A NOTE ON RIESZ FRACTIONAL INTEGRALS IN THE LIMITING CASE α ( x ) p ( x ) ≡ n

We show that the Riesz fractional integration operator I α ( · ) of variable order on a bounded open set in Ω ⊂ R n in the limiting Sobolev case is bounded from L p ( · ) (Ω) into BMO (Ω), if p ( x ) satisﬁes the standard log-condition and α ( x ) is H¨older continuous of an arbitrarily small order.


Introduction
We consider the Riesz fractional integral y ∈ Ω, of variable order, on an open set Ω ⊂ R n , in variable exponent spaces L p(·) (Ω). For the theory of these spaces we refer to the papers [4], [5], [11] and the book [3].
In [7] it was proved that the Sobolev type theorem on the boundedness of the operator I α(·) on bounded sets Ω, from L p(·) (Ω) to L q(·) (Ω), where is bounded in L p(·) (Ω). The boundedness of the latter for log-continuous exponents was later proved in [2]. For unbounded domains the variable exponent Sobolev theorem is known for constant α, see Theorem 6.1.9 in [3].
Meanwhile a question of interest in the variable exponent setting is to cover the case where In the case of constant exponents, it is known that the Riesz fractional integration operator acts in this case from L p to the space This goes back to a result in [13], where it was given in convolution terms; for weighted versions of such a result for Riesz potentials we refer to [6].
In this note we show that the L p(·) → BM O-boundedness holds also in the variable exponent setting when α(x)p(x) ≡ n, at the least in the case of bounded sets Ω.
Observe that the case of unbounded sets needs special treatment not only because of the known problems related to infinity well known in the variable exponent analysis, but also by a reason which goes back to the case of constant exponents. Even when α and p are constant, the operators Ω f (y)|y − x| α−n on an unbounded set Ω is not well defined by this direct definition (as an absolutely convergent integral) for all the functions on the whole space L p (Ω) when αp = n since (n − α)p = n in this case, although it may be treated as a continuous continuation from a dense set in L p , as an operator acting from L p to BM O. The operator may be also treated in this case via distributional interpretation. Since Schwartz test function space is not invariant with respect to the Riesz fractional integration, other space (known as the Lizorkin test function space, see Chapter 2 of [10]) is used. In [9] there was shown that the Riesz fractional operator I α of functions f ∈ L p (R n ) with α ≥ n p , interpreted in the distributional sense is a regular distribution. More precisely, any distribution I α f, 0 < α < ∞, generated by a function f ∈ L p , 1 ≤ p < ∞, is a regular distribution and even belongs to L loc p (R n ). Besides this, finite differences of the distribution I α f are well known to be better globally defined on L p (R n ), which in a sense is reflected in the BMO-language for the range I α (L p (R n )) when αp = n.

Preliminaries
Recall that the variable exponent Lebesgue space The local log-condition well known in the variable exponent analysis has the form where A > 0 does not depend on x and y. In case of a bounded set Ω the condition (2.2) may be also equivalently written in the form The condition in case of unbounded sets, is known as the decay condition. We always assume that

Theorem on L p(·) (Ω) → BM O-boundedness
Let |f (y)| dy be the variable order fractional maximal function. In the following lemma a set Ω is admitted to be unbounded, but the main statement in Theorem 3.1 relates only to bounded sets. In case Ω is unbounded we also suppose that (2.4) holds and p − = p(∞).
when Ω is bounded, and when Ω is unbounded.

P r o o f. By the Hölder inequality for variable exponents, we have
It is known ( [8], Theorem 2.21; see also [12]) that when Ω is bounded, and when Ω is unbounded, where see Corollary 4.5.9 in [3]. Therefore, when Ω is bounded, and when Ω is unbounded. Hence the statement of the lemma follows.

Estimation of F 1 (y):
When y ∈ B(x, r), we have |x − y| < 3r for z ∈ B(x, 2r) so that for y ∈ B(x, r), by the well known inequality. Then 1 |B(x, r)| B(x,r) We apply Hölder inequality and obtain 1 |B(x, r)| B(x,r) whence by (3.1) and the boundedness of the maximal operator in L p(·) (Ω), see for instance [1], we get 1 Estimation of F 2 (y): We denote and have To estimate G 1 , we use the inequality (3.3) and observe that |y − x| < r and |z − y| > 2r imply |x − z| < 3 2 |y − z|, so that

By the inequality
valid for 0 < α(x) < n, we get Therefore,

4) by Lemma 3.1.
For G 2 we use the inequality Since |y − z| ≥ 2 3 |x − z| and Ω\B(x, 2r) ⊆ Ω\B(y, r), we obtain Since y runs the ball B(x, r) centered at x, it suffices to deal only with the term H(y).

Which provides the boundedness of |α(x) − α(y)|[H(y) + H(x)]
provided α(x) has the corresponding Hölder property. Since δ may be chosen arbitrarily small, it is sufficient to suppose that α is Hölderian of an arbitrarily small order.