SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION

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1 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION LOUKAS GRAFAKOS AND NIGEL KALTON Abstract A multilinear version of the Boyd interpolation theorem is proved in the context of quasi-normed rearrangement-invariant spaces A multilinear Marcinkiewicz interpolation theorem is obtained as a corollary Several applications are given, including estimates for bilinear fractional integrals 1 Introduction In this article we give a version of the Boyd interpolation theorem for multilinear operators We will be working with rearrangement invariant quasi-banach spaces, which include all the well-known examples such as Orlicz spaces and Lorentz spaces We will consider the following situation Consider R + = (, ) with Lebesgue measure (which can of course be replaced by any infinite nonatomic measure space) We let L (, ) be the space of all real-valued measurable functions equipped with the topology of local convergence in measure Let E be the space of all measurable functions which are bounded and supported on sets of finite measure Now let T : E n L (, ) be a multilinear map (our results also apply to sublinear maps) We suppose that T is locally continuous ie continuous when restricted to n L (E k ) for every choice of sets E k of finite measure We also suppose that T obeys a finite collection of weak type inequalities T (χ E1,, χ En ) Lp, C E k θ k for every n-tuple of measurable sets (E 1,, E n ) Here L p, is the usual weak L p space and θ k > for every k We then seek to characterize (n + 1)-tuples of rearrangementinvariant spaces (X 1,, X n, Y ) for which T extends to a bounded n-linear map from X 1 X n into Y In general one needs two distinct hypotheses The first consists of an assumption on the Boyd indices of the spaces X 1,, X n, Y, as in the original Boyd interpolation theorem The second hypothesis is that a certain n-linear test map associated with T is continuous Our main result (Theorem 41) gives a necessary (and often sufficient) condition on (X 1,, X n, Y ) in the case when one has n + 1 such conditions which are sufficiently Date: September 17, Mathematics Subject Classification Primary 46B7 Secondary 46E3, 42B99 Key words and phrases Boyd interpolation theorem, multilinear operators Research of both authors was partially supported by the NSF 1

2 2 LOUKAS GRAFAKOS AND NIGEL KALTON independent Note that the original theorem of Boyd [2] corresponds to the case when n = 1 and there are two conditions of the type: T (χ E ) Lp, C E 1/p We deduce Theorem 41 from a similar homogeneous Boyd-type theorem (Theorem 37) which is applicable for example to n-linear forms As a corollary we obtain a multilinear version of the Marcinkiewicz interpolation theorem (Theorem 46) Our work is related to work of Strichartz [17], Janson [5], and Christ [3] Note that as in [5] and [17] (and in contrast to [3]) our multilinear assumptions consist only of a finite number of estimates Our results also develop and extend earlier work of Sharpley (see [15], [16], and [1]) In section 5 we give examples of multilinear interpolation As one of our applications, we characterize the indices (1/p, 1/q, 1/r), < p, q, r, for which the bilinear fractional integral operator I α (f, g)(x) = f(x + t)g(x t) t α n dt R n maps L p (R n ) L q (R n ) L r (R n ) This characterization was also independently obtained by C Kenig and E M Stein [1] 2 Preliminaries In this section we set up the background required to state the multilinear Boyd interpolation theorem Let L (, ) be the space of all complex-valued measurable functions on (, ), with the topology of local convergence in measure We define a quasi-banach function space X on (, ) to be a subspace of L equipped with a quasi-norm X such that: f X = if and only if f = ae αf X = α f X, whenever f X and α C There exists a constant C so that if f, g X then f +g X C( f X + g X ) X is complete (ie a quasi-banach space) for X The injection X L is continuous If E is a set of finite measure then χ E X If f X and g L with g f ae then g X and g X f X If f n f ae and f X then f n X f X By assumption X must contain the space E of all bounded measurable functions supported on sets of finite measure We say that X is minimal if E is dense in X We say that X is maximal if it has the property that if f n f ae with sup f n X <, then f X A quasi-banach function space on (, ) which is either maximal or minimal (cf [12]) is said to be a rearrangement-invariant function space or ri space if f X = f X for all f X, where f is the decreasing rearrangement of f, ie f (t) = inf{x : { f > x} t}

3 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 3 We say that X is r-convex if there is a constant C so that if f 1,, f n X then n n ( f i r ) 1/r X C( f i r X) 1/r i=1 For a discussion of r-convexity in the context of Banach lattices we refer to [12]; we refer to [7] for quasi-banach lattices Every Banach ri space is of course 1-convex, but there are examples of quasi-banach ri spaces which fail to be r-convex for any r >, see [6] However it is very natural to assume r-convexity since all practical spaces are r-convex for some r > Once an ri space X is defined on (, ) it may be transferred to any σ-finite measure space (Ω, µ) by defining X(Ω, µ) to be the space of all measurable f : Ω C such that f X(Ω) = f X(, ) < In general if Ω is a Polish space and µ is an infinite nonatomic Borel measure there is a measure-preserving bijection of Ω onto (, ) Thus there is no loss of generality in treating only the case of Ω = (, ) If X is an ri space then the dilation operators D a : X X given by i=1 (D a f)(x) = f(x/a) are well-defined and bounded We define the Boyd indices by log a p X = lim a log D a and log a q X = lim a log D a Then < p X q X We refer to [12] or [1] for relevant discussion If ɛ > then there is a constant C = C(ɛ, X) so that for all f X we have (1) D a f X C max(a 1 p X +ɛ, a 1 q X ɛ ) It is sometimes useful to have the notion of a carrier space for an ri space X Let X be a maximal quasi-banach function space on (, ) with the property that the dilation operators D a are bounded on X and D a ex Ca κ for some κ > and all a 1 Then we can define an ri space X by requiring f X if and only if f X and by setting f X = f ex It is then easy to show that X is a maximal ri space and that α X κ We will in this case refer to X as a carrier space for X Notice, of course, that X is a carrier space for itself Examples of ri spaces are provided by the usual Lorentz spaces L p,q with (quasi)- norm ( [f (t)t 1/p ] q dt ) 1/q when < q <, (2) f Lp,q = t sup t> f (t)t 1/p when q = for < p, q These spaces are 1-convex (ie normable) when 1 < p < and 1 q or if p = q = 1 In general L p,q is q-convex if q p and s-convex for any s < p if q > p The Boyd indices of L p,q both coincide with p Note that all these spaces have natural carrier spaces which are weighted L p spaces

