Extensions of the Hardy-littlewood Inequalities for Schwarz Symmetrization

For a class of functions H : (0, ∞)×R 2 + → R, including discontinuous functions of Carathéo-dory type, we establish that R N H(|x|, u(x), v(x))dx ≤ R N H(|x|,u * (x), v * (x))dx, where u * (x) and v * (x) denote the Schwarz symmetrizations of nonnegative functions u and v.


Introduction.
We build on the approach to symmetrization, which we presented in [6], to establish the inequality where u * (x) and v * (x) denote the Schwarz symmetrizations of nonnegative functions u and v, for a class of functions H : (0, ∞) × R 2 + → R which includes discontinuous functions of Carathéodory type.Apart from its own role in the calculus of variations, this inequality generalizes simultaneously the inequalities Inequality (1.2) was first proved by Crowe et al. [2] for continuous functions F that satisfy a condition which we call (CZR) below.In the special case, F(s,t) = st, this condition is satisfied and we obtain the classical Hardy-Littlewood inequality.In this paper, we establish (1.2) and its generalization (1.1) for functions F and H which need not be continuous.Concerning extensions of (1.2), the paper [2] ends with the remark that "any proof involving approximations of u and v by step functions or of F by smooth functions is likely to require some additional hypothesis on F ."Note that our Corollary 4.7 and Theorem 5.4 are based on approximation of u or v by simple functions, yet they extend (1.2) to functions F that are not necessarily continuous without introducing any additional hypotheses.Moreover, as Remark 5.2 after Corollary 4.7 and Examples 5.7 and 5.8 show, Theorem 5.4 is optimal in the sense that if any of its hypotheses is not satisfied, then we can construct a triple (F,u,v) satisfying the remaining conditions for which (1.2) is false.
The first proof of the more general inequality (1.1) (for functions defined on a bounded subset of R N rather than all of R N ) seems to be due to Tahraoui [8,9] who requires H to be of class C 3 and who uses rather complicated approximations of H to obtain the result for nonnegative functions u and v in L p under appropriate growth conditions on H.More recently, Brock [1] and Draghici [3] have been able to establish (1.1) for continuous functions H which satisfy similar growth conditions without requiring any analog of the condition ∂ 1 ∂ 2 ∂ 3 H ≤ 0 that was needed by Tahraoui.Motivated by applications in the calculus of variations, our main goal here is to extend (1.1) to cases where H is not necessarily continuous but rather satisfies some conditions of Carathéodory type.Our method requires the assumption that we call (CZR-3) and which corresponds to ∂ 1 ∂ 2 ∂ 3 H ≤ 0 in the case when H is smooth.In this respect, we obtain a result that is less general than the one due to Brock but it has the advantage of dispensing with his assumption of continuity of H and, furthermore, it establishes (1.1) for a bigger class of functions u and v and it does not require any growth conditions like his on H either.
Our main results are Proposition 4.1, which establishes inequality (1.1) for all symmetrizable functions provided that H satisfies appropriate conditions, and Theorem 4.4 which establishes it for a smaller class of symmetrizable functions under weaker conditions on H.As an immediate consequence of Theorem 4.4, we obtain Corollary 4.7 which generalizes the result of [2] to functions F which need not be continuous on R 2 + .In Theorem 5.4, we adapt our approach in order to extend (1.2) to an even bigger class of functions F. Inequalities analogous to (1.1), (1.2), and (1.3) concerning functions defined on a bounded subset of R N can easily be deduced from the case we deal with here by the procedure we used in [6,Section 6].Finally we point out that by following closely the proofs of Proposition 4.1 and Theorem 4.4, we can easily find hypotheses under which the inequality is valid.Indeed, (1.4) is established in [5] using different techniques and some applications of this inequality can also be found there.

