On the Multilinear Fractional Integral Operators with Correlation Kernels
aa r X i v : . [ m a t h . C A ] S e p On the Multilinear Fractional Integral Operatorswith Correlation Kernels
Zuoshunhua Shi ∗ , Di Wu † and Dunyan Yan ‡ Abstract
In this paper, we study a class of multilinear fractional integral operators associated withcorrelation kernels Q ≤ i Fractional integral operators arise frequently in various subjects such as Fourier analysis andpartial differential equations. The Riesz potentials are classical fractional integral operatorswhich were generalized to multilinear variants by many authors; see [9], [2], [11], [16], [13], [26],[3], [27] and [4]. In this paper, we mainly study mapping properties of the multilinear fractionalintegral operators with correlation kernels of the form Q | x i − x j | − α ij . These operators can bewritten as T ( f , · · · , f k )( x k +1 ) = Z R nk k Q i =1 f i ( x i ) Q ≤ i 0. It is clear that T reduces to Riesz potentials when k = 1. It isnatural to assume that the kernel of T is a Schwartz kernel such that T maps ( C ∞ ( R n )) k into D ′ ( R n ) continuously. This requires some restrictions on the exponents α ij . Another issue is todetermine the necessary and sufficient condition under which T is bounded from L p × · · · × L p k into L q . More precisely, we shall establish the following inequality k T ( f , f , · · · , f k ) k L q ≤ C k Y i =1 k f i k L pi (1.2) ∗ 1. School of Mathematics and Statistics, Central South University, Changsha Hunan 410083, P.R. China.2. School of Mathematical Science, Graduate University of the Chinese Academy of Sciences, Beijing 100049, P.R.China. E-mail address: [email protected] . † School of Mathematical Science, Graduate University of the Chinese Academy of Sciences, Beijing 100190, P.R. China. E-mail address: [email protected] . ‡ School of Mathematical Science, Graduate University of the Chinese Academy of Sciences, Beijing 100190, P.R. China. E-mail address: [email protected] . C independent of f i ∈ L p i . The Hardy-Littlewood-Sobolev inequality is aspecial case of this inequality for k = 1. For k ≥ 2, it is more convenient to consider thefollowing multilinear functional:Λ( f , f , · · · , f k +1 ) = Z R n ( k +1) k +1 Q i =1 f i ( x i ) Q ≤ i Let Λ be the multilinear functional defined by (1.3) with all α ij ≥ . Assume < p i < ∞ for ≤ i ≤ k + 1 . Then there exists a constant C such that (1.4) is true if and onlyif the following three conditions hold simultaneously. ( i ) k +1 X i =1 p i + X ≤ i Let T be the multilinear operator as in (1.1) with all α ij ≥ . Assume that < p i < ∞ for ≤ i ≤ k and p k +1 ∈ { , ∞} . Suppose { α ij } and { p i } satisfy the conditions ( i ) and ( ii ) in Theorem and ( a ) of ( iii ) except p k +1 = 1 and I = { k + 1 } , i.e., X I p i + X I α ij n < | I | for any nonempty proper subset I of { , , · · · , k + 1 } unless p k +1 = 1 and I = { k + 1 } . Thenthere exists a constant C such that for all f i ∈ C ∞ k T ( f , f , · · · , f k ) k L ≤ C k Y i =1 k f i k p i if p k +1 = ∞ , k T ( f , f , · · · , f k ) k BMO ≤ C k Y i =1 k f i k p i if p k +1 = 1 . Moreover, if in addition { p i } ki =1 satisfies P ki =1 /p i ≥ when p k +1 = 1 , then the space BM O can be replaced by L ∞ . { α ij } are defined to be symmetric. In other words, weassume α ii = 0 and α ij = α ji for all 1 ≤ i, j ≤ k + 1. For any given subset J of { , , · · · , k + 1 } ,we use the summation conventions P J /p i and P J α ij to denote P i ∈ J /p i and P i 0. We use A ∧ B to denote min { A, B } and | J | to denote the cardinality of an index set J .For a measurable set E in R n , | E | is its Lebesgue measure. For two sets E and F , E − F means E ∩ F c where F c is the complement of F . For brevity, we use S = { , , · · · , k } throughout thepaper.The present paper is organized as follows. The section 2 contains some previously knownresults which will be used in subsequent sections. The necessity part of Theorem 1.1 will beproved in § 3. We shall prove the sufficiency of Theorem 1.1 in § 4. Local integrability of thecorrelation kernel Q | x i − x j | − α ij will be discussed in § 5. In § 6, we shall give the proof ofTheorem 1.2. In the appendix, the finiteness of a class of Selberg integrals on the sphere will beproved by invoking the methods in § In this section, we shall present some results to be used in subsequent sections.For k + 1 points x , x , · · · , x k +1 in R k , we say that these points are affinely independent ifthey do not lie in a hyperplane in R k simultaneously. Theorem 2.1 Assume that T is a k − linear operator which is bounded from L p j , × · · · × L p kj , into L q j , ∞ for < p ij ≤ ∞ and < q j ≤ ∞ with ≤ i ≤ k and ≤ j ≤ k + 1 . Assume alsothat these k + 1 points (1 /p j , · · · , /p kj ) are affinely independent in R k . If there are k + 1 realnumbers λ i with positive λ , · · · , λ k such that q j = k X i =1 λ i p ij + λ k +1 , ≤ j ≤ k + 1 , then T is bounded from L p ,t × · · · × L p k ,t k into L q,t with (1 /p , · · · , /p k , /q ) lying in the openconvex hull of the k + 1 points (1 /p j , · · · , /p kj , /q j ) in R k +1 and < t i , t ≤ ∞ satisfying k X i =1 t i ≥ t . This theorem was previously known; see Janson [14]. We also refer the reader to a similar variantcalled the multilinear Marcinkiewicz interpolation in Grafakos-Kalton [13].The L estimate in Theorem 1.2 implies that Λ( f , · · · , f k +1 ) is bounded by a constant mul-tiple of Q k +1 i =1 k f i k p i with p k +1 = ∞ . Let f k +1 ≡ 1. We see that the integral of (1.1) with respectto x k +1 is a generalization of the beta integral with k = 2. An induction argument requires thatupper bounds of the integral are of the form Q S | x i − x j | − β ij with suitable parameters { β ij } . Inother words, we need the following type estimates: Z R n k Y i =1 (cid:12)(cid:12) x i − x k +1 (cid:12)(cid:12) − α i,k +1 dx k +1 ≤ C Y S | x i − x j | − β ij . (2.6)4he following theorem gives us the desired estimates. Theorem 2.2 Assume α , α , · · · , α k satisfy < α i < n . If P ki =1 α i > n , then the followingestimate Z R n k Y i =1 (cid:12)(cid:12) t − x i (cid:12)(cid:12) − α i dt ≤ C k X u =1 L u ( x , x , · · · , x k ) holds for arbitrary x , x , · · · , x k in R n , where each L u is defined by L u ( x , x , · · · , x k ) = ( d n − P S α i S (cid:16) χ { P S −{ u } α i There are some explicit formulas concerning the integral in the above theorem.These formulas take the form Z R n k Y i =1 (cid:12)(cid:12) t − x i (cid:12)(cid:12) − α i dt = C Y ≤ i Let α , α , · · · , α k be positive numbers satisfying P ki =1 α k = n with k ≥ . For k points x , x , · · · , x k in the unit ball B (0) ⊆ R n , it is true that Z | t |≤ k Y i =1 | t − x i | − α i dt ≤ C log Cd S , (2.7) where C depends on α , · · · , α k and the dimension n . Moreover, the reverse inequality is alsotrue for another constant C depending on n and α , · · · , α k .Proof. Without loss of generality, we may assume | x − x k | = max S | x i − x j | > 0. Recall that d S = P i 5n the case d S ≥ / 3, it follows that (2.7) and its reverse are true. Now assume d S < / 3. It iseasy to see that Z d S ≤| t |≤ k Y i =1 | t − ( x i − x ) | − α i dt ≈ Z d S ≤| t |≤ | t | − n dt = C log 1 d S . Thus (2.7) is also true if d S < / 3. Now we shall prove its reverse form in the case d S < / Z | t |≤ k Y i =1 | t − ( x i − x ) | − α i dt ≥ Z d S ≤| t |≤ k Y i =1 | t − ( x i − x ) | − α i dt ≈ Z d S ≤| t |≤ | t | − n dt ≥ C log 1 d S . Thus we complete the proof of the lemma. ✷ We now turn to the proof of Theorem 2.2. Proof. We assume L = max S | x i − x j | = | x − x k | > 0. Then L ≈ d S . We shall estimate theintegral over B L/ ( x ) and its complement separately. Observe that Z B L/ ( x ) k Y i =1 (cid:12)(cid:12)(cid:12) t − x i (cid:12)(cid:12)(cid:12) − α i dt ≤ Cd − α k S Z B L/ (0) k − Y i =1 (cid:12)(cid:12)(cid:12) t − x i + x (cid:12)(cid:12)(cid:12) − α i dt. In the case P k − i =1 α i ≤ n , it follows from Lemma 2.3 that the integral of Q k − i =1 | t − x i + x | − α i over B L/ (0) is bounded by a constant multiple of d n − P k − i =1 α i S (cid:16) χ { P S −{ k } α i 6e shall see that the proof of Theorem 1.1 and Theorem 1.2 is closely related to existenceof solutions to systems of linear inequalities. A system of linear inequalities in R n is given by(II . ( f i ( x ) = ( v i , x ) < a i , ≤ i ≤ mf i ( x ) = ( v i , x ) ≤ a i , m + 1 ≤ i ≤ k where v i ∈ R n , a i ∈ R and ( · , · ) denotes the standard inner product in R n . It is worthwhilenoting that we may incorporate an linear equality into a system of linear inequalities. Indeed, wemay write g ( x ) = ( v, x ) = a as an equivalent system of two linear inequalities given by g ( x ) ≤ a and − g ( x ) ≤ − a . Lemma 2.4 Suppose that the system f i ( x ) = ( v i , x ) ≤ a i for m + 1 ≤ i ≤ k has at least onesolution. Then there exists a solution x ∈ R n to the system (II.1) if and only if k X i =1 λ i a i > for all nonnegative numbers λ i satisfying P ki =1 λ i f i = 0 with at least one λ i > for ≤ i ≤ m . This lemma is a special case of the existence theorem of systems of convex inequalities in R n .However, it can be proved by a simple method using the concept of elementary vectors of ansubspace of R n . We refer the reader to § 22 (Page 198) in Rockafellar [21]; see also [10] for itsextension to general vector spaces. In this section, we shall prove the necessity of conditions ( i ), ( ii ) and ( iii ) in Theorem 1.1.Indeed the second author obtained these necessary conditions in his thesis [27]. We present herethe details of proof for convenience of the reader.Assume the inequality (1.4) is true