4 4 LOUKAS GRAFAKOS AND NIGEL KALTON The significance of the Boyd indices lies in the fact that they can be used to characterize all rearrangement-invariant Banach spaces X on which certain known operators are bounded For instance the Hardy-Littlewood maximal operator is bounded on X (ri over R n ) if and only if q X <, see [13], [18] The Hilbert transform is bounded on X (ri over R) if and only if 1 < p X q X <, see [2] Let us now recall the Boyd interpolation theorem for (, ) (see [2] or [12], p145): Theorem 21 Suppose 1 p < q < and that T : L p,1 +L q,1 L (, ) is a linear map of weak types (p, p) and (q, q) Suppose X is an ri space with p < p X q X < q Then T is a bounded map from X into itself This result was extended to the case < p < q < in [8] (Theorem 13) with the additional assumption that X is r-convex for some r > The main purpose of this article is to obtain a multilinear version of Theorem 21 This is achieved in the next two sections We first obtain a homogeneous multilinear version of Theorem 21 (Theorem 37), and from this we deduce an inhomogeneous version, Theorem 41 3 The homogeneous multilinear Boyd theorem Let E be the space of all measurable functions on (, ) which are bounded and have support of finite measure We shall say that a map (usually n-linear) T : E n Y in any topological vector space is locally continuous if its restriction to n L (E k ) is continuous for every choice of sets E k of finite measure Now suppose Θ is a finite subset of (R + ) n = [, ) n and Y is a quasi-banach space We say that an n-linear map T : E n Y is Θ admissible if T is locally continuous and there is a constant M so that for every θ = (θ 1,, θ k ) Θ we have (3) T (χ E1,, χ En ) Y M E k θ k, whenever E 1,, E n have finite measure The least such constant M is denoted by T Θ In most of the work that follows, it will be convenient to take Θ R n + ie to require θ k > for all θ, k Let us recall that a quasi-banach space (Y, ) is called s-normed if there is a constant C such that for all y 1,, y m Y we have y y m s C( y 1 s + + y m s ) Now let X = (X 1,, X n ) be an n-tuple of ri spaces We say that X has the interpolation condition (Θ, s), where < s 1, if for every s-normed quasi-banach space Y and every Θ-admissible T : E n Y there is a continuous extension T : X 1 X n Y with norm a constant multiple of T Θ Note here that in the case s = 1 it is sufficient to take Y to be the scalar field R or C and hence we only consider n-linear forms We will need to establish some examples of Θ-admissible multilinear maps We begin with a lemma

5 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 5 Lemma 31 Suppose < u < and < s < r < Then for any measurable set E (, ) we have ( 1/s ( 1/r su x dx) su 1 ru x dx) ru 1 E E In particular if su < 1 then ( 1/s su x dx) su 1 E u E Proof First note that for t > 1, we have (t su 1) 1/s (t ru 1) 1/r and also that (t su 1) 1/s (t ru 1) 1/r is increasing This last fact follows from the observation that t 1 r log(tr 1) 1 s log(ts 1) is monotone decreasing and converges to zero at infinity This implies that if E is an interval we have the desired inequality We now proceed to prove the result for E a disjoint union of m intervals using induction Assume the required inequality is true for all unions of less than m disjoint intervals Now if E is a finite union of m disjoint intervals [v j, w j ) for 1 j m where v 1 < w 1 < < v m < w m, we define h > w m 1 by the condition that If we had h su w su m 1 = (w su m v su m ) + (w su m 1 v su m 1) (4) h ru w ru m 1 (w ru m v ru m ) + (w ru m 1 v ru m 1), then the inductive hypothesis applied to the m 1 intervals [v 1, w 1 ),, [v m 2, w m 2 ), and [v m 1, h) together with (4) would quickly give the desired conclusion It suffices therefore to prove (4) This will follow from the fact that if α, β, γ, and δ are positive numbers satisfying α + γ = β + δ and β < γ < δ, then α r/s + γ r/s β r/s + δ r/s when r > s Indeed, the assumptions above imply that β < α < δ and clearly α r/s + γ r/s max α (β,δ) ( α r/s + (β + δ α) r/s) β r/s + δ r/s Let, denote the usual inner product on R n and the usual Euclidean norm For each θ R n let θ k denote its k th coordinate Suppose Θ is a finite subset of (R + ) n = [, ) n Define a sublinear map associated with Θ as follows (5) a(ξ) = a Θ (ξ) = max ξ, θ θ Θ Let X = (X 1,, X n ) be an n-tuple of ri spaces We have the following theorem See also Sharpley [15] for a somewhat similar result Theorem 32 Let < s 1 Consider the statements: (i) X satisfies the interpolation condition (Θ, s) (ii) There exists a constant C so that if f 1,, f n E, ( 1/s (6) (fk (e ξ k )) exp( sa( ξ))dξ) s C R n f k Xk

6 6 LOUKAS GRAFAKOS AND NIGEL KALTON (iii) There exists a constant C so that if f 1,, f n E, ( 1/s (7) (F k (e ξ k )) exp( sa( ξ))dξ) s C where R n F k (x) = ( 1 x x (f k (t)) s dt) 1/s (iv) There exists a constant C so that if f 1,, f n E, then ( 1/s (8) max (fk (e tξ k )) exp( sta( ξ))dt) s C ξ =1 f k Xk, f k Xk Then (ii) implies (i) Furthermore, if Θ R n + and s is small enough so that sθ k < 1 for every (θ 1, θ k ) Θ and every 1 k n, then (i), (ii), (iii), and (iv) are all equivalent Proof First assume (ii) and that T : E n Y is Θ admissible where Y is s-normed Without loss of generality we assume T Θ 1 We first note that if f k are supported in E k and f k L 1 then we have an estimate: (9) T (f 1,, f n ) Y C min θ Θ E k θ k, where C depends only on s and n To see this it suffices to get an estimate for positive functions f k and then extend to signed and complex functions by additivity But if f k is positive we can write f k = 2 j χ Ajk where A jk E k Expanding out we easily get estimate (9) Now suppose f 1,, f n E We can write each f k in the form f k = f k χ Akm j=1 m= where A km = 2 m and f k χ Akm L f (2 m ) Now by (9) we have T (f 1 χ A1m1,, f n χ Anmn ) Y C min θ Θ 2Pn θ km k n f k (2 m k ) Now since Y is an s norm after summing and making an obvious integral estimate we obtain T (f 1,, f n ) s Y C (f1 (x 1 )) s (fn(x n )) s min x sθ k 1 k dx 1 dx n θ Θ