Notation.
All statements about measurability refer to the Lebesgue measure µ on R N or on [0, ∞), except in Section 5 where we discuss the composition of a Borel measurable function with a Lebesgue measurable function.For r ≥ 0, For a measurable subset A of R N with µ(A) < ∞, Note that A * is open even though A may not be.
The characteristic function of a set A is denoted by χ A .
Let M N denote the set of all extended real-valued functions which are measurable on R N .For u ∈ M N and t ∈ R, let be its distribution function and set the set of Schwarz symmetrizable functions.Following the terminology of [6], we say that an element u ∈ F N is Schwarz symmetric if there exists a nonincreasing function Simple functions can be symmetrized in a very explicit way.Let That is, E N is the set of all functions which can be written as where it follows that where C 0 = B(0,r 0 ) and C i = B(0,r i )\B(0,r i−1 ) for i = 1,...,k.
To deal with functions defined on subsets of R N , we use the following conventions.If ω is a measurable subset of R N , which has finite measure, let F N (ω) denote the set of all extended real-valued functions u such that (i) u is measurable on ω, The Schwarz symmetrization of an element u ∈ F N (ω) is defined as where u is the extension of u to all of R N defined in (2.8).By [6, Lemma 6.1(i)], (2.10) In [6], we make frequent use of the following identity which we refer to again below: where P i = p i − p i+1 for i = 0, 1,...,k− 1, P k = p k , and

Preliminaries.
In an integral where no domain of integration is indicated, the integration extends over all of R N .A measurable function f is said to be integrable provided that (3.1) in this case.
To deal with the case where [f + g] + (x)dx < ∞, we observe that and, similarly, Hence f and g are integrable in this case and the conclusion follows immediately.
The inequalities we deal with involve composite functions.In the calculus of variations, the following definition establishes the standard context for handling the measurability of such compositions.
An important property of such a function is that the composition x G(|x|, u(x)) is measurable on R N for every function u ∈ F N .In the context of inequality (1.1), we introduce the following extension of this notion.
The first part of the next definition gives the property introduced by Crowe et al. [2] in their fundamental paper concerning inequality (1.2).In dealing with (1.1) and (1.3) we require properties of a similar nature.

Definition 3.4. A function F : R 2
+ → R has the property (CZR) when Slight variants of some of our results in [6] concerning inequality (1.3) are useful for our treatment of (1.1) so we present them first.
Then the inequalities and so Proof.By Remark 3.8, we have that G(|x|, u(x))dx > −∞, and, if G(|x|, u * (x))dx = ∞, the conclusion is trivial.Thus we may assume henceforth that We consider the function This is a Carathéodory function which satisfies the hypotheses [6, Proposition 5.
by the monotonicity of G(r , a)− G(r , b) and assumption (iii).Thus we see that Φ satisfies all the conditions of [6, Proposition 5.1] and so for all u ∈ F N .But, by [6, Proposition 4.3(ii)], for all u ∈ F N such that g − (u) is integrable and, by hypothesis, ), it follows that p is integrable and q − (x)dx < ∞.Hence, by (3.17) and Lemma 3.1, from which it follows that q + (x)dx < ∞ and, consequently, that q is integrable.But now Lemma 3.1 enables us to conclude that since we know that G − (|x|, u(x))dx < ∞.This means that (3.16) and (3.17) can be written as and the conclusion follows.
A variant of (1.3) deals with functions defined on subsets of R N .
Theorem 3.9.Let G : (0, ∞) × R + → R + be a Carathéodory function and let ω be a measurable subset of R N with finite measure.Suppose that for all u ∈ F N (ω).
Proof.Consider u ∈ F N (ω).We may suppose that since otherwise the conclusion is trivial.Using (i) and (iii), it follows that Setting Φ(r , s) = G(r , s) − G(r , 0), we find that Φ satisfies the hypotheses of [6, Proposition 5.1] and so completing the proof.
A corollary to the next lemma provides a simple way of ensuring that Theorem 3.9(iii) is satisfied.
for all measurable subsets ω of R N with finite measure.

