PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 132, Number 10, Pages 2929–2938 S 0002-9939(04)07425-8 Article electronically published on May 21, 2004
INTERPOLATION BETWEEN L1 AND Lp, 1 SERGEI V. ASTASHKIN AND LECH MALIGRANDA (Communicated by Jonathan M. Borwein) Abstract. We show that if X is a rearrangement invariant space on [0, 1] that is an interpolation space between L1 and L∞ and for which we have only a one-sided estimate of the Boyd index α(X) > 1/p, 1 1. Introduction Let X be a rearrangement invariant (r. i.) function space on I =[0, 1]. If any linear operator T is bounded in the spaces Lp and Lq (1 ≤ p we obtain that X is an interpolation space between Lp and L∞ (see [M81], Th. 4.6, which is proved even for Lipschitz operators). Received by the editors October 9, 2002. 2000 Mathematics Subject Classification. Primary 46E30, 46B42, 46B70. Key words and phrases. Lp-spaces, Lorentz spaces, rearrangement invariant spaces, Boyd in- dices, interpolation of operators, operators of strong type, operators of weak type, K-functional, Marcinkiewicz spaces. This research was supported by a grant from the Royal Swedish Academy of Sciences for cooperation between Sweden and the former Soviet Union (project 35156). The second author was also supported in part by the Swedish Natural Science Research Council (NFR)-grant M5105- 20005228/2000. c 2004 American Mathematical Society 2929 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2930 SERGEI V. ASTASHKIN AND LECH MALIGRANDA E. M. Semenov posed the following question about the extreme space L1: Let X be an r. i. space on [0, 1] with either the Fatou property or absolutely continuous 1 norm. Is the one-sided estimate on the Boyd index α(X) > p enough for the interpolation property between L1 and Lp, 1 2. Definitions and notation We first recall some basic definitions. If a Banach space X =(X, k·k)ofall (classes of) measurable functions x(t)onI =[0, 1] is such that there exists u ∈ X with u>0a.e.onI and kxk≤kyk whenever |x|≤|y|,wesaythatX is a Banach function space (on I =[0, 1]). A Banach function space X on I =[0, 1] is said to be a rearrangement invariant (r. i.) space provided x∗(t) ≤ y∗(t) for every t ∈ [0, 1] ∗ and y ∈ X imply x ∈ X and kxkX ≤kykX,wherex denotes the decreasing rearrangement of |x|. We always have the imbeddings L∞[0, 1] ⊂ X ⊂ L1[0, 1]. By 0 X we will denote the closure of L∞[0, 1] in X. If χA denotes the characteristic function of a measurable set A in I, then clearly kχAkX depends only on m(A). The function ϕX (t)=kχAkX ,wherem(A)=t, t ∈ I, is called the fundamental function of X. Given s>0, the dilation operator σs given by σsx(t)=x(t/s)χI (t/s),t ∈ I is well defined in every r. i. space X and kσskX→X ≤ max(1,s). The Boyd indices of X are defined by (cf. [KPS], [LT], [BS]) ln kσ k → ln kσ k → α(X) = lim s X X ,β(X) = lim s X X . s→0 ln s s→∞ ln s In general, 0 ≤ α(X) ≤ β(X) ≤ 1. It is easy to see thatϕ ¯X (t) ≤kσtkX→X for any t>0, whereϕ ¯ (t)=sup ϕX (st) . A Banach function space X with a X 0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTERPOLATION BETWEEN L1 AND Lp, 1 (b) absolutely continuous norm if for any x ∈ X and every sequence xn of mea- surable functions on I =[0, 1] satisfying |x|≥xn ↓ 0wehavekxnkX → 0. Note that X is separable if and only if the norm of X is an absolutely continuous norm. From the Calder´on-Mitjagin theorem it follows that the r. i. space X with either the Fatou property or the separable property is an interpolation space with respect to L1 and L∞, i.e., if a linear operator T is bounded in L1 and L∞,thenT k k ≤ k k k k ≥ is bounded in X and T X→X C max( T L1→L1 , T L∞→L∞ )forsomeC 1 ([KPS], [BS]). 0 The associated space X to a Banach function spaceRX is the space of all (classes 1 | | ∞ of) measurable functions x(t)onI =[0, 1] such that 0 x(t)y(t) dt < for every y ∈ X endowed with the norm Z 1 kxkX0 =sup{ |x(t)y(t)|dt : kykX ≤ 1}. 0 0 00 X is a Banach function space. We have the embedding X ⊂ X with kxkX00 ≤ 00 kxkX for every x ∈ X. Moreover, X = X with equality of the norms if and only if X has the Fatou property (cf. [KPS], [LT]). If a Banach function space X is separable, then the embedding X ⊂ X00 is isometric and X0 = X∗.IfX is an r. i. space, then the associated space X0 isalsoar.i.space. Among classical r. i. spaces with the Fatou property we mention Lorentz spaces Lp,q, Lorentz spaces Λϕ and Λp,ϕ, Marcinkiewicz spaces Mϕ and Orlicz spaces LΦ. 0 Typical separable spaces are Mϕ and EΦ, the closures of L∞ in Marcinkiewicz space Mϕ and Orlicz space LΦ, respectively. For other general properties of lattices of measurable functions and r. i. spaces, we refer to the books [LT], [KPS], [BS]. 3. Strong interpolation of L1 and Lp, 1 Proof. We will first show that if T : L1 → L1 is a bounded linear operator such → that T = T|Lp : Lp Lp is bounded, then 0 00 (2) T = T|X0 : X → X 0 is also bounded, where X means the closure of L∞ in X. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2932 SERGEI V. ASTASHKIN AND LECH MALIGRANDA 0 → Let T1 be the associated operator to a linear bounded operator T : L1 L1, 0 → ∈ and let T2 be the associated operator to T|Lp : Lp Lp. Then, for all x Lp and y ∈ L∞,wehave Z Z Z 1 1 1 0 0 Tx(s)y(s)ds = x(s)T1y(s)ds = x(s)T2y(s)ds. 0 0 0 0 0 0 0 0 Hence, T2|L∞ = T1 and we can consider T = T1 = T2.Then 0 0 0 (3) T : Lp0 → Lp0 and T : L∞ → L∞ is bounded, where 1/p +1/p =1. 00 0 − − 1 Since X is isometrically embedded into X , it follows that β(X )=1 α(X) < 1 p (cf. [KPS], Th. 4.11), and by the Boyd interpolation theorem we have that X0 is an interpolation space between Lp0 and L∞. Therefore, (4) T 0 : X0 → X0 is bounded. 00 In view of (3) and Lemma 1 there exists the second associated operator T : Lp → Lp, which is bounded. 