7 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 7 The right-hand side is now estimated by C n f k Xk using (6) We can now extend the definition of T (f 1,, f k ) to X 1 X n by noting that for f k X k the sum m 1,,m n T (f 1 χ A1m1,, f n χ Anmn ) converges in Y It is easy to check that this extends T unambiguously and continuously to X 1 X n Thus (i) holds Now assume (i), Θ R n +, and sθ k < 1 for all θ Θ and 1 k n For each θ Θ Lemma 31 gives that if f k = χ E E, then ( 1/s x sθk 1 f k (x) dx) s (sθ k ) 1/s E θ k, where as usually θ k denotes the k th coordinate of θ It follows that if we define T θ (f 1,, f n )(x 1,, x n ) = x θ k 1 s k f k (x k ), then T θ : E n L s ((, ) n ) is {θ}-admissible and T θ {θ} s n/s n define T by s T (f 1,, f n )(x 1,, x n ) = (min k )f k (x k ) 1 θ Θ xθ k θ 1/s k If we then T is Θ admissible It follows that we can find C so that (6) is valid and thus (ii) holds We now show that (ii) implies (iii) Observe that Hence (1) = R n R n (F k (e ξ k )) s = ξk e η k ξ k (f k (e η k )) s dη k (F k (e ξ k )) s exp( sa( ξ))dξ η ξ (fk (e η k )) s exp( η ξ, 1 sa( ξ)) dη dξ, where η ξ means η k ξ k for 1 k n and 1 denotes the vector (1, 1,, 1) For fixed η pick θ Θ so that a( η) = θ, η Then exp( ξ, 1 sa( ξ))dξ exp( ξ, 1 s θ )dξ C exp( η, 1 sa( η)) ξ η ξ η since for some δ > we have 1 sθ k > δ > for all θ Θ and all k Substituting back into (1) gives the required estimate Next we show (iii) implies (iv) To do this it suffices to note that the map ξ n log F k(e ξ k ) is Lipschitz with a constant depending only on s and n unless some

8 8 LOUKAS GRAFAKOS AND NIGEL KALTON f k is zero This means that if ξ = 1 we have an estimate: (F k (e tξ k )) s exp( sta( ξ))dt C (F k (e ξ k )) s exp( sa( ξ))dξ K where K is a cylinder of radius one with axis {tξ : t } We can then deduce (8) from (7) since f k F k Finally (iv) implies (iii) (which implies (ii)) by just using polar coordinates We can extend this result somewhat to certain multilinear analogues of maximal operators Denote by L + (, ) the set of all nonnegative measurable functions on (, ) Let us say that a positively homogeneous (of degree 1) map T : E + n L + (, ) is n-quasi-sublinear with constant C so that for any k we have (11) T (f 1,, f k 1, (f k +f k), f k+1,, f n ) C(T (f 1,, f k 1, f k, f k+1,, f n )+T (f 1,, f k 1, f k, f k+1,, f n )), for all f j, f j Suppose T is n-quasi-sublinear Then if we choose r so that 2 1/r 1 = C we can use the proof of the Aoki-Rolewicz theorem ([9],[14]) to deduce the existence of a constant C so that for any 1 k n and all m positive integers we have m m T (f 1,, f k 1, g j, f k+1,, f n ) C ( T (f 1,, f k 1, g j, f k+1,, f n ) r ) 1/r, j=1 for all f j and g j Based on this it is easy to show the following, by exactly the same argument as in Theorem 32: Corollary 33 Suppose T : E n L s (, ) is n-quasi-sublinear with constant C = 2 1/r 1 where < s r, and that T is locally continuous Let Θ be a finite subset of R n + and assume that there is a constant M so that T (χ E1,, χ En ) Ls M inf E k θ k, j=1 θ Θ for all E k of finite measure If X is an n-tuple of ri spaces satisfying (6), then there is a constant C so that for f 1,, f n E we have T (f 1,, f n ) Ls CM f k Xk We now consider versions of the Boyd interpolation theorem in this setting Consider the convex hull co Θ; We define the open convex hull co Θ to be the set of all θ Θ α θθ where < α θ < 1 and θ Θ α θ = 1 Then co Θ is the interior of co Θ relative to the affine hyperplane it generates We also define the Boyd cube B X of X to be the set n [1/q X k, 1/p Xk ], where p Xk, q Xk are the Boyd indices of X k It will be convenient to introduce the following sublinear functional associated with B X (12) b(ξ) = b X (ξ) = max φ B X ξ, φ

9 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 9 Let us first note a simple consequence of Theorem 32 Corollary 34 Suppose that Θ is a finite subset of (R + ) n and that X satisfies the (Θ, s)-interpolation condition Then B X co Θ is nonempty Proof Suppose B X, co Θ do not intersect Then we can find η R n so that max η, θ < min η, φ θ Θ φ B X Thus a(η) = b( η) 2δ where δ > Now we refer to (1) to obtain for f E, D e tη k fk Xk C exp(tδ + tb( η)) fk Xk for t It follows from (6) that ( 1/s (fk (e ξ k+tη k )) exp( sa( ξ))dξ) s C exp(tδ + tb( η)) R n = C exp( ta(η) tδ) f k Xk Now a( ξ) + ta(η) a( ξ + tη) Thus we can reorganize to obtain ( 1/s (fk (e ξ k )) exp( sa( ξ))dξ) s C exp( tδ) f k Xk, R n for every t which is absurd f k Xk Theorem 35 Suppose Θ is a finite subset of (R + ) n Suppose X is an n-tuple of ri spaces such that B X co Θ is a nonempty subset of co Θ Then X satisfies the (Θ, s) interpolation condition provided there is a constant C so that if f 1,, f n E, ( 1/s (13) (fk (e ξ k )) exp( sa( ξ))dξ) s C f k Xk, H where H is the subspace of R n of all ξ such that ξ, θ is constant for all θ Θ If Θ R n + and sθ k < 1 for every θ Θ and 1 k n, then inequality (13) is also necessary for X to satisfy the (Θ, s)-interpolation condition Proof Let B ɛ = {ξ : d(ξ, B X ) ɛ} Our assumption on B X and a compactness argument give the existence of ɛ > so that B 2ɛ P co Θ, where P is the affine plane generated by Θ We note that: (14) max η, φ = inf a(ξ) + b(η ξ) + 2ɛ η ξ φ B 2ɛ P ξ H To see this observe that the right hand side obviously dominates the left-hand side and is a sublinear functional It is easy to check that if η, φ is dominated by the right-hand side then we have φ B 2ɛ P