for some constant C independent of f i . We replace f i byits dilation δ λ ( f i )( x ) = f i ( λx ) for λ > 0. By a change of variables, we see that ( i ) must hold byletting λ → λ → ∞ .To show the necessity of (ii), we take all f i = χ B (0) . We shall replace { , , · · · , k + 1 } by S = { , , · · · , k } . We claim that if there were a subset J ⊂ S with | J | ≥ P J α ij ≥ ( | J | − n, we would obtain R ( B (0)) | J | Q J | x i − x j | − α ij dV J = ∞ , where B (0) is the unitball centered at the origin in R n and dV J is the product Lebesgue measure Q J dx i . Since α ij are nonnegative, the argument is essentially the same for different J ′ s . Assume J = S . Then R ( B (0)) k Q S | x i − x j | − α ij dV S is equal to Z B (0) Z ( B ( x )) k − k Y i =2 | x i | − α i Y ≤ i 0. It is clear that B / (0 n ( k − ) is contained in ( B / (0 n )) k − . We may regard7he integrand Q ki =2 | x i | − α i Q ≤ i 1. Assume P J c /p i < 1. We can choose0 < λ i < p i such that P J c /λ i < 1. Let f i ( y ) = | y | − n/p i χ {| y | > } (log | y | ) − /λ i for each i ∈ J c and f i = χ B (0) for i ∈ J . Substituting these functions into the functional Λ, we haveΛ( f , · · · , f k +1 ) ≥ C Z ( B c (0)) | Jc | Y J c | f i ( x i ) | Y J c | x i | − β i Y J c | x i − x j | − α ij dV J c (3.9)= C Z ( B c (0)) | Jc | Y J c | x i | − n/p i (log | x i | ) − /λ i Y J c | x i | − β i Y J c | x i − x j | − α ij dV J c , where β i = P u ∈ J α iu for each i ∈ J c .Since the conditions (i), (ii) and (iii) in Theorem 1.1 are invariant under the permutationgroup on k + 1 letters, we may assume J c = { , , · · · , l } with 1 ≤ l ≤ k . If l = 1, it followsimmediately from the fact n/p + β = n that the right side integral in (3.9) is infinite since1 /λ < . Now we treat the case l > 1. Replacing the region ( B c (0)) l by its proper subset Ω l consisting of all points ( x , · · · , x l ) such that x ∈ B c (0) and | x i | ≥ | x i − | for 2 ≤ i ≤ l , weobtain Z | x l |≥ | x l − | | x l | − n/p l (log | x l | ) − /λ l | x l | − β l | x l | − P l − i =1 α il dx l ≥ C | x l − | − ξ l (log | x l − | ) − /λ l , where β l = P k +1 i = l +1 α il and ξ l = n/p l + P k +1 i =1 α il − n . Substituting this estimate into the integralin (3.9), we see that Λ( f , · · · , f k +1 ) is not less than a constant multiple of Z Ω l − l − Y i =1 | x i | − n/p i − δ ( l − i ξ l (log | x i | ) − /λ i − δ ( l − i /λ l l − Y i =1 | x i | − β i Y ≤ i Assume ≤ p i ≤ ∞ and α ij ≥ for ≤ i, j ≤ k + 1 . Suppose the multilinearfunctional Λ given by (1 . satisfies | Λ( f , f , · · · , f k +1 ) | ≤ C k +1 Y i =1 k f i k p i for some C independent of f i . If J is a nonempty proper subset of { , , · · · , k + 1 } satisfying X J p i + X J α ij n = | J | , then we have Z ( R n ) | J | Y J | f i ( x i ) | Y J | x i − x j | − α ij dV J ≤ C Y J k f i k p i (3.10) and Z ( R n ) | Jc | Y J c | f i ( x i ) | Y J c | x i − x j | − α ij Y J c | x i | − β i dV J c ≤ C Y J c k f i k p i (3.11) with β i = P u ∈ J α iu for each i ∈ J c and both constants C independent of f i . Moreover, it isalso true that X J (cid:18) p i + β i n (cid:19) + X J α ij n ≤ | J | (3.12) for all nonempty subsets J of J c . Conversely, (3.10) and (3.11) imply the boundedness of Λ .Proof. For each ǫ > 0, let f i = ǫ − n/p i χ {| y | <ǫ/ } for each i ∈ J . By P J /p i + P J α ij /n = | J | ,we have Z ( R n ) | J | Y J | f i ( x i ) | Y J | x i − x j | − α ij dV J ≥ C, where C is a constant depending only on n and | J | but not on ǫ . For given nonnegative f i ∈ L p i with i ∈ J c , it follows from the boundedness of Λ that Z ( B cǫ (0)) | Jc | Y J c | f i ( x i ) | Y J c | x i − x j | − α ij Y J c | x i | − β i dV J c ≤ C Y J c k f i k p i ǫ → 0. The first inequality (3.10) can be obtainedby a similar argument. Indeed, put f i = ǫ − n/p i χ { ǫ< | y | < ǫ } for each i ∈ J c . For nonnegativefunctions f i with i ∈ J , we also have Z ( R n ) | Jc | Y J c | f i ( x i ) | Y J c | x i − x j | − α ij Y i ∈ J c Y j ∈ J | x i − x j | − α ij dV J c ≥ C for | x j | < ǫ/ j ∈ J , where the constant C is independent of ǫ and x j ∈ B ǫ/ (0) with j ∈ J . Similarly, we then obtain Z ( B ǫ/ (0)) | J | Y J | f i ( x i ) | Y J | x i − x j | − α ij dV J ≤ C Y J k f i k p i where f i ∈ L p i with i ∈ J and the constant C is independent of f i and ǫ . By letting ǫ → ∞ , thedesired inequality follows. The inequalities in (3.12) can be proved similarly as (3.8) by invoking(3.11). We omit the details here. If (3.10) and (3.11) are true, we first consider the integral in(1.3) with respect to dV J c = Q i ∈ J c dx i . Though the integrand depends on x i with i ∈ J , it isbounded by a constant multiple of Q J c k f i k p i . Hence the boundedness of Λ follows. ✷ Combining above results, the proof of the necessity part of Theorem 1.1 is complete. In this section, we shall prove the sufficiency part of Theorem 1.1. The argument is similar toChrist’s proof of the Brascamp-Lieb inequality for Lorentz spaces; see Perry [20]. Our maintool is the powerful Brascamp-Lieb inequality. For the rank-one case, Barthe [1] applied Lieb’stheorem [18] and the Cauchy-Binet formula to obtain the necessary and sufficient condition forwhich (1.5) holds. Bennett-Carbery-Christ-Tao [5, 6] proved that the general Brascamp-Liebinequality (1.5) is true for some C < ∞ if and only ifdim H = m X i =1 p i dim H i (4.13)and for all subspaces V of H dim V ≤ m X i =1 p i dim( B i V ) . (4.14)It is clear that (4.14) consists of finitely many inequalities. For the rank-one case, Barthe[1] characterized the extreme points of { /p i } for which the Brascamp-Lieb inequality holds.Valdimarsson [25] considered the corank-one and certain mixed rank cases, and constructed aprocedure to find the full list of dimension inequalities in (4.14).Consider the following multilinear functionalΨ( { f i } Ni =1 ; { g ij } ≤ i 2. It is easy to see thatdim span { B i : B i ∈ X } = | J | , dim span { B jk : j, k ∈ J i } = | J i | − , ≤ i ≤ s. Then the decomposition gives us X = span { B i : B i ∈ X } M " s M i =2 span { B jk : j, k ∈ J i } . Hence dim X = | J | + s X i =2 ( | J i | − . Now we can derive (4.18) from (a) and (b). Recall that (4.18) is equivalent to (4.19). Wesee that (a) and (b) imply X B i ∈ X p i + X B jk ∈ X p jk = X J p i + X J p jk + s X i =2 X J i p jk ≤ | J | + s X i =2 ( | J i | − X. . 16) is true if and only if the following threeconditions hold:( α ) The scaling condition is true: N X i =1 p i + X ≤ j 2. In fact, assume Theorem 1.1 is true for dimension one. For general n ≥ x i = ( x (1) i , · · · , x ( n ) i ) and suppose that { p i } and { α ij } satisfy all conditions in Theorem1.1. Then for x i , x j ∈ R n and α ij ≥ 0, it is easy to see that | x i − x j | − α ij ≤ n Y t =1 | x ( t ) i − x ( t ) j | − α ij /n . Hence we have Z R n ( k +1) k +1 Y i =1 | f i ( x i ) | Y ≤ i 2. For this reason, it suffices to show the theorem for dimension one.Now we assume n = 1. For convenience, we first prove Theorem 1.1 in the case when thedata { p i } and { α ij } satisfy ( i ), ( ii ) and ( a ) of ( iii ). This corresponds to simple Brascamp-Liebdata in Bennett-Carbery-Christ-Tao [5]. Define p ij = 1 /α ij . Then the data { p i } and { p ij } satisfy the above conditions ( α ), ( β ) and ( γ ), where inequalities in ( β ) and ( γ ) are strict. Bythis fact and the assumption 1 < p i < ∞ , we can choose k + 1 affinely independent points(1 /p ( i )1 , · · · , /p ( i ) k , /p ( i ) k +1 ) near (1 /p , · · · , /p k , /p k +1 ) from the hyperplane x + · · · + x k + x k +1 = k + 1 − X { , ··· ,k +1 } α ij /n /p ( i )1 , · · · , /p ( i ) k ) and { p ij } still satisfy ( α ), ( β ) and ( γ ). Also, we may as-sume that (1 /p , /p , · · · , /p k +1 ) lies in the open convex hull of points (1 /p ( i )1 , /p ( i )2 , · · · , /p ( i ) k +1 ).Since Ψ satisfies (4.16), T is bounded from L p ( i )1 × · · · L p ( i ) k into L q ( i ) k +1 with q ( i ) k +1 being the con-jugate number of p ( i ) k +1 . Observe that P ki =1 /p i > /p ′ k +1 = 1 − /p k +1 by conditions ( i ) and( ii ) in Theorem 1.1. Therefore we may apply Theorem 2.1 to conclude that T is bounded from L p × · · · L p k into L p ′ k +1 . By duality, we see that Λ is bounded on L p × · · · × L p k +1 .We now prove the remaining case of Theorem 1.1. For the same reason as above, we stillassume n = 1. Suppose that there are proper subsets J of { , , · · · , k + 1 } with | J | ≥ P J /p i + P J α ij = | J | . This implies existence of critical subspaces for Brascamp-Liebdata; see Bennett-Carbery-Christ-Tao [5, 6]. It is worth noting that the fact | J | ≥ < p i < ∞ . Let k + 1 − m be the maximum of | J | over all these propersubsets J for some 1 ≤ m ≤ k − 1. We take a J such that | J | attains the maximum k + 1 − m and P J /p i + P J α ij = | J | . By Theorem 3.1, we can reduce matters to two inequalities. Bythe condition ( iii ), first observe that J c contains at least two elements since P J c /p i ≥ 1. Thechoice of J implies X J p i + X j ∈ J α ij + X J α ij < | J | (4.20)for all nonempty proper subsets J of J c . Indeed, if it were true that X J p i + X j ∈ J α ij + X J α ij = | J | for some nonempty J $ J c , we would obtain X J ∪ J p i + X J ∪ J α ij = | J ∪ J | which contradicts our choice of J since | J ∪ J | > | J | .Now we claim that the inequality (3.11) is true for J . By symmetry, we may assume J = { m + 1 , · · · , k + 1 } for some 2 ≤ m ≤ k − 1. It is clear that J c = { , , · · · , m } . By H¨older’sinequality for Lorentz spaces, we have k f i | · | − β i k L qi ≤ C k f i | · | − β i k L qi, ≤ C k f i k L pi, q i = 1 p i + β i for each 1 ≤ i ≤ m . The datum { q i , α ij : 1 ≤ i < j ≤ m } is simple, i.e., { q i } and { α ij : 1 ≤ i Assume α ij ≥ for ≤ i < j ≤ k + 1 and < p i < ∞ for ≤ i ≤ k + 1 satisfythe conditions ( i ) , ( ii ) and ( a ) of ( iii ) in Theorem . Then there exists a constant C such that Z R nk Y ≤ i Assume α ij ≥ for ≤ i < j ≤ k + 1 satisfy the integrability condition X J α ij < ( | J | − n (5.21) for any subset J of { , , · · · , k + 1 } with | J | ≥ . Then we have I k +1 ( { α ij } ) = Z ( B (0)) k +1 Y { , , ··· ,k +1 } | x i − x j | − α ij dx dx · · · dx k +1 < ∞ , (5.22) where B (0) ⊂ R n is the unit ball centered at the origin.Proof. In the case k = 1, it is clear that the above integral converges absolutely. For k ≥ k = 2 and then make induction for general k . For k = 2, it isconvenient to divide the proof into three cases.Case 1. α + α < n It is clear that Z ( B (0)) Y S | x i − x j | − α ij dx dx dx ≤ C Z ( B (0)) | x − x | − α dx dx which is finite by the assumption α < n .Case 2. α + α = n Using Lemma 2.3, we obtain Z ( B (0)) Y S | x i − x j | − α ij dx dx dx ≤ C Z ( B (0)) | x − x | − α log(4 / | x − x | ) dx dx < ∞ . Case 3. α + α > n Observe that Z B (0) | x − x | − α | x − x | − α dx ≤ C | x − x | n − α − α which implies the integral in (5.22) is finite by the assumption α + α + α < n .We now consider the general case k ≥ 3. Assume all k fold integrals of form (5 . 