Proof. Suppose that (i) is true and set
Clearly, u ∈ F N and v ∈ E N and so by the Hardy-Littlewood inequality (see, e.g., [6, Proposition 2.3]), where Conversely, we suppose that h is not nonincreasing on (0, ∞).We complete the proof by constructing a subset ω with finite measure such that (3.33)By our assumption, there exist P,Q ∈ (0, ∞) with P < Q such that h(P ) < h(Q).Using the right-continuity of h, there exist ε > 0 and s, t as required.
Before giving the proof of this result, we make some comments about the hypotheses.
are equivalent to the following inequalities: (ii) ∂ 2 H(r , •, 0) ≥ 0 for all r > 0, (iii) Furthermore, using (ii) and (iii), we see that and, by (iv) and (v), that (vi) Using (i), we also have that Now using (vi), (iii), and (v), respectively, we find that and H ≥ 0 since we have already noted that ∂ 2 H ≥ 0.
Thus, for smooth functions satisfying (i), the hypotheses (ii)-(v) of Proposition 4.1 are equivalent to Remark 4.3.As we showed at the beginning of the proof, the hypotheses of Proposition 4.1 imply that H ≥ 0. Furthermore, the hypotheses of Proposition 4.1 also imply that H has the following properties: (ii) H(r , •,t) is nondecreasing on R + for all r > 0 and all t ≥ 0, (iv) H(•, •,t) has the property (CZR-2) for all t ≥ 0, and, in fact, all the other monotonicity properties analogous to the conditions (A), (B), and (C) that we have formulated for smooth functions.
As we have already observed, the hypotheses of Proposition 4.1 impose several monotonicity conditions on the function H.As we now show, these hypotheses can be relaxed, although in some cases it may be necessary to restrict the class of functions for which (1.1) holds to achieve this.In this way we obtain conditions on H which seem very natural for dealing with (1.1) in the calculus of variations.
Remark 4.6.We do not claim that the integrals in the conclusion of Theorem 4.4 are finite, but they are well defined in the following sense.By (ii), and hence that for all t ≥ 0 and r ∈ (0, ∞).Since H + (r , 0,t) = 0 whenever H − (r , 0,t) > 0, this implies that for all t ≥ 0 and r ∈ (0, ∞).Therefore But, using (iii), we have that for all s, t ≥ 0 and r ∈ (0, ∞).Hence and so for all s, t ≥ 0 and r ∈ (0, ∞) since H + (r ,s,t) = 0 whenever H − (r ,s,t) > 0. Thus )dx is defined unambiguously by and a similar interpretation applies to H(|x|,u * (x), v * (x))dx.
We close this section with the observation that Theorem 4.4 already contains a generalization of the result by Crowe et al. [2] concerning inequality (1.2), although further extensions will be obtained in the next section.
Remark 4.8.In [2, Theorem 3], inequality (1.2) is proved under similar hypotheses except that (a) is replaced by the stronger assumption that F ∈ C(R 2 + ).We point out that it is claimed in [2] that (1.2) holds for all u, v ∈ F N under these assumptions.However, as the following example shows, this clearly requires some qualification and it seems that the integrability of F(u(•), 0) and + ) and the hypotheses (b) and (c) of Corollary 4.7 are satisfied.Now consider the functions 2) may fail to hold.