10 1 LOUKAS GRAFAKOS AND NIGEL KALTON We will also need (1) which implies that if f 1,, f n E, then D e η k f k Xk C exp(b(η) + ɛ η ) f k Xk for some constant C Now suppose η H Then for a fixed ζ H and f 1,, f n E we have ( ) 1/s (fk (e ξ k+η k )) s exp( sa( ξ η))dξ = ( H H (fk (e ξ k+η k +ζ k )) s exp( sa( ξ ζ) sa( η))dξ exp( a( η) + a(ζ)) exp( a( η) + a(ζ)) ( H ) 1/s (fk (e ξ k+η k +ζ k )) s exp( sa( ξ))dξ D e η k ζ k f k Xk C exp( a( η) + a(ζ) + b( η ζ) + ɛ η + ζ ) f k Xk At this point we use (14) and the fact that B 2ɛ P co Θ to show that ( ) a(ζ) + b( η ζ) + ɛ η + ζ a( η) ɛ η inf ζ H Thus we conclude that ( 1/s (fk (e ξ k+η k )) s exp( sa( ξ η))dξ) C exp( ɛ η ) H ) 1/s f k Xk Raise to the s th power and integrate over η H to obtain (6) Hence X satisfies the (Θ, s)-interpolation condition The last statement follows from (8) We now specialize to the case when Θ is a relatively large subset of (R + ) n Let us define the dimension of Θ, denoted dim Θ, to be the dimension of the affine plane passing through all the points in Θ We say that Θ is affinely independent if the conditions λ θ θ = and λ θ = θ Θ imply λ θ =, θ Θ Obviously if Θ is affinely independent we have Θ = 1+dim Θ Theorem 36 Suppose dim Θ = n (eg if Θ is an affinely independent subset of (R + ) n and Θ = n + 1) Suppose X is an n-tuple of ri spaces such that B X co Θ is a nonempty subset of co Θ Then X satisfies the (Θ, s) interpolation condition θ Θ Proof In this case Theorem 35 applies with H = {}

11 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 11 A more important case is the following: Theorem 37 Suppose that dim Θ = n 1 and that Θ spans R n Let < s 1 and suppose that X is an n-tuple of ri spaces such that B X co Θ is a nonempty subset of co Θ Pick a unique σ = (σ k ) n so that σ, θ = 1 for all θ Θ Consider the following statements: (i) X satisfies the (Θ, s) interpolation condition (ii) There is a constant C such that if f 1,, f n E we have ( n 1/s (15) x s 1 (fk (x σ k )) dx) s C f k Xk Then (ii) implies (i) Moreover, if Θ R n + and sθ k < 1 for every θ Θ and 1 k n, then (i) and (ii) are equivalent Furthermore if Θ R n + and if s( n θ k) 1 for every θ Θ, (i) and (ii) are also equivalent to: (iii) There is a constant C so that if f 1,, f n E, (16) ( x ) 1/s s 1 f k (x σ k ) s dx C f k Xk σ k σ k Remarks The existence of σ follows from the fact that the plane generated by Θ does not contain the origin Note that the indices k for which σ k = become redundant in the sense that that (15) can be rewritten as ( x ) 1/s s 1 (fk (x σ k )) s dx C f k Xk σ k Before we prove Theorem 37, let us illustrate the hypothesis on the Boyd indices, by considering the special but rather typical case when co Θ is the intersection of a cube n [α k, β k ] with the plane n θ k = r 1 In this case σ k = r for all k It may then easily be seen that the hypotheses on the Boyd indices are satisfied if we have both n 1 (17) 1 n q Xk r 1 and (18) α k < 1 q Xk 1 p Xk for all 1 k n However if for some l we have (19) α l + k l 1 q Xk p Xk < β k > 1 r σ k

12 12 LOUKAS GRAFAKOS AND NIGEL KALTON then the lower bound condition on q 1 X l (2) β l + k l then the upper bound condition on p 1 X l in (18) can be removed Similarly if 1 p Xk < 1 r can be removed Proof The fact that (ii) implies (i) is an application of Theorem 35 Indeed, in this case H is one-dimensional, say H = {tσ} t R Then equation (13) becomes ( 1/s + (fk (e tσ k )) s e dt) st C f k Xk which reduces to (15) by substituting x = e t The converse statement follows from Theorem 35 We now prove that (i) implies (iii) under the extra hypothesis s( n θ k) 1 for every θ Θ Define a map T : E n L s ((, ) (, 1)) by setting T (f 1,, f n )(x, y) = x 1 1/s f l (y) σ k f k (x σk ) σ l = We will show that T is Θ admissible Suppose (E k ) are sets of finite measure Let F = {x : x σ k E, σk } and let G = [, 1] σk =E k Then we have and therefore T (χ E1,, χ En )(x, y) = x 1 1/s χ F (x)χ G (y) ( T (χ E1,, χ En ) = F x s 1 dx) 1/s G 1/s Now suppose θ Θ Let r = ( σ k θ k) 1 Clearly s r We have by Lemma 31 ( ) 1/s ( ) 1/r x s 1 dx r 1/r s 1/s x r 1 dx F r 1/r s 1/s j J r 1/r s 1/s k J F ( On the other hand since G 1 and σ k θ k s 1, G 1/s E k θ k σ k F ) θk x σk 1 dx σ k θ k E k θ k Thus T is Θ admissible and hence T extends to a bounded n-linear form on X 1 X n Letting f k = χ [,1] if σ k = and restricting gives (16) Now it is clear that (iii) implies (ii) and so the proof is complete

13 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 13 Corollary 38 Suppose under the hypotheses of Theorem 37 we also have that for some fixed r n θ k = 1 r for every θ Θ Then X has the (Θ, s) interpolation condition if ( n 1/s (21) x s/r 1 (fk (x)) dx) s C f k Xk In particular if r = s then X has the (Θ, s) interpolation condition if and only if X 1 X n L s where X 1 X n = {f 1 f n ; f k X k } Proof In this case σ k = r for all k and (21) is a obtained by a simple change of variables from (15) Finally let us note an unusual case which can arise: Theorem 39 Suppose Θ R n +, dim Θ = n 1 and Θ does not span R n Suppose X is an n-tuple of ri spaces such that B X co Θ is a nonempty subset of co Θ Let σ = (σ k ) n be chosen so that σ, θ = for all θ Θ Assume s > is such that sθ k < 1 for every θ Θ and 1 k n Then the following are equivalent: (i) X satisfies the (Θ, s) interpolation condition (ii) There is a constant C so that for f 1,, f n E we have ( n 1/s (22) x 1 (fk (x σ k )) dx) s C f k Xk We omit the proof which is similar to the that of Theorem 37 4 The inhomogeneous multilinear Boyd theorem and applications Suppose that Θ is a finite subset of (R + ) n and that θ r θ is a map from Θ to R + We denote φ θ = (θ, r θ ) and Φ = {φ θ : θ Θ} Clearly Φ R n+1 Now consider the case when we are given a map T : E n L (, ), which we assume to be locally continuous We will say that T satisfies a weak-type (θ, r θ ) estimate if there exists M > so that if E 1,, E n are sets of finite measure then (23) T (χ E1,, χ En ) Lr, M E k θ k We now give a version of the Boyd interpolation theorem for this setting which follows almost immediately from Theorem 35 For simplicity we shall only treat the most important case Theorem 41 Suppose Θ is a subset of (R + ) n with Θ = n + 1 and dim Θ = n Suppose for each θ we have < r θ Let σ R n be the unique solution of the equation σ, θ = 1 r θ + τ