22) convergeunder the assumption (5.21). We shall prove that the k + 1 fold integral is absolutely convergent.Let Θ = { i : 1 ≤ i ≤ k, α i,k +1 > } . 15y simple calculations, it is easy to verify our claim in the case P Θ α i,k +1 ≤ n . Indeed, if P Θ α i,k +1 is less than n , we first take integration with respect to x k +1 and then the matterreduces to a k − multiple integral. If P Θ α i,k +1 = n , it follows from Lemma 2.3 that Z B (0) Y Θ | x i − x k +1 | − α i,k +1 dx k +1 ≤ C X Θ | x i − x j | ! − ε where ε > i , j ∈ Θ with i < j . Let α ij = α ij + δ i i δ j j ǫ, ≤ i < j ≤ k, where δ ts is the Kronecker symbol. In other words, δ ts = 1 if s = t and δ ts = 0 otherwise. If ǫ is sufficiently small, then { α ij } still satisfies the integrability condition. Therefore the k + 1-multiple integral in (5.22) is less than a constant multiple of a k -multiple integral with { α ij } replaced by { α ij } . The desired result follows by induction.The crux of the proof lies in the case P Θ α i,k +1 > n . The argument depends on the numberof elements in Θ. The simplest case is | Θ | = 2 in which the argument is direct. For 3 ≤ | Θ | ≤ k ,we shall reduce the matter to the case | Θ | = 2 by invoking a useful procedure. Indeed, if | Θ | = 2,we may assume Θ = { , } by the symmetry of parameters. Then by a similar treatment inCase 3 for k = 2, put α ij = α ij + δ i δ j X Θ α i,k +1 − n ! , ≤ i < j ≤ k. (5.23)It is easy to verify that (5.21) are still true for { α ij } . Indeed, it suffices to show that (5.21)holds for those J containing both 1 and 2 since we obviously have X J α ij = X J α ij for all subsets J of S = { , , · · · , k } satisfying { , } * J . For J ⊆ S containing Θ = { , } asa subset, it follows from the definition of { α ij } that X J α ij = X J ∪{ k +1 } α ij − n < ( | J | − n by the assumption (5.21). Hence by induction the integral in (5.22) converges in the case | Θ | = 2.If | Θ | = m with 3 ≤ m ≤ k , our idea is to show that the k + 1 − multiple integral isdominated by summation of two similar kinds of integrals by distributing some powers into { α ij : 1 ≤ i < j ≤ k } appropriately. The first type of these integrals has a k − point correlationintegrand. The other type is the same as the integral in (5.22) with | Θ | = m − 1. Withoutloss of generality, we may assume Θ = { , , · · · , m } . By the assumption P Θ α i,k +1 > n , we useTheorem 2.2 to obtain Z R n Y Θ | x i − x k +1 | − α i,k +1 dx k +1 ≤ C X Θ L i , (5.24)where L i are defined as in Theorem 2.2. 16eplacing the integral relative to x k +1 by each L i , we shall prove that each k + 1 − multipleintegral is dominated by integrals of the above two types.If P mi =2 α i,k +1 < n , we shall prove that Z ( B (0)) k Y S | x i − x j | − α ij ! X Θ | x i − x j | ! n − P Θ α i,k +1 dx · · · dx k ≤ C Z ( B (0)) k Y S | x i − x j | − α ij − δ ij dx · · · dx k , where { α ij = α ij + δ ij } satisfies (5.21). Here P Θ δ ij = P Θ α i,k +1 − n for δ ij ≥ δ ij = 0 ifeither i or j does not lie in Θ. Now we turn our attention to the existence of such a solution { δ ij } . In other words, we need solve the following system of linear inequalities:( V. ( i ) δ ij ≥ , ≤ i < j ≤ m ;( ii ) P Θ δ ij = P Θ α i,k +1 − n ;( iii ) P J ∩ Θ δ ij < ( | J | − n − P J α ij for J ∈ F m , where the class F m consists of all subsets J of { , · · · , k } satisfying | J T Θ | ≥ 2. Note that wehave assumed Θ = { , , · · · , m } . Here we use the notation F m instead of F Θ for simplicity.Now we shall apply Lemma 2.4 to show the existence of solutions δ ij . Obviously, ( i ) and ( ii )in ( V. 1) have solutions and so Lemma 2.4 is applicable. For arbitrary nonnegative numbers λ ij , θ , θ and µ J with at least one µ J > J ∈ F m satisfying λ ij + ( θ − θ ) − X J ∋ i,j µ J = 0 (5.25)for 1 ≤ i < j ≤ m , we must show that( θ − θ ) (cid:16) X Θ α i,k +1 − n (cid:17) − X J ∈F m µ J (cid:16) ( | J | − n − X J α ij (cid:17) < . (5.26)It suffices to prove this inequality when θ − θ > µ J > J and P J α ij < ( | J | − n . Now assume θ − θ > 0. By dilation, put θ − θ = 1. Then µ J and λ ij satisfy X J ∋ i,j µ J = 1 + λ ij (5.27)for 1 ≤ i < j ≤ m . To prove this inequality, a basic idea is to determine the supremum of theobjective function in (5.26). Though the supremum cannot be attained generally, µ J and λ ij have simple forms when the value of the objective function is sufficiently close to its supremun.More precisely, for any { µ J (0) , λ ij (0) } , we shall construct a sequence { µ J ( N ) , λ ij ( N ) } suchthat the objective function for { µ J ( N ) , λ ij ( N ) } increases. By taking N → ∞ , the sign of theobjective function will be easily verified.Now we turn to construct such a process. Suppose { µ J ( N − } are given. For convenience,we define the following conditions for two subsets J and J :( a ) µ J ( N − ) µ J ( N − ) > ; ( b ) J ∩ J = ∅ ; ( c ) J * J and J * J . (5.28)17f { µ J ( N − } and { λ ij ( N − } are known, then we choose two subsets J and J satisfyingsuitable restrictions and set µ J ( N ) as follows. We shall explain later why these restrictions on J and J are required. In the following, the notation A ∧ B means min { A, B } . Case I : J , J ∈ F m satisfy the above conditions ( a ) , ( b ) , ( c ) and J T J ∈ F m . µ J ( N ) = µ J ( N − − µ J ( N − ∧ µ J ( N − µ J ( N ) = µ J ( N − − µ J ( N − ∧ µ J ( N − µ J ∩ J ( N ) = µ J ∩ J ( N − 1) + µ J ( N − ∧ µ J ( N − µ J ∪ J ( N ) = µ J ∪ J ( N − 1) + µ J ( N − ∧ µ J ( N − µ J ( N ) = µ J ( N − , J / ∈ { J , J , J ∩ J , J ∪ J } . (5.29) Case II : J , J ∈ F m satisfy ( a ) , ( b ) , ( c ) and J T J / ∈ F m . µ J ( N ) = µ J ( N − − µ J ( N − ∧ µ J ( N − µ J ( N ) = µ J ( N − − µ J ( N − ∧ µ J ( N − µ J ∪ J ( N ) = µ J ∪ J ( N − 1) + µ J ( N − ∧ µ J ( N − µ J ( N ) = µ J ( N − , J / ∈ { J , J , J ∪ J } . (5.30)We also define λ ij ( N ) by (5.27) with { µ J ( N ) } in place of { µ J } .The motivation for construction of such a process is the following inequality, (cid:16) ( | J ∩ J | − n − X J ∩ J α ij (cid:17) + (cid:16) ( | J ∪ J | − n − X J ∪ J α ij (cid:17) ≤ X s =1 (cid:16) ( | J s | − n − X J s α ij (cid:17) (5.31)for all J and J in the class F m . Here we use the summation convention P J α ij = 0 if | J | ≤ λ ij ( N ) increases as N . Assumetwo subsets J and J are chosen in the N − th step. For each pair i and j , there are severalpossible cases. If either i / ∈ J ∪ J or j / ∈ J ∪ J , then λ ij ( N ) = λ ij ( N − i, j ∈ J ∪ J and divide this into three subcases. (1) If { i, j } ⊆ J ∩ J , then it is easy to seethat λ ij ( N ) = λ ij ( N − (2) If i / ∈ J ∩ J or j / ∈ J ∩ J but either { i, j } ⊆ J or { i, j } ⊆ J ,then λ ij remains unchanged in the N step. (3) The remaining subcase is that { i, j } * J and { i, j } * J in which λ ij increases in the N -th step. Thus we have established our claim.The key observation is that the objective function also increases as N . Equivalently, we have − X J ∈F m µ J ( N − ( | J | − n − X J α ij ! ≤ − X J ∈F m µ J ( N ) ( | J | − n − X J α ij ! (5.32)for any N ≥ 1. Indeed, this observation is an immediate consequence of the inequality (5.31)and its simple variant( | J ∪ J | − n − X J ∪ J α ij ≤ X s =1 (cid:16) ( | J s | − n − X J s α ij (cid:17) | J ∩ J | ≥ 1. This explains why the recursion (5.30) only appliesto those J and J having nonempty intersection; condition ( b ) in (5.28).Now we shall introduce some subclasses of F m . Let A m , B m and C m be defined by A m = { J ∈ F m : µ J > } , B m = { J ∈ A m : Θ * J } , C m = { J ∈ A m : Θ ⊆ J } . (5.33)For convenience, we also use A m ( N ), B m ( N ) and C m ( N ) defined similarly as above to keep trackof the above process. It is clear that A m = B m S C m and B m T C m = ∅ . Definition 5.1 If { µ J : J ∈ F m } is invariant under any possible process described as in (5.29) and (5.30) , then we say that { µ J : J ∈ F m } is stable.Let F ( { µ J } ) be a function of µ J ≥ for J ∈ F m . Assume { µ J (0) : J ∈ F m } is a setof nonnegative numbers. We say that F is stable with respect to { µ J (0) } if for all N ≥ F ( { µ J (0) } ) = F ( { µ J ( N ) } ) , where { µ J ( N ) } is obtained by an arbitrary process of N steps. By this definition, { µ J : J ∈ F m } is stable if and only if for all J , J ∈ A m one of the threerelations holds: (i) J ∩ J = ∅ ; (ii) J ⊂ J ; (iii) J ⊂ J . Further