00 We can extend T to the whole space L1. In fact, if x ∈ L∞ and y ∈ Lp,then by (3), Z 1 00 00 kT yk1 =sup x(s)T y(s)ds kxk∞≤1 0 Z 1 0 ≤k 0k k k =sup T x(s)y(s)ds T L∞→L∞ y 1. kxk∞≤1 0 00 Since Lp is dense in L1, it follows that T : L1 → L1 is bounded. The uniqueness 00 0 0 shows that T = T . On the other hand, X ⊂ Lp0 and for all y ∈ X and x ∈ Lp we have by (4) that Z 1 kTxkX00 =sup Tx(s)y(s)ds kyk 0 ≤1 0 X Z 1 0 0 =sup x(s)T y(s)ds ≤kT kX0→X0 kxkX . k k ≤ y X0 1 0 Thus, 0 kTxkX00 ≤kT kX0→X0 kxkX for all x ∈ Lp ⊂ X, which gives (2). Now, let X be a separable r. i. space. For every x ∈ X we can find a sequence 00 {xn}⊂Lp such that xn → x in X. From (2) we obtain Txn → Tx in X . 00 Moreover, {Txn}⊂Lp ⊂ X.Since{Txn} is a Cauchy sequence in X and X is 00 isometrically embedded in X , it follows that {Txn} is a Cauchy sequence in X and so Tx∈ X. Suppose that the r. i. space X has the Fatou property or, equivalently, X = X00. Let y = Tx,wherex ∈ X and T : L1 → L1 is a bounded linear operator such that → T = T|Lp : Lp Lp is bounded. Then (5) K(t, y; L1,Lp) ≤ CK(t, x; L1,Lp) ∀ t ∈ (0, 1]. There is a sequence of step-functions {ym} such that ym ↑|y| a.e. on [0, 1]. We can take, also, the truncations xn(s)=min{|x(s)|,n},n=1, 2,.... License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTERPOLATION BETWEEN L1 AND Lp, 1 ↑| | ∗ ↑ ∗ Since xn x a.e. on [0, 1], it follows that xn x a.e. on [0, 1] (cf. [KPS], p. 67). By using (5) and the Holmstedt formula (cf. [H], Th. 4.1), Z 0 Z tp 1 1/p ∗ ∗ p K(t, x; L1,Lp) ≈ x (s)ds + t x (s) ds 0 0 tp with x ∈ L1 and 0 0 1 with a constant C1 > 0 independent of m. Since α(X )=α(X) > p (cf. [KPS], p. 143), then, by above, the separable space X0 is an interpolation space for the couple (L1,Lp). But (L1,Lp)isaK-monotone couple (see [C], Th. 4). Therefore, from the last inequality we have k k k k 0 ≤ k k 0 ≤ k k ym X = ym X C2 xnm X C2 x X 00 with some constant C2 > 0. Since X = X , it follows that kykX ≤ C2kxkX , and this completes the proof. Remark 1. Using results from [Ms], the proof of Theorem 1 can be a little shorter. 0 Once we have proved that the associated operator T is bounded in the spaces L∞ 00 and Lp0 , and the second associated operator T exists, then from Theorem 3.5(b) in [Ms] it follows that T 00 = T is bounded in X00. In this paper we can also find some sufficient conditions under which from the interpolation property of the space X between spaces X0,X1 follows the interpolation property of its associated space 0 0 0 X between the associated spaces X0,X1. Theorem 2. Let 1 1−1/p 1−1/p 1−1/p K(t ,Tx; L1,Lp) ≤ C3K(t ,x; L1,Lp) ≤ C3K(t ,x; L1,Lp,1) for any x ∈ L1 and all 0 for any x ∈ L1 and all 0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2934 SERGEI V. ASTASHKIN AND LECH MALIGRANDA we obtain, by using the Fubini theorem, Z Z Z t t 1 (Mx∗)∗(s)ds = s−1/p u1/p−1x∗(u)du ds 0 Z0 Z s t u = s−1/pds u1/p−1x∗(u)du 0 0 Z Z 1 t + s−1/pds u1/p−1x∗(u)du Z t 0 Z t 1 = p0 x∗(u)du + t1−1/p u1/p−1x∗(u)du 0 t 1−1/p ≈ K(t ,x; L1,Lp,1). Therefore, in view of (6), Z Z t t ∗ ≤ C4 ∗ ∗ (7) (Tx) (s)ds 0 (Mx ) (s)ds 0 p 0 for all 0 ∗ (8) kMx k≤C5kxk for x ∈ X. Note that