14 14 LOUKAS GRAFAKOS AND NIGEL KALTON where τ is independent of θ Let X be an n-tuple of ri spaces and suppose Y is a maximal ri space which is s-convex for some s > Suppose the Boyd cube B X [1/q Y, 1/p Y ] intersects co Φ in a non-empty subset of co Φ Then in order that every locally continuous n-linear T : E n L (, ), which satisfies the weak type (θ, r θ ) estimate (23) for θ Θ, extends to a bounded n-linear map T : n X k Y (with norm a multiple of M), it is sufficient that there is a constant C so that if f 1,, f n E then n (24) x τ fk (x σ k ) Y C If in addition we have < 1 r θ n f k Xk for every θ Θ, then (24) is also necessary and is equivalent to the condition that there exists a constant C so that for f 1,, f n E we have (25) x τ f k (x σ k ) Y C f k Xk σ k Remark The same conclusions can be obtained if Θ > n + 1 provided there is a solution of the equation σ, θ = r 1 θ + τ Proof We first choose s > small enough so that Y is s-convex, s < r θ for all θ Θ, and τ + 1 > Next define X s n+1 to be the space of all f L (, ) so that fg L s (, ) for all g Y with the quasi-norm f Xn+1 = θ k σ k sup fg Ls g Y 1 Since Y is both maximal and s-convex we obtain that g Y if and only if sup{ fg Ls : f Xn+1 1} is finite and furthermore there is a constant C so that f Y sup{ fg Ls : f Xn+1 1} (If Y is s-convex with constant one then C = 1; this is easily seen by noting that Y s is a Banach ri space and X s n+1 is simply the Köthe dual space; in general we can always renorm Y to have s-convexity constant one) It easy to calculate the Boyd indices of X n+1 ; these are given by 1 = 1 p Xn+1 s 1, q Y 1 = 1 q Xn+1 s 1 p Y We refer to [12] for similar calculations for dual spaces Now if T satisfies the weak-type (θ, r θ ) estimate (23) for every θ Θ, then we consider the map T : E n+1 L s (, ) defined by T (f 1,, f n+1 ) = T (f 1,, f n )f n+1

15 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 15 If E 1,, E n+1 are sets of finite measure then ( T (χ E1,, χ En+1 ) Ls T (χ E1,, χ En ) s dx E n+1 M ( r θ ) 1/s En+1 r θ 1/s 1/r θ s ) 1/s E k θ k for every θ Θ Thus if we let ψ θ = (θ, 1 1 s r θ ) and Ψ = {ψ θ : θ Θ} then T is (Ψ, s)-admissible It is clear from our discussion of X n+1 that we only need to show that T extends to a bounded n-linear map on X 1 X n+1 We now use Theorem 37 We first argue that Ψ is linearly independent in R n+1 Indeed from the definition of τ this is equivalent to the linear independence of the points (θ, τ + 1 ) which follows from the affine independence of Θ We will show that s the (n + 1)-tuple (X 1,, X n+1 ) has the (Ψ, s)-interpolation condition We note that our hypotheses on the Boyd indices of X and Y imply that the hypotheses on the Boyd indices for Theorem 36 hold Define σ k = σ k(τ + 1 s ) 1 for k n and σ n+1 = (τ + 1 s ) 1 Then σ, ψ θ = (τ + 1 s ) 1 (τ + 1 r θ ) + (τ + 1 s ) 1 ( 1 s 1 r θ ) = 1 Now if f 1,, f n+1 E n+1, we have ( n+1 1/s ( x s 1 fk (x σ k ) dx) s = (τ + 1 s )1/s x sτ fn+1(x) s n f k (x σ k ) s dx) 1/s Now it is clear that if we assume (24) then we obtain (15) in Theorem 37 and so (X 1,, X n+1 ) satisfies the interpolation condition (Θ, s) For the second part we construct the map T : E n L ((, ) (, 1)) by T (f 1,, f n )(x, y) = x τ f j (y) σ k f k (x σk ) σ j = Let u θ = ( σ k θ k) 1 so that r θ u θ Then by arguments similar to those for Theorem 37 we have that if E 1,, E n are measurable sets of finite measure, T (χ E1,, χ En ) Lrθ r 1/r θ θ u 1/u θ θ E k θ k Our hypotheses then guarantee that T maps X 1 X n into Y ie we have (25) and hence also (24) Let us isolate a simple special case: Corollary 42 Suppose that in the preceding theorem we have n θ k = 1 r θ for every θ Θ Then (25) is equivalent to the inclusion X 1 X n X 1 X n is the set of all products f 1 f n with f k X k Y, where

16 16 LOUKAS GRAFAKOS AND NIGEL KALTON Proof We need only to observe that in this case σ k = 1 for every k and τ = We next point out that under certain hypotheses, we can replace (24) with an alternative criterion: Corollary 43 Suppose that in Theorem 41, Ỹ is a carrier space for Y with the property that D a ey C a ρ for all < a < 1 where ρ > and that Ỹ is s-convex for some s > Then the sufficent condition (24) can be replaced by: n (26) x τ fk (x σ k ) ey C f k Xk, for f 1,, f n E Proof We note that in the proof of Theorem 24 we can take s small enough so Ỹ is s-convex Suppose f 1,, f n E Let ϕ(x) = x τ n f k (xσ k ) By assumption ϕ ey C n f k Xk Now let ψ be defined by ( ψ(x) = ϕ(y) s dy ) 1/s y x By the s-convexity of Ỹ we obtain that ( 1 1/s ψ ey M a da) sρ 1 φ ey so that we have an estimate ψ ey C 1 ϕ ey However ψ is decreasing and so ψ Y C 1 ϕ ey Now if f n+1 E we have that ( ) 1/s n+1 (fn+1(x)) s ϕ(x) s dx CC 1 f k Xk and the proof is completed in the same way At this point we note that we can use Corollary 33 to extend this result to n- quasi-sublinear maps Corollary 44 Assume that X 1,, X n, Y satisfy (24) Suppose T : E n L (, ) is n-quasi-sublinear, locally continuous, and satisfies the weak-type (θ, r θ )-inequality (23) for every θ Θ Then we have the estimate T (f 1,, f n ) Y CM f k Xk for f 1,, f n E We omit the details of the proof The key point to note is that we should choose s in the argument for Theorem 41 above sufficiently small so that 2 1/s 1 C where is the constant in (11) It is also worth noting that we can give a similar result to Theorem 41 in the case when Θ fails to be affinely independent This case is somewhat degenerate For