observation also shows P J ∈C m µ J = 1 + min Θ λ ij when { µ J } is stable. This observation will be proved later. Forconvenience, we introduce the notation Ω m to denoteΩ m = X J ∈C m µ J . (5.34)And Ω m ( N ) is defined as above with C m and µ J replaced by C m ( N ) and µ J ( N ) respectively.We do not know whether arbitrary { µ J (0) : J ∈ F m } and { λ ij (0) } satisfying (5.27) canreach a stable state by a process consisting of finite steps. However, by a passage to the limit,we can arrive at a special state, not stable generally, which is enough for our purpose. Let { µ ∗ J ( N ) : J ∈ F m } be obtained by a process of N steps described as in (5.29) and (5.30) suchthat the supremum Ω ∗ m ( N ) = sup Ω m ( N )is attained. Here the supremum is taken over all possible processes consisting of N steps. Itis possible that these processes are not unique. We can take one of these processes by whichΩ ∗ m ( N ) is obtained. It should be pointed out that { µ ∗ J ( N ) } is not obtained by a continuousprocedure with respect to N . Therefore in general we cannot obtain { µ ∗ J ( N ) } from { µ ∗ J ( N − } by one step. On the one hand, it is clear that Ω ∗ m ( N ) increases as N . On the other hand, wealso have X J ∋ i µ ∗ J ( N ) ≤ X J ∋ i µ J (0) ≤ max ≤ i ≤ k X J ∋ i µ J (0) ! , (5.35)for any i ∈ S = { , , · · · , k } . Suppose at the k -th step with k ≤ N − J and J in F m . Then we see that P J ∋ i µ J ( k ) = P J ∋ i µ J ( k + 1) unless i ∈ J ∩ J and J ∩ J / ∈ F m . In the latter case, we have P J ∋ i µ J ( k ) > P J ∋ i µ J ( k + 1). There is anothersimilar observation as (5.35). In the system (V . δ ij is assumed to be zero if i / ∈ Θ or j / ∈ Θ.If { µ J (0) } and { λ ij (0) } satisfy (5.27), we may assume that µ J (0) = 0 for nonempty J ⊂ S but J / ∈ F m . Of course we have µ J ( N ) = 0 if J / ∈ F m for any N . Then it is clear that X J ⊂ S µ J ( N ) ≤ X J ⊂ S µ J (0) , (5.36)19here µ J ( N ) are obtained by any possible process with N steps. Both (5 . 35) and (5 . 36) implythat { Ω ∗ m ( N ) } has a uniform upper bound. PutΩ ∗ m ( ∞ ) = lim N →∞ Ω ∗ m ( N ) . (5.37)This limit is well defined since { Ω ∗ m ( N ) } is a bounded increasing sequence. It is possible thatΩ ∗ m ( ∞ ) can be obtained by a process of finite steps, i.e., Ω ∗ m ( ∞ ) = Ω ∗ m ( N ) for N ≥ N for some N . In this situation, Ω ∗ m ( N ) becomes stable. To calculate Ω ∗ m ( ∞ ) explicitly, we are going toestablish a necessary and sufficient condition under which Ω m is stable. It will be convenient tointroduce a concept related to a sequence of sets. Definition 5.2 If { J i } ai =1 with a ≥ is a sequence of sets with the property that each inter-section (cid:0) S it =1 J t (cid:1) T J i +1 is nonempty for ≤ i ≤ a − , then we call { J i } ai =1 a continuouschain. Lemma 5.2 Assume { µ J : J ∈ F m } is a set of nonnegative numbers and Ω m is defined as in (5.34) . Then Ω m is stable with respect to { µ J } if and only if Θ " S ai =1 J i for any continuouschain { J i } ai =1 in B m .Proof. We first show the necessity part. Assume the converse, i.e., Ω m is stable and there existsa continuous chain { J i } ai =1 in B m (0) such that Θ ⊆ S ai =1 J i . We shall see that there is a processsuch that Ω m ( a − > Ω m (0). First applying the recursion in (5.29) and (5.30) to J and J ,we obtain { µ J (1) } . In the second step, we repeat the recursion with respect to J ∪ J and J . Likewise, in the i -th step we apply the recursion to S ≤ t ≤ i J t and J i +1 . Here we assume S ≤ t ≤ i J t and J i +1 satisfy ( c ) in (5.28). If this is not the case, we can skip over to apply therecursion to S ≤ t ≤ i +1 J t and J i +2 . After at most a − J i " Θ thatΩ m ( a − ≥ Ω m (0) + min ≤ i ≤ a µ J i (0) ≥ Ω m (0) + min J ∈A m µ J (0) (5.38)which contradicts the assumption that Ω m (0) is stable.The proof of the sufficiency part is intricate. We first establish a useful property of B m underthe assumption that for all continuous chains { J i } ai =1 in B m , the union S ai =1 J i does not containΘ as a subset. Proposition 5.3 Suppose that Θ " S ai =1 J i for all continuous chains { J i } ai =1 in B m . Then thereexists a nonempty proper subset Θ of Θ such that for all J and J in B m with J T J = ∅ , itis true that: either (cid:16) J [ J (cid:17) \ Θ ⊂ Θ or (cid:16) J [ J (cid:17) \ Θ ⊂ Θ − Θ . (5.39) Generally, this property also holds for all continuous chains { J i } ai =1 in B m . More precisely, wehave either (cid:16) a [ i =1 J i (cid:17) \ Θ ⊂ Θ or (cid:16) a [ i =1 J i (cid:17) \ Θ ⊂ Θ − Θ . Now we turn to prove this proposition. Define η by η = sup (cid:12)(cid:12)(cid:12)(cid:16) a [ i =1 J i (cid:17) \ Θ (cid:12)(cid:12)(cid:12) , (5.40)20here the supremum is taken over all continuous chains { J i } in B m . Since η is an integer, wesee that η can be achieved for some continuous chain { I i } ai =1 in B m . Let Θ = (cid:0) S ai =1 I i (cid:1) T Θ.Then Θ is a proper subset of Θ. For all J in B m , we claim that either J T Θ ⊆ Θ or J T (cid:0) S ai =1 I i (cid:1) = ∅ is true. Otherwise, if there were some I a +1 ∈ B m such that I a +1 ∩ Θ " Θ and I a +1 T (cid:0) S ai =1 I i (cid:1) = ∅ , it would follow that { I i } a +1 i =1 is a continuous chain in B m and satisfies (cid:12)(cid:12)(cid:12)(cid:16) S a +1 i =1 I i (cid:17) T Θ (cid:12)(cid:12)(cid:12) > η . This contradicts the definition of η .Define B (1) m and B (2) m by B (1) m = n J ∈ B m : J \ (cid:0) a [ i =1 I i (cid:1) = ∅ o , B (2) m = B m − B (1) m . Then J T Θ ⊆ Θ for all J ∈ B (2) m . Moreover, we also have that J T J = ∅ if J ∈ B (1) m and J ∈ B (2) m . Actually, if there were J ∈ B (1) m and J ∈ B (2) m with J T J = ∅ , it would follow that { I i } a +2 i =1 is a continuous chain in B m and (cid:12)(cid:12)(cid:12)(cid:16) S a +2 i =1 I i (cid:17) T Θ (cid:12)(cid:12)(cid:12) > η where I a +1 = J and I a +2 = J .This contradicts our choice of η . Therefore one of two relations in (5.39) is true if J T J = ∅ .It remains to prove that the same conclusion holds for any continuous chain in B m . Withoutloss of generality, we may assume a = 3. Since the intersection of J and J is nonempty, wesee that either J , J ∈ B (1) m or J , J ∈ B (2) m is true. By the assumption that J , J , J is acontinuous chain, we have either J T J = ∅ or J T J = ∅ . Thus either J , J , J ∈ B (1) m or J , J , J ∈ B (2) m holds. We conclude the proof of Proposition 5.3.Now we shall invoke Proposition 5.3 to give a complete proof of the sufficiency part of Lemma5.2. For initial data { µ J (0) } and { λ ij (0) } satisfying (5.27), it follows from Proposition 5.3 thatthere is a nonempty proper Θ ⊆ Θ such that for all J , J ∈ B m (0) satisfying J T J = ∅ ,one of two relations in (5.39) is true. We shall prove that Ω m (0) is stable. In other words,Ω m (0) = Ω m ( N ) for any process of N steps. Here { µ J ( N ) } is obtained from { µ J (0) } by anyprocess consisting of N steps as in (5.29) and (5.30). At the k − th step in which { µ J ( k ) } isobtained, assume that we apply the recursion to J and J in A m ( k − J T J = ∅ . Wedivide the k -th step into four cases:(a) J and J are in C m ( k − J ∈ B m ( k − 1) and J ∈ C m ( k − J ∈ C m ( k − 1) and J ∈ B m ( k − J and J are in B m ( k − m ( k − 1) = Ω m ( k ). For J and J in the cases (a), (b) and (c), this statementis easily verified. In the case (d), we need an additional property of B m ( k ). In other words,Proposition 5.3 is still true for all B m ( k ) with the same Θ . Since B m (0) has the property inProposition 5.3, it suffices to show that B m (1) shares this property. Assume in the first step, therecursion applies to J and J in F m . Observe that J ∩ J and J ∪ J are the only two subsetswhich are possibly contained in B m (1) but not in B m (0). It follows from the above propositionthat both intersections ( J ∩ J ) ∩ Θ and ( J ∪ J ) ∩ Θ are subsets of Θ or Θ − Θ . Hence theassumption in the proposition is also valid for B m (1). By induction, it follows that the sameresult holds for all B m ( k ). Thus in the case (d), we still have Ω( k ) = Ω( k − ✷ We now turn our attention to prove that Ω ∗ m ( ∞ ) is stable. Recall that both (5.35) and (5.36)imply the uniform boundedness of µ ∗ J ( N ) and λ ∗ ij ( N ). Thus by passing { N } to a subsequence,denoted by { N t } , such that µ ∗ J ( N t ) and λ ∗ ij ( N t ) converge to µ ∗ J ( ∞ ) and λ ∗ ij ( ∞ ) respectively. It21s easy to see that the limits µ ∗ J ( ∞ ) and λ ∗ ij ( ∞ ) still satisfy (5.27). By the definition of Ω ∗ m ( ∞ ),we have Ω ∗ m ( ∞ ) = X J ∈ C m ( ∞ ) µ ∗ J ( ∞ )with C m ( ∞ ) = { J ∈ F m : Θ ⊆ J, µ ∗ J ( ∞ ) > } . Let A m ( ∞ ) be the class of all J ∈ F m satisfying µ ∗ J ( ∞ ) > 0. Similarly, B m ( ∞ ) can be defined as (5.33). With these notations, we claim thatΩ ∗ m ( ∞ ) is stable with respect to µ ∗ J ( ∞ ). This means that any application of above processesto A m ( ∞ ) does not change the value of Ω ∗ m ( ∞ ). Otherwise, by Lemma 5.2, there exists acontinuous chain { J i } ai =1 in B m ( ∞ ) satisfying Θ ⊆ S ai =1 J i . For all t ≥ t with some sufficientlylarge t , first observe that { µ ∗ J ( N t ) } has a uniform positive lower bound for each J ∈ A m ( ∞ ).By (5.38) and related remarks, it follows thatΩ ∗ m ( N t + a − ≥ Ω ∗ m ( N t ) + min J ∈A ( ∞ ) µ ∗ J ( N t ) ≥ Ω ∗ m ( N t ) + 12 min J ∈A ( ∞ ) µ ∗ J ( ∞ ) (5.41)for all sufficiently large N t . This contradicts the fact that Ω ∗ m ( N ) converges to Ω ∗ m ( ∞ ).Another simple but key observation isΩ ∗ m ( ∞ ) = 1 + min Θ λ ∗ ij ( ∞ ) ≥ Θ λ ij (0) . (5.42)In fact, if B m ( ∞ ) is empty, then A m ( ∞ ) = C m ( ∞ ) and Ω ∗ m ( ∞ ) = 1. Assume B m ( ∞ ) isnonempty. As in the proof of Proposition 