Boyd proved (see [B68], Th. 1, [B69], Lemma 2, and [BS], Th. 5.15) the boundedness of the operator M in r. i. spaces X with the Fatou property. He showed that the operator M is bounded in X if and only if the lower Boyd index α(X) > 1/p. In particular, this result gives the estimate (8) but only for r. i. spaces with the Fatou property. To show (8) in general we first note that the assumption α(X) > 1/p is equivalent to the property that there exist ε>0andA>0 such that 1/p+ε (9) kσskX→X ≤ As for all 0 For t ∈ (0, 1) we have Z Z 1 1/t Mx∗(t)=t−1/p s1/p−1x∗(s)ds = s1/p−1x∗(st)ds t 1 Z ∞ 1/p−1 ∗ = s x (st)χ(1,1/t)(s)ds 1 Z X∞ 2n 1/p−1 ∗ n−1 ≤ s x (2 t)χ(1,1/t)(s)ds 2n−1 n=1 Z X∞ 2n (n−1)(1/p−1) ∗ n−1 ≤ 2 x (2 t) χ(1,1/t)(s)ds n−1 n=1 2 X∞ (n−1)(1/p−1) ∗ n−1 n−1 n−1 ≤ 2 x (2 t)2 χ(0,1)(2 t) n=1 X∞ (n−1)/p ∗ n−1 n−1 = 2 x (2 t)χ(0,1)(2 t). n=1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTERPOLATION BETWEEN L1 AND Lp, 1 The space X is complete. So we can apply the triangle inequality to infinite sums in X and using (9) we obtain X∞ ∗ (n−1)/p ∗ n−1 n−1 kMx k≤k 2 x (2 t)χ(0,1)(2 t)k n=1 X∞ (n−1)/p ∗ n−1 n−1 ≤ 2 kx (2 t)χ(0,1)(2 t)k n=1 X∞ (n−1)/p ≤ 2 kσ2−n+1 kX→X kxk n=1 X∞ ≤ A 2(n−1)/p2(−n+1)(1/p+ε)kxk n=1 ∞ X 2ε = A 2−nε+εkxk = A kxk, 2ε − 1 n=1 and the proof of the estimate (8) is complete. Recall now the Calder´on-Mitjagin theorem (see [KPS], Th. 4.3) which says that an r. i. space X is an interpolation space between L1 and L∞ if and only if the following condition is satisfied: there exists a constant B>0 such that if y ∈ X, x ∈ L1 and Z Z t t (10) x∗(s)ds ≤ y∗(s)ds for all 0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2936 SERGEI V. ASTASHKIN AND LECH MALIGRANDA Remark 2. We can similarly, as in the proof of Theorem 2, show a more general result: Let 1 ≤ r for any x ∈ Lr and for all 0 4. Strong type (1, 1) and weak type (p, p) interpolation We observed, after the proof of Theorem 2 was completed, that an even more general result is possible to prove. A linear operator T is said to be of weak type (p, p), 1 ≤ p<∞ if T is bounded from Lp,1 into Lp,∞, where the spaces Lp,1 and Lp,∞ on I =[0, 1] are generated by the functionals Z 1/p ∗ 1/p−1 ∗ kxkp,∞ =supt x (t), kxkp,1 = t x (t)dt. t∈I I A bounded linear operator T : Lp → Lp is said to be of strong type (p, p) and, of course, every operator of strong type (p, p)isalsoofweaktype(p, p) but not vice versa since Lp,1 ⊂ Lp ⊂ Lp,∞. Boyd [B69] proved in 1969 that any linear operator T that is of weak types (p, p) ≤ ∞ 1 ≤ and (q, q), 1 p Theorem 3. Let 1 Proof. The sufficiency follows immediately from the proof of Theorem 2 since the 1 ⊂ estimate (6) is still true. We must only show that if α(X) > p ,thenLp,∞ X. By Theorem 5.5 in [KPS], it is enough to prove that Z 1 ∗ −1 x (s)ϕX (s)s ds ≤ C7kxkp,∞ 0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTERPOLATION BETWEEN L1 AND Lp, 1 ∈ 1 for all x Lp,∞. The assumption α(X) > p gives (9) and since the estimate kσsk≥ϕX (s)/ϕX (1) is clear, we obtain Z Z 1 1 ∗ −1 ∗ 1/p+ε−1 x (s)ϕX (s)s ds ≤ AϕX (1) x (s)s ds 0 0 Z 1 ε−1 AϕX (1) ≤ AϕX (1)kxkp,∞ s ds = kxkp,∞. 