17 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 17 example in the case n = 1 it applies to linear operators which satisfy weak type estimates (p, q 1 ) and (p, q 2 ) where q 1 q 2 Theorem 45 Suppose Θ is an affinely dependent subset of (R + ) n with Θ = n + 1 Suppose for each θ we have < r θ and that the set Φ = {(θ, r 1 θ ) : θ Θ} is linearly independent in R n+1 Choose σ R n so that σ, θ = 1 for all θ Θ Let X be an n-tuple of ri spaces and suppose Y is a maximal ri space Let r = min θ Θ r θ and suppose < s 1 is such that s < 1 if r = 1 and s r otherwise Suppose also the Boyd cube B X [1/q Y, 1/p Y ] intersects co Φ in a non-empty subset of co Φ Then, in order that every locally continuous n-linear T : E n L (, ), which satisfies the weak type (θ, r θ ) estimate (23) for θ Θ, extends to a bounded n-linear map T : n X k Y (with norm a multiple of M), it is sufficient that there exists a constant C so that ( n 1/s (27) x s 1 (fk (x σ k )) dx) s C f k Xk, for f 1,, f n E Remark The existence and uniqueness of σ is a consequence of our hypotheses, since Θ generates a plane of dimension n 1 which cannot be a linear subspace Proof Our hypotheses are such that the space L r, is s-normable In this case the convex set Φ generates a plane containing the line in the direction parallel to the basis vector e n+1 We first prove the result when Y = L t, for some t By the above remark we have t > r Let X n+1 = L u, where = 1 Let ψ t u r θ = (θ, 1 1 r r θ ) R n+1 and Ψ = {ψ θ : θ Θ} Now it is clear that (27) implies that the (n + 1)-tuple (X 1,, X n+1 ) satisfies the conditions of Theorem 37 for the (Ψ, s)-interpolation condition We apply this to the map T : E n+1 L r, where T (f 1,, f n+1 ) = T (f 1,, f n )f n+1 A routine calculation gives T (χ E1,, χ En+1 ) Lr, CM E 1/r 1/r θ E k θ k Hence we have the estimate This implies the estimate T (f 1,, f n )f n+1 Lr, CM f n+1 Lu, T (f 1,, f n ) Lt, CM n f k Xk f k Xk by simply considering f n+1 = χ E for E a set of finite measure We have now proved our claim Next we consider the general case By our assumptions on Φ we may find t < p Y q Y < u so that both (X 1,, X n, L t, ) and (X 1,, X n, L u, ) satisfy the

18 18 LOUKAS GRAFAKOS AND NIGEL KALTON interior condition on the Boyd indices Hence T maps X 1 X n boundedly into L t, L u, with norm a multiple of M But it is easy to calculate from the Boyd indices that L t, L u, Y The theorems below extend the classical Marcinkiewicz interpolation theorem to the multilinear setting Theorem 46 Let < p jk for 1 j n + 1 and 1 k n, and also let < p j for 1 j n + 1 Suppose that a locally continuous n-linear map T : E n L satisfies T (χ E1,, χ En ) Lpj, M E 1 1/p j1 E n 1/p jn for all sets E j of finite measure and all 1 j n + 1 Assume that the system below 1/p 11 1/p 12 1/p 1n 1 σ 1 1/p 1 1/p 21 1/p 22 1/p 2n 1 σ 2 1/p 1/p n1 1/p n2 1/p nn 1 σ = 2 n 1/p, n 1/p (n+1)1 1/p (n+1)2 1/p (n+1)n 1 τ 1/p n+1 has a unique solution (σ 1,, σ n, τ) R n+1 with not all σ j = Suppose that (1/q 1,, 1/q n, 1/q) lies in the open convex hull of the points (1/p j1,, 1/p jn, 1/p j ) in R n+1 and let < t k, t satisfy 1 (28) 1 t k t 1 k n σ k Then T extends to a bounded n-linear map T : n L q k,t k L q,t with norm a multiple of M Remark We remark that the existence of the unique solution in the linear system of Theorem 46 is equivalent to the condition that the n + 1 points θ j = (1/p jk ) n are affinely independent in R n We also note that as in Corollary 44 the result above is valid for n-quasi-sublinear maps Proof We clearly only need to consider the case of equality in (28) It is clear that the Boyd index assumption of Theorem 41 is satisfied Clearly we have n σ k = τ + 1 q k q Hence if f 1,, f n E we have x 1/q+τ n f k (x σ k ) = x σ k/q k fk (x σ k ) Let F (x) = x τ σ k f k (xσ k ) Then ( (x 1/q F (x)) t dx ) 1/t C ( (x σ k/q k fk (x σ k )) t dx ) 1/tk k x x σ k

19 Now if σ k SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 19 ( (x σ k/q k fk (x σ k )) t dx ) 1/tk k = σ k 1 f Lqk,t x k Thus we have an estimate ( (x 1/q F (x)) t dx x In view of Corollary 43 this completes the proof ) 1/t C f k Lqk,t k σ k There is a version of the above result for the degenerate case corresponding to Theorem 45: Theorem 47 Let < p jk for 1 j n + 1 and 1 k n, and let < p j for 1 j n + 1 Suppose that a locally continuous n-linear map T : E n L satisfies T (χ E1,, χ En ) Lpj, M E 1 1/p j1 E n 1/p jn for all subsets E k of finite measure and all 1 j n+1 Assume that the n+1 points θ j = (1/p jk ) n are affinely dependent in Rn, but the points (θ j, 1/p j ) are linearly independent in R n+1 Suppose that (1/q 1,, 1/q n, 1/q) lies in the open convex hull of the points (1/p j1,, 1/p jn, 1/p j ) in R n+1 Let r = min 1 j n+1 p j and < t k, t satisfy { 1 > 1 if r = 1, (29) 1 if r 1, r where {σ k } n 1 k n σ k t k are the unique solutions of the system n σ k = 1, 1 j n + 1 p jk Then T extends to a bounded n-linear map T : n L q n,t n L q,t with norm a multiple of M Proof This is deduced from Theorem 45 It is clear our hypotheses guarantee the appropriate conditions on the Boyd indices Pick any < s 1 so that s r if r 1 and s < 1 otherwise with 1 s 1 t k σ k It then suffices to verify (27) in Theorem 45 To do this we can clearly suppose that 1 s = 1 t k σ k

20 2 LOUKAS GRAFAKOS AND NIGEL KALTON Suppose f 1,, f n E and set F (x) = x fk (x σ k ) σ k Then F (x) = x σ k/q k fk (x σ k ) and so σ k ( F (x) s dx ) 1/s σ k 1 f k Lqk,t x k σ k This establishes (27) and completes the proof 5 Examples and applications In this section we discuss some examples of multilinear interpolation For simplicity we restrict ourselves to bilinear and trilinear examples Example 51 (Young s inequality and O Neil s inequality) On a locally compact abelian group consider the bilinear operator (f, g) f g, where denotes convolution Let H denote the closed triangle in R 3 with vertices (1,, ), (, 1, ), and (1, 1, 1) The well known Young s inequality says that (3) f g Lr C f Lp g Lq holds if the point (1/p, 1/q, 1/r) lies in the closure of the triangle H The three trivial estimates f g L1 f L1 g L1, f g L f L1 g L, and f g L f L g L1 give (3) on the interior of H The estimates on the sides follow from bilinear complex interpolation Applying Theorem 46 in the situation above we obtain O Neil s inequality If the point (1/p, 1/q, 1/r) lies in the interior of the triangle H and < s 1, s 2 and 1/s = 1/s 1 + 1/s 2, then (31) f g Lr,s C f Lp,s1 g Lq,s2 The special case s 1 = p, s = s 2 = is of particular interest Observe that if (1/p, 1/q, 1/r) lies in the interior of H, then 1/p + 1/q = 1/r + 1, from which it follows that p < r, which in turn implies that f g Lr C f g Lr,p C f Lp g Lq, The inequality above provides a sharpening of Young s inequality since the space L q is replaced by L q, More generally we can use Theorem 41 to obtain the following result: Theorem 52 Suppose X, Y, Z are ri spaces whose Boyd indices satisfy the conditions 1 < p X, p Y, p Z, q X, q Y, q Z <, , p X p Y q Z