5.3, we can choose a continuous chain { I i } ai =1 from B m ( ∞ ) such that (cid:12)(cid:12)(cid:0) S ai =1 I i (cid:1) T Θ (cid:12)(cid:12) is the largest over all continuous chains in B m ( ∞ ). By Proposi-ton 5.3, (cid:0) S ai =1 I i (cid:1) T Θ is a nonempty proper subset of Θ. Hence we can choose i ∈ (cid:0) S ai =1 I i (cid:1) T Θand j ∈ Θ but j / ∈ (cid:0) S ai =1 I i (cid:1) T Θ. Then it follows from Proposition 5.3 that µ ∗ J ( ∞ ) = 0 forall J ∈ F m satisfying i , j ∈ J and Θ " J . Then (5.27) becomes P J ⊃ Θ µ ∗ J ( ∞ ) = 1 + λ ∗ ij ( ∞ )with ( i, j ) = ( i , j ). This implies our claim.Now the desired inequality (5.26) is easily verified. In fact, we have that X Θ α i,k +1 − n ! − X J ∈F m µ J (0) ( | J | − n − X J α ij ! ≤ X Θ α i,k +1 − n ! − X J ∈F m µ ∗ J ( ∞ ) ( | J | − n − X J α ij ! . By the assumption (5.21), we observe that P Θ α i,k +1 − n < ( | J | − n − P J α ij for all J ∈ F m satisfying Θ ⊆ J . Then it follows from (5.42) that X Θ α i,k +1 − n ! − X J ∈F m µ ∗ J ( ∞ ) ( | J | − n − X J α ij ! ≤ X Θ α i,k +1 − n ! − X Θ ⊂ J ∈F m µ ∗ J ( ∞ ) ( | J | − n − X J α ij ! < . By Lemma 2.4, there exists a solution { δ ij } to the system ( V. k − 1, we have completed the proof in the first case P mi =2 α i,k +1 < n .22ow we treat the second case P mi =2 α i,k +1 = n . By Lemma 2.2, it suffices to show that thefollowing integral is finite, Z ( B (0)) k Y S | x i − x j | − α ij ! X Θ | x i − x j | ! n − P Θ α i,k +1 log 2 P Θ | x i − x j | P ≤ i 1) depends only on the conditions (5.21) and the assumption P Θ α i,k +1 > n , we can find a solution { δ ij : 1 ≤ i < j ≤ m } to the system ( V. δ ij = 0 forother ( i, j ). It follows from P Θ α i,k +1 > n that there exist i and j with 1 ≤ i < j ≤ m suchthat δ i ,j > 0. Choose any pair i and j with 2 ≤ i < j ≤ m . For small ε > 0, we put α ij = α ij + δ ij − δ i i δ j j ε + δ i i δ j j ε (5.43)for 1 ≤ i < j ≤ k , where δ ts denotes the Kronecker symbol. It is easily verified that { α ij } satisfies the integrability condition (5.21) with sufficiently small ε . By the induction hypothesis,the above integral converges.The final case is P mi =2 α i,k +1 > n . The integral in (5.22) with L ( x , · · · , x m ) in place of theintegral with respect to x k +1 is bounded by a constant multiple of Z ( B (0)) k Y S | x i − x j | − α ij ! X Θ | x i − x j | ! − α ,k +1 Z R n m Y i =2 | x i − x k +1 | − α i,k +1 dx k +1 ! dV S . (5.44)Likewise, our task is to distribute α ,k +1 into powers { α ij : 1 ≤ i < j ≤ k } appropriately. Thenwe will obtain new parameters { α ij : 1 ≤ i < j ≤ k + 1 } with | Θ | = m − 1. By the integrabilitycondition (5.21), we shall solve the following system of linear inequalities,( V. ( i ) δ ij ≥ , i < j in Θ;( ii ) P Θ δ ij = α ,k +1 ;( iii ) P J ∩ Θ δ ij < ( | J | − n − P J (cid:16) α ij − δ i δ k +1 j α ,k +1 (cid:17) , J ∈ F m ;where F m consists of all subsets J ⊆ { , , · · · , k + 1 } such that | J T Θ | ≥ 2. The existenceof solutions can be proved similarly. Indeed, by Lemma 2.4, for nonnegative λ ij , θ , θ and µ J with at least one µ J > J in the class F m satisfying λ ij + ( θ − θ ) − X J ∋ i,j µ J = 0 , ≤ i < j ≤ m, (5.45)we must prove( θ − θ ) α ,k +1 − X J ∈F m µ J (cid:16) ( | J | − n − X J (cid:0) α ij − δ i δ k +1 j α ,k +1 (cid:1)(cid:17) < . We may assume θ − θ = 1. To follow the argument in the first case P mi =2 α i,k +1 < n , weshall make some remarks. The first observation is that the inequality (5.31) is still true with α ij − δ i δ k +1 j α ,k +1 in place of α ij . In other words, we have X s =1 X J s (cid:16) α ij − δ i δ k +1 j α ,k +1 (cid:17) ≤ X J ∩ J (cid:16) α ij − δ i δ k +1 j α ,k +1 (cid:17) + X J ∪ J (cid:16) α ij − δ i δ k +1 j α ,k +1 (cid:17) J and J of { , , · · · , k + 1 } .The process in (5.29) and (5.30) is also applicable here with F m replaced by F m in Case Iand Case II there. For arbitrary { µ J (0) ≥ J ∈ F m } and { λ ij (0) ≥ ≤ i < j ≤ m } satisfying equations (5.45), we can use the above argument to obtain µ ∗ J ( ∞ ) = lim N t →∞ µ ∗ J ( N t ) , λ ∗ ij ( ∞ ) = lim N t →∞ λ ∗ ij ( N t ) , where { µ ∗ J ( N ) } is obtained by one of those processes such that Ω ∗ m ( N ) = P Θ ⊂ J ∈F m µ ∗ J ( N ) isthe maximum over all possible processes of N steps. All symbols can be defined parallel to thefirst case. Similarly, we haveΩ ∗ m ( ∞ ) = lim N →∞ Ω ∗ m ( N ) = 1 + min Θ λ ∗ ij ( ∞ ) . Note that for any J in F m satisfying Θ ⊂ J , α ,k +1 − (cid:16) ( | J | − n − X J (cid:0) α ij − δ i δ k +1 j α ,k +1 (cid:1)(cid:17) < . Actually, this inequality is obvious if k + 1 ∈ J by the integrability condition (5.21). Assume k + 1 / ∈ J . Then the left side of this inequality equals α ,k +1 + X J α ij − (cid:0) | J | − (cid:1) n = X J ∪{ k +1 } α ij − m X i =2 α i,k +1 − (cid:0) | J | − (cid:1) n< X J ∪{ k +1 } α ij − | J | n < P mi =2 α i,k +1 > n .Since | x i | ≤ ≤ i ≤ m , it is clear that the integral of Q mi =2 | x i − x k +1 | − α i,k +1 withrespect to x k +1 over R n is bounded by a constant multiple of the integral over B (0). Thus theintegral in (5.44) is less than a constant multiple of Z ( B (0)) k +1 Y { , , ··· ,k +1 } | x i − x j | − α ij dx dx · · · dx k +1 , where α ij = (cid:0) α ij − δ i δ k +1 j α ,k +1 (cid:1) + δ ij and { δ ij } is a solution to the system ( V. 2) with δ ij = 0when i / ∈ Θ or j / ∈ Θ. Thus we have reduced the integral in (5.44) to an integral of form (5.22)but with | Θ | ≤ m − 1. Since the k + 1 fold integral in the theorem converges for | Θ | = 2, weneed at most m − | Θ | = m to | Θ | = 2. Hence the integral (5 . 44) isalso finite.Until now, we have obtained the desired conclusion for L . For other terms L i , the treatmentis the same as above. Therefore the proof of Theorem 5.1 is complete. ✷ Remark 5.1 The integrability criterion in the theorem is also true for Selberg integrals on thesphere S n . More precisely, for symmetric and nonnegative exponents α ij , the following Selbergintegral Z ( S n ) k +1 Y ≤ i Assume { α ij } and { p i } with p k +1 = ∞ satisfy the assumptions in Theorem .Then there exists a finite set ∆ . For each t ∈ ∆ we have { β ij ( t ) ≥ ≤ i < j ≤ k } suchthat the datum { p i , β ij ( t ) } satisfies ( i ) , ( ii ) and ( a ) of ( iii ) in Theorem , and the followingestimate holds: Z R n ( k +1) k Y i =1 | f i ( x i ) | Y ≤ i 0. By assumptions inTheorem 1.2 for I = S , we have X Θ α i,k +1 = np k +1 + X S α i,k +1 > n. To reduce mapping properties of the k + 1 − linear functional Λ to that of a k − linear one, weneed estimate the following integral with respect to x k +1 , I ( x i : i ∈ Θ) = Z R n Y Θ | x i − x k +1 | − α i,k +1 dx k +1 . By the symmetry of parameters, we may assume Θ = { , , · · · , m } with m ≥ 2. By Lemma2.2, we claim that (5.46) is still true if I ( x i : i ∈ Θ) is replaced by each term L i . We firstprove this statement for m = 2. Observe that the integral I ( x , x ) equals a constant multipleof | x − x | n − α ,k +1 − α ,k +1 . Put β ij = α ij + δ i δ j ( α ,k +1 + α ,k +1 − n ) for 1 ≤ i < j ≤ k . Thenwe see that { β ij } and { p i } satisfy (i), (ii) and (a) of (iii) in Theorem 1.1. Thus our claim is truein the case m = 2.For 3 ≤ m ≤ k , we claim that there exist finite families { β ij ( t ) } such that Z R nk k Y i =1 | f i ( x i ) | Y S | x i − x j | − α ij ! X Θ L i ( x , · · · , x m ) ! dx dx · · · dx k ≤ C X t ∈ ∆ Z R nk k Y i =1 | f i ( x i ) | Y S | x i − x j | − β ij ( t ) dx dx · · · dx k , (5.47)25here ∆ is a finite set and { p i , β ij ( t ) } satisfy conditions ( i ), ( ii ) and ( a ) of ( iii ) in Theorem1 . 1. Since arguments for different L i ’s are similar, we need only establish the desired estimateconcerning L . As shown in Lemma 2.2, there are three possible cases for L . The treatment oftwo previous cases P mi =2 α i,k +1 ≤ n will be reduced to the following system of linear inequalities:( V. ( i ) δ ij ≥ , ≤ i < j ≤ m ;( ii ) P Θ δ ij = P Θ α i,k +1 − n ;( iii ) P J ∩ Θ δ ij < (cid:18) ( | J | − n − P J α ij (cid:19) ∧ (cid:18) | J | n − P J α ij − P J np i (cid:19) for J ∈ F m and J = S ;( iv ) P S ∩ Θ δ ij ≤ (cid:18) ( k − n − P S α ij (cid:19) ∧ (cid:18) kn − P S α ij − P S np i (cid:19) . Recall that F m is the class of all subsets J of S which contains at least two members in { , , · · · , m } . We shall point out that ( iv ) is an equality. By ( ii ) in the system ( V. P S ∪{ k +1 } α ij < kn and the condition ( i ) in Theorem 1.1. For thisreason, the system ( V. 3) is equivalent to the system of inequalities (i), (ii) and (iii). We add ( iv )for convenience of notations. As in the previous theorem, we shall also use Lemma 2.4 to proveexistence of a solution. Assume λ ij , θ , θ and µ J are nonnegative numbers satisfying λ ij − X J ∋ i,j µ J + ( θ − θ ) = 0 , ≤ i < j ≤ m. (5.48)Here there is at least one µ J > J of S in the class F m . Under theseconditions, we have to prove( θ − θ ) X Θ α i,k +1 − n ! < X J ∈F m µ J ( | J | − n − X J α ij ! ∧ | J | n − X J α ij − X J np i ! . (5.49)This inequality is obviously true if θ − θ ≤ µ J > J ∈ F m of S .Thus it suffices to show the inequality for θ − θ > 0. By dilation, we may assume θ − θ = 1.There is an important observation like (5.31) given as follows: (cid:16) ( | J ∩ J | − n − X J ∩ J α ij (cid:17) ∧ (cid:16) | J ∩ J | n − X J ∩ J α ij − X J ∩ J np i (cid:17) + (cid:16) ( | J ∪ J | − n − X J ∪ J α ij (cid:17) ∧ (cid:16) | J ∪ J | n − X J ∪ J α ij − X J ∪ J np i (cid:17) ≤ X s =1 ((cid:16) ( | J s | − n − X J s α ij (cid:17) ∧ (cid:16) | J s | n − X J s α ij − X J s np i (cid:17)) (5.50)for all subsets J and J of S . To show this inequality, we have to verify it in all possible cases.Case (i): P J ∪ J