A Weak-Type Expression of the Orlicz Modular

An equivalent expression of Orlicz modulars in terms of measure of level sets of difference quotients is established. The result in a sense complements the famous Maz’ya–Shaposhnikova formula for the fractional Gagliardo–Slobodeckij seminorm and its recent extension to the setting of Orlicz functions.


Introduction
The fractional order Sobolev spaces W s,p (R N ), p ∈ [1, ∞), s ∈ (0, 1), endowed with the Gagliardo-Slobodeckij seminorm, which is defined for smooth compactly supported functions u as dy, have played an important role in the theory of partial differential equations and its applications for a long time (see the introductory section of [5]).Much as it is tempting to think that or the Gagliardo-Slobodeckij seminorm notoriously fails to capture these limiting cases-to that end, it is sufficient to consider any nonconstant u ∈ C ∞ 0 (R N ) and observe that |u| p s,p converges to ∞ as s → 1 − or s → 0 + .Nevertheless, it was discovered around 20 years ago that these "defects" can be, in a sense, "fixed" by introducing certain compensatory factors.Namely, for every u ∈ C ∞ 0 (R N ), a special case of what is now often called the Bourgain-Brezis-Mironescu formula [3] Moreover, V.G.Maz'ya and T. Shaposhnikova proved in [8] that Recently, a completely different approach, not involving integration of fractional difference quotients at all, to repairing (1) was taken by H. Brezis, J. Van Schaftingen and P.-L.Yung.They proved in [4] that, instead of introducing a compensatory factor, the limit as s → 1 − can be recovered if the strong L p norm of fractional difference quotients is replaced by the weak L p,∞ quasi-norm.More precisely, they obtained the following result.Let p ∈ [1, ∞) and u ∈ C ∞ 0 (R N ) and define where | • | 2N stands for the Lebesgue measure on R 2N .In [4] it was shown that c(N, p) Following this innovatory approach, Q. Gu and P.-L.Yung established in [7] other, possibly even more unanticipated, formulae.They complement the Maz'ya-Shaposhnikova formula (3) in the same way the result of Brezis, Van Schaftingen and Yung complements the Bourgain-Brezis-Mironescu formula (2).Namely the result of [7] asserts that, for every and where The classical results ( 2) and (3) were recently considerably strengthened in [1,2,6] by replacing the p-th power in the integrals with Orlicz functions, thus allowing for nonpolynomial growth.
The aim of this short paper is to similarly extend the new developments of [7]; i.e. ( 4) and ( 5), by replacing the p-th power with a general Orlicz function globally satisfying the ∆ 2 condition.Therefore, we express the Orlicz modular in terms of measures of certain level sets, without using any integral.Our proof technique is based on the argument presented in [7], appropriately extended to the Orlicz framework.
In what follows, we introduce some basic notations and definitions, needed for understanding the setting we will be considering in our main result.A proper detailed treatment of Orlicz functions and classes may be found e.g. in [9].
is any continuous convex function vanishing at 0. Note that Young functions are nondecreasing.We say that a Young function Φ (globally) satisfies the ∆ 2 condition if there is k > 0 such that Φ(2t) ≤ kΦ(t), for every t > 0. Then necessarily k ≥ 2, which follows from the convexity of Φ.We denote by ∆ 2 (Φ) the infimum over all such k.
Given a Young function Φ, we say that a measurable function u : R N → R belongs to the Orlicz class L Φ , and write u ∈ L Φ , if for every γ > 0.
As usual, ω N denotes the volume of the unit ball in R N .

Main result
Theorem 2.1.Let Φ be a Young function satisfying the ∆ 2 condition.Let u ∈ L Φ and for every t > 0 define Furthermore, Proof.For every t > 0 define the set and observe that, thanks to symmetry, it satisfies At first, we are going to suppose that u has compact support; i.e., there exists R > 0 such that supp u ⊂ B R .
Notice that, if (x, y) ∈ H t , then necessarily x ∈ B R , otherwise we would have x, y ∈ R N \B R and thus u(x) = u(y) = 0, which would imply (x, y) / ∈ H t .For a fixed x ∈ B R define the sets and The first inclusion together with the definition of H t,x,R implies while the second inclusion in (8) implies Since x ∈ B R was arbitrarily chosen, we may integrate ( 9) and (10) over B R with respect to x and multiply by Φ(t) to get Recalling that u is supported in B R and , we may further rewrite this as Letting t → 0 + , we obtain (6).Now we are going to extend the result beyond compactly supported functions.Suppose that u : R N → R is measurable.For any fixed t > 0, the set E t satisfies Indeed, if (x, y) ∈ R N is not contained in either of the two sets on the right-hand side, then, by monotonicity and convexity of Φ, hence (x, y) / ∈ E t .This shows (12).Using the symmetry of the two sets on the right-hand side of (12), we obtain Notice that neither the assumption nor the ∆ 2 condition of Φ has been used yet, so this estimate in fact holds for any measurable u and any Young function Φ.If Φ satisfies the ∆ 2 condition, (13) readily implies the second inequality in (7).Assume that u ∈ L Φ .Choose R > 0 and define Furthermore, choose t > 0, λ ∈ (0, 1) and define and .
Similarly as before, this can be seen by using monotonicity and convexity of Φ to get from which the inclusion follows.
Observe that the set A 1 is obtained by replacing u with u R λ in the definition of E t .Since u R λ is compactly supported and belongs to L Φ (since Φ satisfies the ∆ 2 condition), we may use the previously obtained estimate (11) with A 1 and u R λ in place of E t and u, respectively, to get Analogously, applying (13) to the function v R 1−λ in place of u (A 2 plays the role of E t for this function ), we get and Φ satisfies the ∆ 2 condition, both integrals above are finite regardless the choice of λ, and the second integral (with a fixed λ) vanishes as R → ∞ by the dominated convergence theorem.Hence, consecutively letting t → 0 + , R → ∞ and λ → 1 − , we finally obtain lim sup It remains to show the opposite inequality for the lower limit.Fix R > 0, λ ∈ (0, 1) and define u R , v R as in (14).Then |u| −|v R | = |u R | and, by convexity of Φ, for any (x, y) ∈ R 2N we have For any t > 0 define and where the last inequality follows from (16).This shows that (x, y) ∈ E t .Thus, we have We proceed analogously as before, realizing that A 3 plays the role of E t for the compactly supported function λu R , we use (11) to obtain Similarly, an appropriate interpretation of (13) yields Hence, Letting t → 0 + , R → ∞ and λ → 1 − , in this order, now yields Once again, the ∆ 2 condition of Φ as well as the assumption u ∈ L Φ are both required in this step.This clearly implies the first inequality in (7).Finally, combining (17) with (15), we arrive at (6), and so the proof is complete.