21 and SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 21 1 q X + 1 q Y p Z Assume that Z is maximal and s convex for some s > Then (f, g) f g maps X Y to Z provided the map (f, g) xf(x)g(x) maps X(, ) Y (, ) to Z(, ) Remark Of course we can state this theorem with less stringent requirements on the Boyd indices, namely that the Boyd cube intersects H in a subset of its relative interior As in the discussion in the remarks after Theorem 37 this can be illustrated We can allow for example p X 1 provided q Y < p Z, and q X = is permissible provided p 1 Y < 1 + q 1 Z is permissible if q 1 X Similarly p Z 1 is permissible if p 1 X Y > 1 + q 1 + p 1 Y < 2 and q Z = Example 53 Fix three numbers < α, β, γ < n such that α + β > n, β + γ > n and γ + α > n Consider now the trilinear fractional integral form I α,β,γ (f, g, h) = f(x)g(y)h(z) x y α y z β z x γ dxdydz R n R n R n We claim that the following inequality is valid if and only if I α,β,γ (f, g, h) C f Lp g Lq h Lr (32) 1 p + 1 q + 1 r + α + β + γ n = 3, 1 < p, q, r <, and 1 p + 1 q + 1 r > 1 Note that (32) requires α + β + γ < 2n Examples can be given to prove the necessity of the conditions on the indices above Let us prove here the sufficiency The assumptions α+β > n, β+γ > n, and γ+α > n imply α + β + γ > 3n/2 and hence it follows from (32) that 1/p + 1/q + 1/r < 3/2 Therefore the plane given by the first equation in (32) cuts the unit cube [, 1] 3 at the six points A 1 = (1, (2n α β γ)/n, ), A 2 = ((2n α β γ)/n, 1, ), A 3 = (, 1, (2n α β γ)/n) A 4 = (, (2n α β γ)/n, 1) A 5 = ((2n α β γ)/n,, 1), and A 6 = (1,, (2n α β γ)/n) These six points form the vertices of a hexagon It suffices to prove Lorentz space estimates at these vertices for characteristic functions For instance at the vertex A 1 the estimate we need to establish is (33) E 1 E 2 E 3 x y α y z β z x γ dxdydz C E 1 E 2 2 α+β+γ n First integrate in z We have (34) y z β z x γ dz E 3 y z β z x γ dz = C x y n β γ, R n

22 22 LOUKAS GRAFAKOS AND NIGEL KALTON 1/r (,,1) A4 A5 A3 A6 (,1,) 1/q A2 (,1,) A1 1/p Figure 1 The set of all (1/p, 1/q, 1/r) such that I α,β,γ (f, g, h) C f Lp g Lq h Lr for all x y since β + γ > n The last equality above can be easily shown by a translation, a dilation, and a rotation Using (34) we obtain E 1 C C E 1 E 1 E 2 E 3 x y α y z β z x γ dxdydz x y n α β γ dydx E 2 y c E 2 1/n y n α β γ dydx C E 1 E 2 (2n α β γ)/n, which proves the required estimate (33) This example can be found in [3] when n = 1 and α = β = γ In this example we have a trilinear form and it is appropriate to apply Corollary 38 Again simplifying our conditions on the Boyd indices gives:

23 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 23 Theorem 54 Suppose X 1, X 2, X 3 are ri spaces on R n Suppose the Boyd indices satisfy the conditions 1 < p Xi q Xi < for i = 1, 2, 3 and α + β + γ 3 1 n p Xi i=1 q Xi Then I α,β,γ is bounded on X 1 X 2 X 3 provided the trilinear form (f, g, h) α+β+γ 2 x n f(x)g(x)h(x) is bounded on X 1 (, ) X 2 (, ) X 3 (, ) Remark Here as in the preceding example we can relax the conditions on the Boyd indices with the right extra hypotheses For example if p X1 1 it is necessary that q X2 q X3 Example 55 Consider the operator I(f, g)(x) = i=1 > 2 α + β + γ n t 1 f(x + t)g(x t) dt We will show I maps L p (R n ) L q (R n ) into L r (R n ) when (1/p, 1/q, 1/r) lies in the closed convex hull of the points (1,, ), (, 1, ), (1,, 1), (, 1, 1), (1, 1, 1), and (1, 1, 1/2) By interpolation it suffices to establish boundedness estimates at these six points Five of these estimates are trivial We only prove that I maps L 1 L 1 L 1/2 Suppose that we have established the estimate (35) I(f, g) L1/2 C f L1 g L1 for all f and g supported in two cubes of sidelength one Then we prove (35) (with a larger constant) for all f and g integrable For each k Z n, let Q k be the cube of sidelength one whose sides are parallel to the axes and whose lower left corner is k Z n let f k = fχ Qk and g m = gχ Qm Then for each k Z n there exist at most finitely many m Z n such that I(f k, g m ) is nonzero This is because the intersection of the sets {t : t 1} and 1(Q 2 k Q m ) has to be nonempty Now write I(f, g) = I(f k, g m ) k Z n m Z n as a sum of a finite number of terms of the form I(f k, g k+d ) k Z n where d Z n lies in a ball of radius at most a dimensional constant Now ( ) 2 I(f, g) L1/2 I(f k, g k+d ) 1/2 dx R n C ( k Z n k Z n f k 1/2 L 1 g k+d 1/2 L 1 ) 2 C f L1 g L1,