n/p i ≤ n .The above inequality is just (5.31). We also note that (5.31) becomes an equality if and onlyif α ij = 0 , ( i, j ) ∈ { ( s, t ) : s < t, s, t ∈ J ∪ J } − [ i =1 { ( s, t ) : s < t, s, t ∈ J i } . (5.51)26ase (ii): P J ∩ J n/p i ≥ n .It is true that (cid:16) | J ∩ J | n − X J ∩ J α ij − X J ∩ J np i (cid:17) + (cid:16) | J ∪ J | n − X J ∪ J α ij − X J ∪ J np i (cid:17) ≤ X s =1 (cid:16) | J s | n − X J s α ij − X J s np i (cid:17) , where the equality is valid if and only if (5.51) holds.Case (iii): P J ∩ J n/p i ≤ n and P J ∪ J n/p i ≥ n , but P J n/p i ≤ n and P J n/p i ≥ n .We have (cid:16) ( | J ∩ J | − n − X J ∩ J α ij (cid:17) + (cid:16) | J ∪ J | n − X J ∪ J α ij − X J ∪ J np i (cid:17) ≤ (cid:16) ( | J | − n − X J α ij (cid:17) + (cid:16) | J | n − X J α ij − X J np i (cid:17) . The equality is true if and only if (5.51) holds and p i = ∞ for i in J but not in J . Theremaining cases of (5.50) can be proved similarly.Case (iv): P J ∩ J n/p i ≤ n and P J ∪ J n/p i ≥ n , but P J n/p i ≥ n and P J n/p i ≤ n .As in Case (iii), it is clear that (cid:16) ( | J ∩ J | − n − X J ∩ J α ij (cid:17) + (cid:16) | J ∪ J | n − X J ∪ J α ij − X J ∪ J np i (cid:17) ≤ (cid:16) | J | n − X J α ij − X J np i (cid:17) + (cid:16) ( | J | − n − X J α ij (cid:17) . Combining above results, we see that if (5.50) becomes an equality then we must have (5.51).This fact will be used later.To show the inequality (5.49), we also need the recursion in (5.29) and (5.30). By theinequality (5.50), the objective function also increases as the recursion continues. In fact, wehave an analogue of (5.32) in the present situation, − X J ∈F m µ J ( N − ( | J | − n − X J α ij ! ∧ | J | n − X J α ij − X J np i ! ≤ − X J ∈F m µ J ( N ) ( | J | − n − X J α ij ! ∧ | J | n − X J α ij − X J np i ! (5.52)where µ J ( N ) are obtained by the recursion in (5.29) and (5.30).We shall follow some notations in the proof of Theorem 5.1. Let { µ ∗ J ( N ) } be obtained byone of processes consisting of N steps such that Ω ∗ m ( N ) achieves its maximum. By a similarargument as in the proof of Theorem 5.1, we can obtain { µ ∗ J ( N ) } and { λ ∗ ij ( N ) } for initial data { µ J (0) } and { λ ij (0) } satisfying (5.48) with θ − θ = 1. By passing to a subsequence { N t } , wealso use µ ∗ J ( ∞ ) and λ ∗ ij ( ∞ ) to denote the limits of µ ∗ J ( N t ) and λ ∗ ij ( N t ), respectively. It is clearthat (5.42) is also true and hence Ω ∗ m ( ∞ ) ≥ 1. The following argument is somewhat different27epending on whether µ ∗ S ( ∞ ) = 1 and µ ∗ J ( ∞ ) = 0 for all proper subsets J of S in F m . Observethat X Θ α i,k +1 − n = ( | S | − n − X S α ij ! ∧ | S | n − X S α ij − X S np i ! and the right side equals | S | n − P S α ij − P S n/p i . And we also have X Θ α i,k +1 − n < ( | J | − n − X J α ij ! ∧ | J | n − X J α ij − X J np i ! for all proper subsets J of S satisfying Θ ⊂ J . By (5.52), we have − X J ∈F m µ J (0) (cid:16) ( | J | − n − X J α ij (cid:17) ∧ (cid:16) | J | n − X J α ij − X J np i (cid:17) ≤ − X J ∈F m µ ∗ J ( ∞ ) (cid:16) ( | J | − n − X J α ij (cid:17) ∧ (cid:16) | J | n − X J α ij − X J np i (cid:17) . Let H ( λ ij , µ J ) be the objective function X Θ α i,k +1 − n ! − X J ∈F m µ J ( | J | − n − X J α ij ! ∧ | J | n − X J α ij − X J np i ! . Recall that we have Ω ∗ m ( ∞ ) = 1 + min Θ λ ∗ ij ( ∞ ).Now we divide the proof into three cases.Case (i): Ω ∗ m ( ∞ ) > H ( λ ij (0) , µ J (0)) ≤ H ( λ ∗ ij ( ∞ ) , µ ∗ J ( ∞ )), H ( λ ∗ ij ( ∞ ) , µ ∗ J ( ∞ )) < Ω ∗ m ( ∞ ) (cid:16) X Θ α i,k +1 − n (cid:17) − X J ∈ C m ( ∞ ) µ ∗ J ( ∞ ) (cid:16) ( | J | − n − X J α ij (cid:17) ≤ , where C m ( ∞ ) = { J ∈ F m : Θ ⊆ J, µ ∗ J ( ∞ ) > } .Case (ii): Ω ∗ m ( ∞ ) = 1 and µ ∗ J ( ∞ ) > J of S .As in Case (i), we also have H ( λ ij (0) , µ J (0)) < m (0) < λ ij (0) = 0. In fact, this implies that there exists a positive µ J (0) for some J ∈ F m satisfying Θ " J . Now we claim that there exist J and J in B m (0) with nonemptyintersection such that J \ Θ * J \ Θ and J \ Θ * J \ Θ . (5.53)Assume the converse. Then either J T Θ ⊂ J T Θ or J T Θ ⊂ J T Θ is true for all J and J in B m (0) with J T J = ∅ . Choose a J ∈ B m (0) such that (cid:12)(cid:12)(cid:12) J \ Θ (cid:12)(cid:12)(cid:12) = max J ∈B m (0) (cid:12)(cid:12)(cid:12) J \ Θ (cid:12)(cid:12)(cid:12) . Then J T Θ ⊂ J T Θ for all J ∈ B m (0). Since J T Θ is a proper subset of Θ, we may choose i ∈ Θ but i / ∈ J T Θ. Choose a j ∈ J T Θ arbitrarily. Then by the choice of J , we see28hat all µ J (0) = 0 for all J ∈ F m satisfying i , j ∈ J and Θ " J . Then we have Ω m (0) = 1by the equation P J ∋ i,j µ J (0) = 1 with ( i, j ) = ( i , j ) or ( i, j ) = ( j , i ). As a consequence, B m (0) becomes an empty class which contradicts our assumption Ω m (0) < 1. Recall that wehave assumed Θ = { , · · · , m } . Choose J , J ∈ B m (0) with J T J = ∅ satisfy J \ { , , · · · , m } * J \ { , , · · · , m } J \ { , , · · · , m } * J \ { , , · · · , m } . Then we apply the recursion (5.29) or (5.30) to J and J and then obtain { µ J (1) } and { λ ij (1) } .By this process, we will obtain at least one λ ij (1) > 0. Indeed, we may choose i ∈ J \ { , , · · · , m } but i / ∈ J \ { , , · · · , m } j ∈ J \ { , , · · · , m } but j / ∈ J \ { , , · · · , m } . It follows from J , J ∈ A m (0) that λ ij (1) = µ J (0) ∧ µ J (0) > 0. Thus the datum { µ J (1) , λ ij (1) } has been considered in Case (i) and (ii). Since the objective function H ( λ ij ( N ) , µ J ( N )) increasesas N , we see that H (cid:16) λ ij (0) = 0 , µ J (0) (cid:17) ≤ H (cid:16) λ ij (1) , µ J (1) (cid:17) < . Case (iii): Ω ∗ m ( ∞ ) = 1 and µ ∗ J ( ∞ ) = 0 for all proper subsets J ∈ F m of S .By previous analysis, it suffices to consider Ω m (0) = 1. Recall Ω m (0) = P J ∈C m (0) µ J (0). Byequation P J ∋ i,j µ J (0) = 1, we have µ J (0) = 0 for all J ∈ F m satisfying Θ " J . The choice of { µ J (0) } implies the existence of µ J (0) > J of S in the class C m (0).Thus H (cid:16) λ ij (0) = 0 , µ J (0) (cid:17) < . Combining above results, we conclude that there exists at least one solution to the system( V. { δ ij } be a solution to the system ( V. δ ij = 0 if either i or j does notlie in Θ. If P mi =2 α i,k +1 < n , then our claim (5.47) is true by setting β ij = α ij + δ ij . In the case P ki =2 α i,k +1 = n , the treatment is similar as (5.43) and we omit the details here.Now we turn our attention to the final case P mi =2 α i,k +1 > n . Then L equals X Θ | x i − x j | ! − α ,k +1 Z R n m Y i =2 (cid:12)(cid:12) x i − x k +1 (cid:12)(cid:12) − α i,k +1 dx k +1 . In this case, we shall adapt the treatment of the system ( V. V. ( i ) δ ij ≥ , ≤ i < j ≤ m ;( ii ) P Θ δ ij = α ,k +1 ;( iii ) P J ∩ Θ δ ij < (cid:16) ( | J | − n − B J (cid:17) ∧ (cid:18) | J | n − B J − P J np i (cid:19) , J ∈ F m , J = S ∪ { k + 1 } ;( iv ) P S ∪{ k +1 }∩ Θ δ ij ≤ (cid:0) kn − B S ∪{ k +1 } (cid:1) ∧ ( k + 1) n − B S ∪{ k +1 } − P S ∪{ k +1 } np i ! ;29here F m is the class of all subsets J ⊂ { , , · · · , k + 1 } with | J ∩ { , , · · · , m }| ≥ B J is given by, for each J ∈ F m , B J = X J (cid:16) α ij − δ i δ k +1 j α ,k +1 (cid:17) . The existence of a solution to ( V. 4) can be proved similarly. The argument can be outlinedas follows. For nonnegative λ ij , µ J , θ and θ satisfying λ ij − X J ∋ i,j µ J + ( θ − θ ) = 0 , ≤ i < j ≤ m, (5.54)where there exists one µ J > J of S ∪ { k + 1 } in the class F m , it isenough to show that( θ − θ ) α ,k +1 − X J ∈F m µ J (cid:16) ( | J | − n − B J (cid:17) ∧ (cid:16) | J | n − B J − X J np i (cid:17) < . (5.55)The above inequality is obvious for θ − θ ≤ 0. By scaling, we may assume θ − θ = 1. In thissetting, the argument is the same as the proof of existence of solutions to the system ( V. 2) and( V. δ ij = 0 for i / ∈ Θ or j / ∈ Θ. Let { δ ij : 1 ≤ i < j ≤ m } be a solution to the system ( V. α ij = (cid:0) α ij − δ i δ k +1 j α ,k +1 (cid:1) + δ ij for 1 ≤ i < j ≤ k + 1. Then the datum { α ij , p i } still satisfies assumptions in the theorem.Moreover, the left side integral in (5.47) with L in place of P Θ L i is bounded by a constantmultiple of Z R n ( k +1) k Y i =1 | f i ( x i ) | Y ≤ i 2. Of course, the existence of such nonnegative numbers { α ij } is obvious. For any given { α ij } satisfying this integrability condition, a natural question ariseswhether there is a set of positive numbers { p i } such that { α ij } and { p i } satisfy conditions ( i ),( ii ) and ( iii ) in Theorem 1.1. If this is true, we will obtain the local integrability of the integral(1.3). In fact, the answer is affirmative except some trivial cases. For example, the simplest casein which all α ij = 0 should be ruled out since the boundedness of Λ is valid only if all p i = 1.Moreover, we may further assume that for any given i not all α ij are zero with j ranging over { , , · · · , k + 1 } . More precisely, the existence of { p i } can be stated as follows.30 heorem 6.1 Assume α ij ≥ satisfy the system of inequalities ( ii ) in Theorem and X i ∈ J X j ∈ J c α ij > for any nonempty proper subset J of { , , · · · , k + 1 } . Then there exist infinitely many { p i } such that ( V I. ( i ) 1 < p i < ∞ , ≤ i ≤ k + 1;( ii ) k +1 P i =1 1 p i + P ≤ i We begin with discussing the necessity of the additional assumption (6.56) which doesnot lose generality. This assumption is only necessary to ensure the existence of a solution to thesystem ( V I. { α ij } satisfying P J α ij < ( | J | − n for all subsets J with | J | ≥ i there exists some α ij > j ∈ { , , · · · , k + 1 } . Bythis weaker assumption, we can divide { , , · · · , k + 1 } into disjoint subsets J , J , · · · , J l with | J i | ≥ α ij is positive with i < j only