0 ε Necessity. The operator M is of strong type (1, 1) and of weak type (p, p). 0 Moreover, kMxkp,∞ ≤kxkp,1 for every x ∈ Lp,1 and kMxk1 ≤ p kxk1 for every x ∈ L1. Then M is bounded in X by assumption. If 0 <ε<1/kMkX→X, then the −1 operator (IX − εM) exists and is bounded in X.Moreover, X∞ −1 n n (IX − εM) x(t)= ε M x(t), n=0 where the series converges in the operator norm and M n denotes the nth iteration of M.Wehave Z 1 n−1 s n −1/p 1/p−1 ln t M x(t)=t s − x(s)ds, t (n 1)! and the operator Mf given by X∞ f −1 n n+1 Mx(t)=M(IX − εM) x(t)= ε M x(t) n=0 Z ∞ 1 X εn lnn s = t−1/p s1/p−1 t x(s)ds t n! Z n=0 1 s = t−1/p s1/p−1( )εx(s)ds t t is bounded in X. Since, for 0 <τ<1, Z t/τ s Mxf ∗(t) ≥ t−1/p s1/p−1( )εx∗(s)ds χ (t/τ) t (0,1) t Z t/τ ∗ −1/p 1/p−1 s ε ≥ x (t/τ)χ(0,1)(t/τ)t s ( ) ds t t p = σ x∗(t) τ −1/p−ε(1 − τ 1/p+ε), τ 1+εp it follows that 1 τ 1/p+ε kσ xk = kσ x∗k≤( + ε) kMxf ∗k τ τ p 1 − τ 1/p+ε 1 τ 1/p+ε f ≤ ( + ε) kMk → kxk, p 1 − τ 1/p+ε X X and so α(X) ≥ 1/p + ε>1/p. We should also show that the r. i. space X must be an interpolation space → → between L1 and L∞.IfT : L1 L1 is bounded and T = T|L∞ : L∞ L∞ is also bounded, then by the Riesz-Thorin interpolation theorem T is of strong type (p, p), License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2938 SERGEI V. ASTASHKIN AND LECH MALIGRANDA which implies that T is of weak type (p, p). The assumption on X gives that T is also bounded in X, and the proof is complete. Remark 3. Theorem 3 can be generalized (cf. Remark 2): Let 1 ≤ r [BS] C. Bennett and R. Sharpley, Interpolation of Operators, Academic Press, Boston, 1988. MR 89e:46001 [B67] D.W.Boyd,Spaces between a pair of reflexive Lebesgue spaces, Proc. Amer. Math. Soc. 18 (1967), 215-219. MR 35:3427 [B68] D. W. Boyd, The spectral radius of averaging operators, Pacific J. Math. 24 (1968), 19-28. MR 36:4360 [B69] D.W.Boyd,Indices of function spaces and their relationship to interpolation, Canad. 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Math. Soc. 159 (1971), 207-221. MR 52:1347 [M81] L. Maligranda, A generalization of the Shimogaki theorem, Studia Math. 71 (1981), 69- 83. MR 83d:46034 [M85] L. Maligranda, Indices and interpolation, Dissertationes Math. 234 (1985), 1-54. MR 87k:46059 [Ms] M. Masty lo, Interpolation of linear operators in the K¨othe dual spaces, Ann. Mat. Pura Appl. 154 (1989), 231-242. MR 91f:46105 [R] G.I.Russu,Symmetric spaces of functions that do not have the majorization property, Mat. Issled. 4 (1969), 82-93 (Russian). MR 43:2490 Department of Mathematics, Samara State University, Akad. Pavlova 1, 443011 Samara, Russia E-mail address: [email protected] Department of Mathematics, Lulel˚a University of Technology, se-971 87 Lule˚a, Sweden E-mail address: [email protected] License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use