Estimates of Fractional Integral Operators on Variable Exponent Lebesgue Spaces
By some estimates for the variable fractional maximal operator, the authors prove that the fractional integral operator is bounded and satisfies the weak-type inequality on variable exponent Lebesgue spaces.
It is well known that the boundedness of fractional integral operators is the focus of study in the classical Lebesgue spaces. Sobolev  showed that the fractional integral operator is bounded from the classical Lebesgue space to . In 1999, Kenig and Stein  have obtained the boundedness of multilinear fractional integral operator from to . During the recent three decades, variable exponent function spaces have been studied extensively; see, for example, [3–15]. The characterization and the boundedness of the classical operators on variable exponent function spaces were systemically studied; see [5, 6, 16–21]. For example, Capone et al.  proved that Riesz fractional integral satisfies the weak-type inequality on variable exponent Lebesgue spaces in 2007.
Motivated by the aforementioned results, we will consider the variable fractional integral operator in this paper. Before stating our results, we need to recall some notions firstly.
Given a bounded open set and a measurable function The variable exponent Lebesgue space is equipped with the norm
Denote by the set of measurable functions on with values in satisfyingwhere we use the standard notation
Definition 1. (i) If satisfies then we say satisfies the Log-Hölder condition.
(ii) If satisfiesthen we say satisfies the Log-Hölder decay condition.
Given , , the variable fractional operator can be defined by Now the first result is the following theorem.
Theorem 2. Given an open set and , let be a function from to such that and . Suppose further that satisfies condition (5) and satisfies conditions (3), (5), and (6). Define by Then, the fractional integral operator satisfies the weak-type inequality
Given with , the multilinear variable fractional integral operators can be defined byThe following is our second result.
Theorem 4. Let and satisfy the Log-Hölder condition, If satisfies the Log-Hölder condition and , for , and then if each ,
2. Proof of Theorem 2
All cubes are assumed to have their sides parallel to the coordinate axes; denotes that a cube is centered at with side length . is the Lebesgue measure of . denotes that a ball is centered at with radius
Define the variable fractional maximal operator by If , becomes which is the usual Hardy-Littlewood maximal function.
Given a function , denote
Lemma 5 (see ). Given such that , then if and only if . In particular, if either constant equals 1, we can take the other equal to 1 as well.
Lemma 6 (see ). Given a set with finite measure and exponent functions such that ,
The next lemma is the generalized Hölder inequality on variable exponent Lebesgue spaces.
Lemma 7 (see ). If , there is a constant such that, for all and all , where .
Lemma 8 (see ). Given an open set and a function which satisfies the Log-Hölder condition, then, for any ball such that ,
Lemma 9 (see ). Given an open set and a function which satisfies the Log-Hölder condition, then, for any ball such that and ,
Here and below, for , let
Lemma 10 (see ). Given a set and two nonnegative functions and , suppose that, for each , Then, there exists a constant such that, for every function obeying ,
Lemma 11. Given , fix . Then, there exists a constant such that, for all and all ,
Proof. Let be given. Then, Let ; we have where . Hence, Similarly, By choosing , one finds
Lemma 12. Given an open set and , let be such that , , and such that satisfies the Log-Hölder condition. Define by Then, for all such that and such that or ,
Proof. Fix , and fix a ball containing . Then, by the definition of , To complete the proof we will show that We consider it in two cases depending on the size of Suppose first that . Let . By Chebyschev’s inequality and Lemma 5, Therefore, since , by Lemma 6, Now suppose . If , then and satisfies the Log-Hölder condition, so by Lemmas 7 and 9The argument is the same when except that instead of applying Lemma 9 we note that, by Lemma 6, Then, by Lemma 8, This completes the proof.
Lemma 13. Given an open set and , let be such that , , and satisfy the Log-Hölder decay condition. Define by Then, for all such that and , ,where .
Proof. Fix and let be any ball containing . It will suffice to show that where is independent of .
It follows at once from the definition of that . Therefore, by Hölder’s inequality, Since , , so . Therefore, to complete the proof we only need to show that Forit suffices to prove Define the setsThen, The integral in the right-hand side is easy to estimate. If , then . Thus, since , by Lemma 5, To estimate the first integral, we will apply Lemma 10. If , then, by the Log-Hölder decay condition, Therefore, for any and , is integrable, andThis completes the proof.
Lemma 14 (see ). Given an open set , let satisfy the Log-Hölder decay condition. Suppose such that . Then, for every ball and every , where and
Proof of Theorem 2. Fix ; without loss of generality we may assume that . Since , by Lemma 5 it will suffice to prove that Fix , such that Define by Since and , by (23) and Young’s inequality, for each , Therefore,Denote , where and .
Therefore, by (29) and (38), Therefore, We estimate each integral in turn.
For each , define , where . Then, for each the set is bounded, and by the monotone convergence theorem, the first integral is immediate: So, it will suffice to prove that where is independent of Fix ; for each there exists a ball containing such that Therefore, by the Besicovitch covering lemma (see ), there exists a sequence of such that and such that the balls have finite overlap uniformly bounded by a constant depending only on the dimension. Hence, Forby Lemma 14, we have where, in the third inequality, since , we used the boundedness of maximal operator on , and in the last inequality, we used the fact that the balls have uniformly bounded finite overlap.
For , ; then, it will suffice to prove We consider the two cases and to estimate it.
If , since , therefore If , since , by Hölder’s inequality To estimate the second term, we need to show that We first need to control the quantity Fix ; then, there exists , such that Then, by the Log-Hölder decay condition, we have Since is arbitrary and , it follows that For , is integrable, and by Lemma 10The proof now proceeds as it does for the first term.
We estimate as we did before; here we omit the detail. Then, we finish the proof of Theorem 2.
3. Proof of Theorem 4
To prove Theorem 4, we need the following lemma.
Proof of Theorem 4. Since , some . If , , since , we have . Then, integration in reduces matters to the case when all are finite (and ). Thus, we can assume that all .
Now, observe that ; we can find such that . Let . Since and it follows that where is the variable exponent fractional integral operator.
Because , by Hölder’s inequality and (72), we obtain This finishes the proof.
The authors declare that there is no conflict of interests regarding the publication of this paper.
The first author is supported by the National Natural Science Foundation of China (Grant no. 11021043) and the Fundamental Research Funds for the Central Universities (no. 3132014324). The third author is supported by the National Natural Science Foundation of China (Grant no. 11361020) and the Natural Science Foundation of Hainan Province (Grant no. 20151011).
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