if( i, j ) ∈ l [ u =1 { ( s, t ) : s < t, s, t ∈ J u } . If each J i can not be decomposed further as above, i.e., P u ∈ I P v ∈ J i − I α uv > I of J i , then we can reduce matters to l multilinear functionals { Λ J i } of form(1.3). Therefore we may assume that { , , · · · , k + 1 } cannot be decomposed as above. This isequivalent to P i ∈ J P j ∈ J c α ij > J of { , , · · · , k + 1 } . Thusthe additional assumption (6.56) does not lose generality.Now we turn to verify the existence of { p i } . Define δ i = 1 /p i . As required in System ( V I. { δ i } should satisfy the following system of linear inequalities:( V I. ( i ) k +1 P i =1 δ i = k + 1 − P ≤ i 1. Our proof of the existence of { p i } needs Lemma2.4. It is clear that the system consisting of ( i ) and ( iv ) does have infinitely many solutions. Toapply Lemma 2.4, we assume that θ , θ , π i , µ J are nonnegative numbers satisfying( θ − θ ) + π i − X J ∋ i µ J = 0 , ≤ i ≤ k + 1 , (6.57)where either π i > i or µ J > J . By Lemma 2.4,it is enough to show( θ − θ ) (cid:16) k + 1 − X ≤ i 0, then the above inequality follows immediately by our choice of parameters. Nowwe consider the case θ − θ > 0. By scaling, we may assume θ − θ = 1. For any initial data { µ J (0) } and { π i (0) } satisfying X J ∋ i µ J = 1 + π i , we apply a process to obtain new data { µ J ( N ) } and { π i ( N ) } for which the objective functionin (6.58) increases as N with µ J ( N ) in place of µ J . Now we describe this process. Choose twononempty subsets J and J satisfy conditions ( a ) and ( c ) in (5 . b ) on J and J . Then put µ J t ( N ) = µ J t ( N − − µ J ( N − ∧ µ J ( N − , t = 1 , µ J ∩ J ( N ) = µ J ∩ J ( N − 1) + µ J ( N − ∧ µ J ( N − 1) if J ∩ J = ∅ µ J ∪ J ( N ) = µ J ∪ J ( N − 1) + µ J ( N − ∧ µ J ( N − 1) (6.59)and µ J ( N ) = µ J ( N − 1) for other nonempty subsets J of { , , · · · , k +1 } . Though this process isnot unique, it does not change the value of π i , i.e., π i ( N ) = π i (0), and equations in (6.57) are stilltrue with { µ J (0) } replaced by { µ J ( N ) } . Let µ ∗{ , ··· ,k +1 } ( N ) be the supremum of µ { , ··· ,k +1 } ( N )among all possible processes consisting of N continuous steps. Let { µ ∗ J ( N ) } be obtained by oneof those N continuous steps. Then µ ∗{ , ··· ,k +1 } ( N ) increases as N and { µ ∗ J ( N ) } has a uniformupper bound for each nonempty subset J . By passing to a subsequence { N t } , we obtain Cauchysequences { µ ∗ J ( N t ) } . Let µ ∗ J ( ∞ ) = lim t →∞ µ ∗ J ( N t ) , J ⊆ { , , · · · , k + 1 } . (6.60)Now we can show that µ ∗{ , ··· ,k +1 } ( ∞ ) is stable, i.e., any process described in (6.59) does notchange the value of µ J for the new initial data e µ J (0) = µ ∗ J ( ∞ ). Indeed, the union of all propersubsets J satisfying µ ∗ J ( ∞ ) > { , · · · , k + 1 } . Thus we get µ ∗{ , ··· ,k +1 } ( ∞ ) = 1 + min ≤ i ≤ k +1 π i . (6.61)Notice that the objective function in (6.58) increases as N with µ J replaced by µ J ( N ). Forthis reason, we see that the inequality (6.58) is true when there exists a positive π i . Indeed, ifsome π i is positive and µ ∗{ , ··· ,k +1 } ( ∞ ) = 1, then we obtain a µ ∗ J ( ∞ ) > J $ { , · · · , k + 1 } .Assume now all π i are zero. In this case, the argument is somewhat different and (6.56) willbe used. By our choice of the parameters in (6.57), it follows that there is a positive µ J = µ J (0)with J being a nonempty proper subset of { , · · · , k + 1 } . Thus µ { , ··· ,k +1 } (0) < 1. For all N , itis true that X J ∋ i µ ∗ J ( N ) = 1 , ≤ i ≤ k + 1 . Recall that µ ∗{ , ··· ,k +1 } ( N ) tends to 1. This implies that each sequence µ ∗ J ( N ) tends to zero forall proper subsets J . Choose a sufficiently large N such that µ ∗{ , ··· ,k +1 } ( N ) > − ε > µ { , ··· ,k +1 } (0)for small enough ε > 0. This observation shows that in the process, consisting of N continuoussteps, by which we obtain { µ ∗ J ( N ) } , there is a M -th step such that µ { , ··· ,k +1 } ( M ) is larger than32 { , ··· ,k +1 } ( M − X J µ J ( M ) | J | − X J α ij n ! < X J µ J ( M − | J | − X J α ij n ! . The argument may vary depending on whether µ { , ··· ,k +1 } ( N ) = 1. If µ { , ··· ,k +1 } ( N ) is equalto 1, we will obtain, using the notation A k = k + 1 − P { , ··· ,k +1 } α ij /n , A k − X J µ J (0) | J | − X J α ij n ! ≤ A k − X J µ J ( M − | J | − X J α ij n ! < A k − X J µ J ( M ) | J | − X J α ij n ! . Notice that we also have A k − X J µ J ( M ) (cid:16) | J | − X J α ij n (cid:17) ≤ A k − X J µ J ( N ) (cid:16) | J | − X J α ij n (cid:17) , where the process used to obtain µ J ( N ) is one of those N continuous steps such that µ { , ··· ,k +1 } ( N )equals µ ∗{ , ··· ,k +1 } ( N ) and µ { , ··· ,k +1 } ( M ) > µ { , ··· ,k +1 } ( M − 1) for some 1 ≤ M ≤ N . The de-sired inequality (6.58) follows from the assumption µ { , ··· ,k +1 } ( N ) = 1.If µ { , ··· ,k +1 } ( N ) < 1, we only have A k − X J µ J (0) | J | − X J α ij n ! < A k − X J µ J ( N ) | J | − X J α ij n ! . However, we may regard { µ ∗ J ( N ) } as new initial data satisfying (6.57) with all π i = 0 and θ − θ = 1 since there exist at least two positive µ ∗ J ( N ) for two nonempty proper subsets J .This implies A k − X J µ J ( N ) | J | − X J α ij n ! = A k − X J µ ∗ J ( N ) | J | − X J α ij n ! ≤ . Combining above results, we have completed the proof of (6.58). Thus the system ( V I. 2) has asolution.It remains to show that there are infinitely many solutions to the system ( V I. H bethe hyperplane in R k +1 given by k +1 X i =1 x i = k + 1 − X ≤ i 2) forms an open convex subset of H . Let { δ (1) i } and { δ (2) i } be twosolutions. Then it is clear that their convex combinations { λδ (1) i + (1 − λ ) δ (2) i } are also solutionsfor all 0 < λ < 1. Assume { δ i } is a solution to the system ( V I. ε > 0, all33oints in the ε -neighborhood of { δ i } in H are also solutions. Indeed, if ρ = ( ρ , · · · , ρ k +1 ) ∈ H and ( P k +1 i =1 | ρ i − δ i | ) / < ε, then ( ii ) and ( iii ) in ( V I. 2) are also true for { ρ i } with sufficientlysmall ε > 0. Thus we have established our claim. The proof is therefore concluded. ✷ As a corollary, we can apply Theorem 1.1 to obtain Theorem 5.1. Corollary 6.2 Assume α ij are nonnegative numbers for ≤ i < j ≤ k + 1 . Then for all f i ∈ C ∞ Z R n ( k +1) k +1 Y i =1 | f i ( x i ) | Y ≤ i BM O does not require this assumption. Now we prove the L ∞ estimate for T under theadditional assumption P ki =1 /p i ≥ 1. It suffices to show that there exists a constant C suchthat Z R nk k Y i =1 (cid:16) f i ( x i ) | x i | − α i,k +1 (cid:17) Y ≤ i 4, it is easy to see that the inequality (6.63) holds.For f i ∈ C ∞ , both Theorem 5.1 and Theorem 6.1 imply that T ( f , · · · , f k ) is locally inte-grable. For each cube Q with sides parallel to the axes, we use ∗ Q to denote the cube whichis concentric to Q but has the side length twice as long as that of Q . We first decompose T ,corresponding to Q , as a major term T S ( f , · · · , f k )( x k +1 ) = Z ( ∗ Q c ) k Y S f i ( x i ) Y { , ··· ,k +1 } | x i − x j | − α ij dV S and k terms of the form T i ( f , · · · , f k ) = Z ∗ Q Z ( R n ) k − Y S f i ( x i ) Y { , ··· ,k +1 } | x i − x j | − α ij dV S −{ i } dx i for i ∈ { , · · · , k } . Here dV J is the product Lebesgue measure Q j ∈ J dx j . As in § 4, for 1 ≤ i ≤ k we claim that Z Q | T i ( f , · · · , f k )( x k +1 ) | dx k +1 ≤ C | Q | Y S k f i k p i . i = 1 for example. Put p = (cid:16) ǫ ǫ + p (cid:17) − , p k +1 = 1 + ǫ, and p i = p i for 2 ≤ i ≤ k . Then { p i } and { α ij } satisfy conditions ( i ), ( ii ) and ( a ) of ( iii ) in Theorem 1.1 with small ǫ > 0. Thus Z Q | T ( f , · · · , f k ) | dx k +1 ≤ Λ( | f | χ ∗ Q , | f | , · · · , | f k | , χ Q ) ≤ C | Q | Y S k f i k p i . Similarly, we can show that the same estimate still holds for each T i ( f , · · · , f k ) with 2 ≤ i ≤ k .This also implies the local integrability of T ( f , · · · , f k ).Now it remains to show the average of | T S ( f , · · · , f k ) − T S ( f , · · · , f k ) Q | over Q is boundedby a constant multiple of Q S k f i k p i . Let Θ consist of those i ∈ { , · · · , k } such that α i,k +1 > | Q | Z Q (cid:12)(cid:12)(cid:12) T S ( f, · · · , f k )( x k +1 ) − T S ( f , · · · , f k ) Q (cid:12)(cid:12)(cid:12) dx k +1 ≤ C | Q | /n X t ∈ Θ Z ( ∗ Q c ) k k Y i =1 | g f ( t ) i ( x i ) | Y ≤ i 0, i.e., i ∈ Θ; δ i ≤ /p i + α i,k +1 /n, if α i,k +1 = 0, i.e., i / ∈ Θ;( iii ) k P i =1 δ i = k − P S α ij /n ;( iv ) P J δ i < | J | − P J α ij /n for nonempty proper J ⊆ { , · · · , k } . For i / ∈ Θ, we indeed have δ i = p i and include it here for convenience of notations. For i ∈ Θ,we see that g f ( t ) i ∈ L δ − i ( ∗ Q c ). To prove the existence of solutions to ( V I. X i ∈ J X j ∈ J c α ij > J of { , , · · · , k } . Here J c is the complement of J relative to { , , · · · , k } . If there were some nonempty proper J ⊆ { , · · · , k } such that all α ij = 0 for i ∈ J and j ∈ J c , we would obtain a contradiction. By assumptions in Theorem 1.2, we have X J p i + X J ∪{ k +1 } α ij n + 1 p k +1 < | J | + 135ith J = J and J = J c . Recall p k +1 = 1. The above two inequalities contradict the homogene-ity condition ( i ) in Theorem 1.2. Thus (6.65) is true.Suppose that u i , v i , θ , θ and µ J are nonnegative numbers satisfying u i − v i + ( θ − θ ) − X J ∋ i µ J = 0 , ≤ i ≤ k, (6.66)where either v i > i ∈ Θ or µ J > J of { , · · · , k } .Our task is to show X S u i p i − X S v i (cid:18) p i + α i,k +1 + δ ti n (cid:19) + ( θ − θ ) k − X S α ij n ! − X J ⊂ S µ J | J | − X J α ij n ! = − X J ⊂ S µ J | J | − X J p i − X J α ij n ! + ( θ − θ ) B S − X S v i α i,k +1 n − v t n < , (6.67)where B J = | J | − X J /p i − X J α ij /n for nonempty subsets J of { , · · · , k } . For convenience, we use H to denote the objectivefunction in the above inequality. Similarly, we may assume θ − θ = 1.Now we also need a process described as in (6.59). For nonempty subsets J and J of { , , · · · , k } satisfying conditions ( a ) and ( c ) in (5 . µ J i ( N ) = µ J i ( N − − µ J ( N − ∧ µ J ( N − , i = 1 , .