{m=1}^{∞} 2^{−m(α+2|λ|−n)} \frac{1}{(2^{−m}k |x|_{λ})^{2|λ|}} ∫_{|x−y|_{λ}<2^{−m+1}k|x|_{λ}} \frac{|f(y)|}{|y|_{λ}^{β}}dy

= 2^{2|λ|} |x|_{λ}^{2|λ|−n} k^{α+2|λ|−n} ∑_{m=1}^{∞} 2^{−m(α+2|λ|−n)}M_{β,λ}f(x)

Therefore

J^{1} ≤ c |x|_{λ}^{2|λ|−n} M_{β,λ}f(x)  (19)

where c = 2^{n−α}k^{α+2|λ|−n}.

On the other hand, it remains to prove the boundedness of the operator J^{2}. Obviously, |x − y|_{λ} ≥ 2k |x|_{λ} implies that

|x − y|_{λ} ≤ k(|x|_{λ} + |y|_{λ})
|y|_{λ} ≥ k^{−1} |x − y|_{λ} − |x|_{λ}
|x − y|_{λ} ≤ 2k |y|_{λ} .

Therefore we have

J^{2} = |x|_{λ}^{β−α} ∫_{|x−y|_{λ}≤2k|y|_{λ}} \frac{|f(y)|}{|y|_{λ}^{β}|x−y|_{λ}^{n−α}} dy := J_{1}^{2}

The operator conjugate to J_{1}^{2} has the form

J_{1}^{2∗} = |x|_{λ}^{β} ∫_{|x−y|_{λ}≤2k|x|_{λ}} \frac{|g(y)|}{|y|_{λ}^{α−β}|x−y|_{λ}^{n−α}} dy

which is nothing else but the operator of the familiar type J^{1}.

According to (19) and Theorem 2 the operator J_{1}^{2∗} is bounded in conjugate space L_{q(.)}(Ω) if and only if −\frac{n}{q(0)} < α − β < \frac{n}{p(0)}, that is α − \frac{n}{p(0)} < β < α + \frac{n}{q(0)}. Therefore, the operator J_{1}^{2} is bounded in L_{p(.)}(Ω) and J^{2} is bounded in this space.

Remark. Analysis of the proof of Theorem 3 shows that it is also valid in the case when order α is variable as well, in the form

∥|x − x_{0}|_{λ}^{β−α(x_{0})} ∫_{Ω} _{\frac{|f(y)|}{|y−x_{0}|_{λ}^{β}|x−y|_{λ}^{n−α(x_{0})}}} dy∥_{L_{p(.)}} ≤ c ∥f∥_{L_{p(.)}}

for all β in the interval

α(x_{0}) − \frac{n}{p(x_{0})} < β < \frac{n}{q(x_{0})}

if inf_{x∈Ω}α(x) > 0 and α(x) satisfies the same logarithmic condition as p(x)

in (3)

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[7] Sarikaya, M,Z. and Yıldırım,H.: The Restriction and the Continuity Properties of Potentials Depending On λ-Distance, Turk. J. Math.(in press).

[8] Sarikaya, M.Z. and Yıldırım, H.: On the β-spherical Riesz potential generated by the β-distance, Int. Journal of Contemp. Math. Sciences, Vol. 1,2006, no. 1-4, 85 - 89.

[9] Sarikaya, M.Z. and Yıldırım, H.: On the non-isotropic fractional integrals generated by the λ-distance, Selçuk Journal of Appl. Math. Vol. 1, 2006.

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[11] Yıldırım, H.: On Generalization of The Quasi Homogeneous Riesz Potential, Turk. J. Math., (2005), 381-387.

Department of Mathematics, Faculty of Science and Arts, Kocatepe University, Afyon-TURKEY E-mail address: sarikaya@aku.edu.tr E-mail address: hyildir@aku.edu.tr

Received June 5, 2006