24 24 LOUKAS GRAFAKOS AND NIGEL KALTON by Cauchy-Schwarz, where the penultimate inequality above follows from the asumption that (35) holds for the functions f k and g m Summing over d we obtain the required estimate I : L 1 L 1 L 1/2 with a larger constant We now prove (35) for f and g supported in cubes of sidelength one (Think of f = f k and g = g k+d ) Now observe that I(f, g) is supported in a cube of sidelength two Hölder s inequality gives I(f, g) L1/2 C I(f, g) L1 C f(x + t) g(x t) dtdx C f L1 g L1 R n t 1 Example 56 We now consider the bilinear fractional integral I α (f, g)(x) = f(x + t)g(x t) t α n dt, R n where < α < n Homogeneity considerations imply that I α can map L p (R n ) L q (R n ) L r (R n ) only when 1 p + 1 q = 1 r + α n We will now show that I α maps L p L q L r when the point (1/p, 1/q, 1/r) lies in the open convex hull of the pentagon with vertices ( α,, ), (1,, 1 α 2n α ), (1, 1, n n n (, 1, 1 α), and (, α, ) More precisely we will show that a weak-type estimate n n holds at each vertex of the pentagon below We first consider the vertex ( α,, ) Take f = χ n A and g = χ B, where A and B are measurable sets of finite measure We have I α (χ A, χ B ) L sup x R t α n dt t α n dt = C A α/n x+a t c A Likewise we obtain the required estimate at the vertex (, α, ) n The estimates at the vertices (1,, 1 α) and (, 1, 1 α ) follow from the estimates at the vertices ( α,, ) and (, α, ) respectively via duality Alternatively, just n n n n observe that I α (χ A, χ B ) J α (χ A ), where J α is the usual fractional integral (J α f)(x) = f(x y) y α n dy, R n and thus the estimate I α (χ A, χ B ) Ln/(n α), C A directly follows from the corresponding estimate for the linear operator Finally we are left with the estimate at the vertex (1, 1, 2n α ) For j Z we n introduce operators I j (f, g)(x) = f(x + t)g(x t)dt t 2 j and we note that for f, g we have I α (f, g) C j Z 2 j(α n) I j (f, g)

25 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 25 1/r (1,1, (2n- )/ n) (,1,1) (,1, (n- )/ n) (1,,1) (1,, (n- )/ n) ( / n,,) (, / n,) (,1,) 1/q (1,,) (1,1,) 1/p Figure 2 The set of all (1/p, 1/q, 1/r) such that I α : L p L q L r Next we observe that by a easy dilation argument I j maps L 1 L 1 L 1/2 with norm bounded by a constant times 2 jn This fact together with the observation E ( (I j (f, g)(x)) 1/2 dx E ) 1/2 I j (f, g)(x) dx E 1/2 C f 1/2 L 1 g 1/2 L 1 E 1/2, implies that for any measurable set E with finite measure we have (36) E (I j (f, g)(x)) 1/2 dx f 1/2 L 1 g 1/2 L 1 min(2 jn, E ) 1/2

26 26 LOUKAS GRAFAKOS AND NIGEL KALTON Now pick E = E λ = {x : I α (f, g)(x) > λ} Then Chebychev s inequality and (36) give 1/2 λ 1/2 E λ 2 j(α n) I j (f, g)(x) dx E λ j Z j Z j Z 2 j(α n)/2 E λ I j (f, g)(x) 1/2 dx 2 j(α n)/2 f 1/2 L 1 g 1/2 L 1 min(2 jn, E λ ) 1/2 =C f 1/2 L 1 g 1/2 L 1 E λ α/2n This implies that λ E λ 2n α n C f L1 g L1 which is the required weak type estimate at the vertex (1, 1, 2n α ) This example n was studied in [4] when r 1 and should be contrasted with the main result in [11] The same result was independently obtained in [1] To use the full strength of our results we apply Theorem 42 and the succeeding remark to obtain the following generalization for ri spaces Theorem 57 Suppose X, Y, Z are ri spaces on R n with Z maximal and s-convex for some s > Suppose the Boyd indices of X, Y, Z satisfy the condition that the Boyd cube intersects the pentagon generated by ( α,, ), (1,, 1 α 2n α ), (1, 1, ), (, 1, 1 α) n n n n and (, α, ) in a nonempty subset of the interior Then in order that I n α maps X Y to Z it is sufficient that (f, g) x α f(x)g(x) maps X(, ) Y (, ) to Z(, ) References [1] C Bennett and R Sharpley, Interpolation of operators, Pure and Applied Mathematics 129, Academic Press, Orlando FL, 1988 [2] D W Boyd, The Hilbert transform on rearrangement-invariant spaces, Canad J Math 19 (1967), [3] M Christ, On the restriction of the Fourier transform to curves: Endpoint results and the degenerate case, Trans Amer Math Soc 287 (1985), [4] L Grafakos, On multilinear fractional integrals, Studia Math 12 (1992), [5] S Janson, On interpolation of multilinear operators, Function spaces and applications (Lund 1986), 29 32, Lecture Notes in Math, 132, Springer, Berlin-New York 1988 [6] W B Johnson and G Schechtman, Sums of independent random variables in rearrangement invariant function spaces, Ann Probab 17 (1989), [7] N J Kalton, Convexity conditions on non-locally convex lattices, Glasgow J Math 25 (1984), [8] N J Kalton,Endomorphisms of symmetric function spaces, Indiana Univ Math J 34 (1985), [9] N J Kalton, J W Roberts, and N T Peck, An F -space sampler, Lon Math Soc Lec Note Series, 89, Cambridge University Press, Cambridge-New York, 1984 [1] C Kenig and E M Stein, Multilinear estimates and fractional integration, Math Res Let 6 (1999), 1 15 [11] M T Lacey and CM Thiele, L p bounds for the bilinear Hilbert transform, p > 2, Ann Math 146 (1997),

27 SOME REMARKS ON MULTILINEAR MAPS AND INTERPOLATION 27 [12] J Lindenstrauss and L Tzafriri, Classical Banach spaces II Function spaces, Ergebnisse der Mathematik und ihrer Grenzgebiete 97 Springer-Verlag, Berlin-New York, 1979 [13] G G Lorentz, Majorants in spaces of integrable functions, Amer J Math 77 (1955), [14] S Rolewicz, Metric linear spaces, Second edition, Mathematics and its Applications (East European Series), 2 D Reidel Publishing Co, Dordrecht-Boston, Mass; PWN Polish Scientific Publishers, Warsaw, 1985 [15] R Sharpley, Interpolation of n pairs and counterexamples employing indices, J Approx Theory 13 (1975), [16] R Sharpley, Multilinear weak type interpolation of mn-typles with applications, Studia Math 6 (1977), [17] R Strichartz, A multilinear version of the Marcinkiewicz interpolation theorem, Proc Amer Math Soc 21 (1969), [18] T Shimogaki, Hardy-Littlewood majorants in function spaces, Proc Japan Acad 17 (1965), [19] E M Stein and G Weiss, An extension of a theorem of Marcinkiewicz and some of its applications, J Math Mech 8 (1959), [2] M Zafran, A multilinear interpolation theorem, Studia Math 62 (1978), Loukas Grafakos, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA address: loukas@mathmissouriedu Nigel Kalton, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA address: nigel@mathmissouriedu