µ J ∩ J ( N ) = µ J ∩ J ( N − 1) + µ J ( N − ∧ µ J ( N − 1) if J ∩ J = ∅ µ J ∪ J ( N ) = µ J ∪ J ( N − 1) + µ J ( N − ∧ µ J ( N − 1) (6.68)and µ J ( N ) = µ J ( N − 1) for other nonempty subsets J of { , · · · , k } . Here we also do not require J ∩ J = ∅ in the recursion. For any initial data { µ J (0) } , let { µ ∗ J ( N ) } be obtained by one ofthose processes consisting of N steps such that µ S ( N ) attains its maximum µ ∗ S ( N ). Since theprocess does not change the values of u i and v i , by passing to a subsequence { N t } , we can denotethe limit of µ ∗ J ( N t ) by µ ∗ J ( ∞ ). We can prove that µ ∗ S ( ∞ ) is stable with respect to the process(6.68). However, µ ∗ J ( ∞ ) may not be stable generally for J $ { , · · · , k } . Similarly, if { µ ∗ J ( ∞ ) } is regarded as new initial data, then we can apply the process (6.68) again to µ ∗ J ( ∞ ) for allproper subsets J . This procedure will continue if not all µ ∗ J ( ∞ ) are stable. For this reason, wecan assume that all µ ∗ J ( ∞ ) are stable. By this process, we also have − X J ⊂ S µ J (0) | J | − X J p i − X J α ij n ! ≤ − X J ⊂ S µ ∗ J ( ∞ ) | J | − X J p i − X J α ij n ! (6.69)which becomes a strict inequality if µ ∗ S ( ∞ ) > µ S (0) by the property (6.65). Let w i = u i − v i for1 ≤ i ≤ k . Let σ (1) , σ (2) , · · · , σ ( k ) be a permutation of 1 , , · · · , k such that w σ (1) ≥ w σ (2) ≥ · · · ≥ w σ ( k ) . Since all µ ∗ J ( ∞ ) are stable, it follows from (6.66) that we have µ ∗ J k ( ∞ ) = µ ∗{ , ··· ,k } ( ∞ ) = 1 + w σ ( k ) ,µ ∗ J k − i ( ∞ ) = w σ ( k − i ) − w σ ( k − i +1) , ≤ i ≤ k − , J k = { σ (1) , σ (2) , · · · , σ ( k ) } and J k − i = { σ (1) , σ (2) , · · · , σ ( k − i ) } for 1 ≤ i ≤ k − 1. Since p k +1 = 1, it follows from ( i ) in Theorem 1.2 that B S = P S α i,k +1 /n .Then the objective function H is equal to H = − k − X i =1 ( w σ ( i ) − w σ ( i +1) ) B J i − (1 + w σ ( k ) ) B S + B S − X S v i α i,k +1 n − v t n = − k − X i =1 ( w σ ( i ) − w σ ( i +1) ) B J i − (1 + w σ ( k ) ) B S + X S (cid:16) ( u i − v i ) + 1 (cid:17) α i,k +1 n − X S u i α i,k +1 n − v t n = − k − X i =1 ( w σ ( i ) − w σ ( i +1) ) B J i − (1 + w σ ( k ) ) B S + X S X J ∋ i µ ∗ J ( ∞ ) ! α i,k +1 n − X S u i α i,k +1 n − v t n = − k − X i =1 ( w σ ( i ) − w σ ( i +1) ) B J i − (1 + w σ ( k ) ) B S + X J ⊂ S µ ∗ J ( ∞ ) X J ∋ i α i,k +1 n ! − X S u i α i,k +1 n − v t n . Hence we have H = − k − X i =1 ( w σ ( i ) − w σ ( i +1) ) B J i + k − X i =1 ( w σ ( i ) − w σ ( i +1) ) X J i ∋ j α j,k +1 n − X S u i α i,k +1 n − v t n = − k − X i =1 ( w σ ( i ) − w σ ( i +1) ) B J i ∪{ k +1 } − X S u i α i,k +1 n − v t n . Since w σ ( i ) − w σ ( i +1) ≥ B J i ∪{ k +1 } > ≤ i ≤ k − 1, the desired inequality H < w σ ( i ) − w σ ( i +1) > ≤ i ≤ k − 1. Assume now w i = w j for all i, j in { , , · · · , k } . Then we have H = − X { , ··· ,k } u i α i,k +1 n − v t n . If w = u − v > 0, then all u i are positive and hence H < 0. If w = u − v = 0, then u i = v i for i ∈ { , · · · , k } . If there is a positive u i for those i such that α i,k +1 > 0, we also get H < 0. Otherwise all u i and v i are zero. By the choice of u i , v i and µ J , we see that there existsa µ J (0) > J $ { , · · · , k } . However, (6.69) becomes a strict inequality in thiscase. Therefore we also obtain H < 0. 37or each t ∈ Θ, let { δ i ( t ) : 1 ≤ i ≤ k } be a solution of the system ( V I. | Q | Z Q (cid:12)(cid:12)(cid:12) T S ( f, · · · , f k )( x k +1 ) − T S ( f , · · · , f k ) Q (cid:12)(cid:12)(cid:12) dx k +1 ≤ C | Q | /n X t ∈ Θ Y S k g f ( t ) i χ ∗ Q c k δi ( t ) ≤ C Y S k f i k p i . This proves that T has a bounded extension from L p × · · · × L p k into BM O . In this section, we shall prove the claim in Remark 5.1 in § 5. Indeed, there is an analogue ofTheorem 2.2 on the sphere S n = { x ∈ R n +1 : | x | = 1 } . For ξ, η ∈ S n , we shall use | ξ − η | todenote the standard Euclidean metric between ξ and η in R n +1 . Theorem 7.1 Assume that < α i < n for ≤ i ≤ k satisfy P ki =1 α i > n . We have thefollowing estimate: Z S n k Y i =1 (cid:12)(cid:12) ξ − ξ i (cid:12)(cid:12) − α i dσ ( ξ ) ≤ C k X u =1 L u ( ξ , ξ , · · · , ξ k ) , ξ i ∈ S n for 1 ≤ i ≤ k, where dσ is the Lebesgue measure on the S n . Here L u is defined by L u ( ξ , ξ , · · · , ξ k ) = ( d n − P S α i S (cid:16) χ { P S −{ u } α i Proof. If there exist two points ξ i and ξ j such that | ξ i − ξ j | ≥ 1, then the integral in the lemma isfinite with a bound depending only on α i and the dimension n . Now we assume that | ξ i − ξ j | ≤ ≤ i < j ≤ k . By rotation, we assume ξ = (0 , · · · , , − Z ξ ∈ S n , | ξ − ξ |≤√ k Y i =1 | ξ − ξ i | − α i dσ ( ξ ) ≤ C log Cd S . (7.71)38et π be the stereographic projection from R n onto the sphere S n minus the north pole. Then π ( x , x , · · · , x n ) = (cid:18) x | x | , x | x | , · · · , x n | x | , | x | − 11 + | x | (cid:19) , x = ( x , x , · · · , x n ) ∈ R n . The inverse of π is given by π − ( ξ , ξ , · · · , ξ n +1 ) = (cid:18) ξ − ξ n +1 , ξ − ξ n +1 , · · · , ξ n − ξ n +1 (cid:19) for ( ξ , ξ , · · · , ξ n +1 ) ∈ S n − { e n +1 } , where e n +1 is the north pole of S n . By direct computations,we have the following properties: | π ( x ) − π ( y ) | = 2 | x − y | p | x | p | y | , x, y ∈ R n , | π − ( ξ ) − π − ( η ) | = | ξ − η | p − ξ n +1 √ − η n +1 = 2 | ξ − η || e n +1 − ξ || e n +1 − η | , ξ, η ∈ S n − { e n +1 } . Define π ( x ) = ξ and π ( x i ) = ξ i for 1 ≤ i ≤ k . Then the integral in (7.71) is equal to Z B (0) k Y i =1 | x − x i | p | x | p | x i | ! − α i (cid:18) 21 + | x | (cid:19) n dx ≈ Z B (0) k Y i =1 | x − x i | − α i dx since all x i ∈ B (0) ⊂ R n . By previous assumptions, we see that | ξ i − ξ j | are comparable to | x i − x j | . Thus we can apply Lemma 2.3 to obtain the desired estimate. The proof is complete. ✷ By Lemma 7.2, we can prove Theorem 7.1 now. Proof. Without loss of generality, we assume L = max S | ξ i − ξ j | = | ξ − ξ k | > 0. Since thedesired estimate is rotation invariant, we may assume that ξ is the south pole on S n . Dividethe integral in the theorem into two parts as in the proof of Theorem 2.2. We first consider theintegral over | ξ − ξ | ≤ L/ Estimate of the integral over | ξ − ξ | ≤ L/ It is easy to see that Z ξ ∈ S n , | ξ − ξ |≤ L/ k Y i =1 | ξ − ξ i | − α i dσ ≤ Cd − α k S Z ξ ∈ S n , | ξ − ξ |≤ L/ k − Y i =1 | ξ − ξ i | − α i dσ. Case (i): P k − i =1 α i < n .We claim that the right side of the above inequality is bounded by a constant multiple of L n − P S α i . For L ≥ 1, the claim is obvious since Q k − i =1 | ξ − ξ i | − α i is integrable on S n . Now assume L < 1. Then we can use the stereographic projection π to change variables as in the proof ofLemma 7.2. Our claim follows from the observation that | π − ( ξ ) − π − ( η ) | is comparable to | ξ − η | for all ξ, η ∈ { ξ ∈ S n : | ξ − ξ | ≤ L } .Case (ii) P k − i =1 α i = n .We shall prove that the integral of Q k − i =1 | ξ − ξ i | − α i over { ξ : | ξ − ξ | < L/ } is bounded by aconstant multiple of log(2 d S /d S −{ k } ). Assume | ξ i − ξ | ≤ L/ ≤ i ≤ k − 1. Otherwise39f | ξ i − ξ | > L/ ≤ i ≤ k − 1, then it is clear that Z S n : | ξ − ξ |≤ L/ k − Y i =1 | ξ − ξ i | − α i dσ ≤ CL − α i Z S n : | ξ − ξ |≤ L/ k − Y i =1 ,i = i | ξ − ξ i | − α i dσ ≤ CL − α i L n − P i = i ,k α i which is bounded by a constant since P k − i =1 α i = n . Therefore we can now assume | ξ i − ξ | ≤ L/ ≤ i ≤ k − 1. Note that 2 L/ ≤ / < √ 2. Using the stereographic projection π again, wedefine π ( x ) = ξ and π ( x i ) = ξ i for 1 ≤ i ≤ k − 1. Then Z S n : | ξ − ξ |≤ L/ k − Y i =1 | ξ − ξ i | − α i dσ ≤ C Z | x |≤ CL k − Y i =1 | x − x i | − α i dx ≤ C log CL P S −{ k } | x i − x j | ! . Since L ≈ d S and | x i − x j | ≈ | ξ i − ξ j | , the desired estimate follows.Case (iii): P k − i =1 α i > n .In this case, the desired estimate is obvious. Estimate of the integral over | ξ − ξ | > L/ Case (i): P ki =2 α i > n .As above, the desired estimate is obvious. Indeed, we have Z S n : | ξ − ξ | >L/ k Y i =1 | ξ − ξ i | − α i dσ ≤ CL − α Z S n k Y i =2 | ξ − ξ i | − α i dσ. Case (ii): P ki =2 α i ≤ n and L ≥ ≤ L ≤ Z S n : | ξ − ξ | >L/ k Y i =1 | ξ − ξ i | − α i dσ ≈ Z S n : | ξ − ξ | >L/ k Y i =2 | ξ − ξ i | − α i dσ. Then the desired estimate is true since the integral of Q ki =2 | ξ − ξ i | − α i over { S n : | ξ − ξ | > L/ } is bounded by a constant. If P ki =2 α i < n , then the desired estimate is true. In the case P ki =2 α i = n , our estimate is also true by Lemma 7.2.Case (iii): P ki =2 α i ≤ n and L < < α < n , we see that P S α i < n . Then by the stereographic projection π , wedefine π ( x ) = ξ and π ( x i ) = ξ i for 1 ≤ i ≤ k . Since ξ is the south pole and | ξ i − ξ | ≤ L < c and c , independent of L , such that | x i | ≤ c L and | x | ≥ c L for | ξ − ξ | > L/ 2. Then Z S n : | ξ − ξ | >L/ k Y i =1 | ξ − ξ i | − α i dσ ≤ Z | x |≥ c L k Y i =1 | x − x i | p | x | p | x i | ! − α i (cid:18) 21 + | x | (cid:19) n dx ≤ C Z | x |≥ c L k Y i =1 | x − x i | − α i dx ≤ CL − α Z c L ≤| x |≤ L k Y i =2 | x − x i | − α i dx, P ki =1 α i < n in the second inequality. By a similar argument asin our treatment of the integral over { S n : | ξ − ξ | ≤ L/ } , we can also obtain the desiredestimate. 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