Borel functions.
In this section, we establish inequality (1.2) for a class of functions F which are not even separately continuous.2) are satisfied, since in the case where (3) and ( 4) hold, we can replace F by F(s,t) = F(t,s) and recover the former situation.
Remark 5.3.Suppose that u, v ∈ F N and that F : R 2 + → R is an H-Borel function.Then F(u(•), v(•)) is Lebesgue measurable on R N .In fact, there is a sequence {u m } ⊂ E N such that lim m→∞ u m (x) = u(x) for all x ∈ R N and consequently lim m→∞ F(u m (x), v(x)) = F (u(x), v(x)).Hence it is enough to show that F(u m (•), v(•)) is measurable.But, in the notation (2.5), and so where Since F(a i , •) is Borel measurable and v is Lebesgue measurable, [7,Theorem 19.B] (with footnote (2) on page 82) shows that F(a i ,v(•)) is Lebesgue measurable on R N and our assertion follows easily from this.
Remark 5.5.As in Theorem 4.4, these integrals may not be finite.However, using (b) and (c), we find that, for all s, t ≥ 0, F(s,t)− F(s,0) − F(0,t) ≥ 0 (5.4) and hence that (5.7) whereas (5.9) The following example was inspired by a similar one given by Draghici [3] in the context of polarization inequalities.(5.10) Let A and B be measurable subsets on R N with (5.11) Using the notation of (2.5), we now define two simple functions u and v ∈ E N as follows: where where (5.15) Proof.We set F(s,t) = F(s,t) − F(0,t) − F(s,0) + F(0, 0) and begin by considering u ∈ E N and v ∈ F N .Clearly, F(0,t) ≡ 0 and it follows from (c) that F ≥ 0. Using the notation (2.5)-(2.7),(5.16)where we have used (2.11) with (5.17) and a k+1 = 0. Clearly G i : R + → R is a Borel measurable function with G i (0) = 0 and it follows from (CZR) that G i is nondecreasing on R + .Thus [G i ] − ≡ 0, so by [6, Theorem 6.2], 0 On the other hand, still using the notation (2.5)-(2.7)and then (2.11), By Remark 5.5 preceding the proof, we know that for all such functions u and v, )dx = ∞, the conclusion holds without further discussion.Hence we assume from now on that and by (5.23), (q − p)(x)dx ≤ 0. (5.28) Hence we see that q(x)dx ≤ p(x)dx < ∞ and so q is integrable.
Example 5.9.Let h : R + → R be a continuous function and let k : R + → R be a monotone function.Then F(s,t) = h(s)k(t) is an H-Borel function.If h and k are both nondecreasing on R + , it follows that F has the property (CZR).Furthermore, F(0, 0) = 0 provided that h(0)k(0) = 0.This gives a very simple type of function satisfying the hypotheses of Theorem 5.4 and these functions F are continuous on R 2 + only when k is continuous on R + or h ≡ 0. Note that examples of this kind do not have the continuity property with respect to rectangles that is discussed in the remarks about extensions of (1.2) in [2, Section 6], unless k is continuous on R + or h is constant.Since, if R is the rectangle in R

Variants and extensions.
First of all we observe that in Definitions 3.2, 3.3, and 5.1, the assumption of continuity can be replaced by left-continuity without changing the conclusions of our results since our method uses the approximation of a function in F N from below.
Our results concerning inequalities (1.1) and (1.2) are presented in the case of functions defined on all R N .However, it is easy to deduce analogous results for functions defined on subsets of R N by the procedure which we used in [6, Section 6], so we do not formulate such results here.Proposition 4.1 requires rather restrictive monotonicity properties of function H but yields inequality (1.1) for all u, v ∈ F N .Our proofs of Theorems 4.4 and 5.4 begin by introducing auxiliary functions Φ and F which have additional monotonicity properties not enjoyed by H and F. To obtain conclusions concerning H and F from those involving Φ and F , we impose some integrability assumptions on the functions u and v.A variant of this device is to assume that H (or F ) is monotone with respect to one of the variables and to modify its dependence on the other variable.This leads to a result requiring monotonicity of H (or F ) with respect to u and some additional assumption of integrability concerning v.Here is one example of what we mean.

Theorem 5 . 4 .
Let F : R 2 + → R be an H-Borel function which satisfies conditions (b) and (c) of Corollary 4.7.Then the inequalities

Theorem 6 . 1 .
Let F : R 2 + → R be an H-Borel function which satisfies conditions (b) and (c) of Corollary 4.7 and also (d) F(•,t) is nondecreasing on R + for all t ≥ 0.
and, since we have already observed that G i (r , •) is nondecreasing on R + , it follows that To extend the conclusion to all u ∈ F N , we recall that for any u ∈ F N , there is a sequence {u k } ⊂ E N such that u k ≤ u k+1 and u = lim k→∞ u k .By [6, Proposition 2.4], we have that u * * = B i with B i = B(0,r i ) in the notation(2.6).Recalling (2.8), (v| S i ) = vχ S i ≤ v on R N and so [ (v| S i )] * ≤ v * by [6, Proposition 2.4(iv)].Hence, by (2.10),(v| S i ) * ≤ v * on B i k ≤ u * k+1 and u * = lim k→∞ u * k .Since we have already shown that 0