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In [31], also using a combinatorial approach, the Lorentz spaces isomorphic to a subspace of L1 were characterized by Schütt. In the works [2, 27, 29] embeddings of certain matrix spaces into L1 were obtained. It is natural to ask whether similar embedding results or characterizations can be shown for more general classes of Banach sequence spaces (with a symmetric basis), keeping in mind that already the cases of Orlicz and Lorentz spaces required sophisti- cated ideas and technical finesse. This paper is a contribution towards this goal and we shall show that certain Orlicz-Lorentz spaces, hybrids combining both Orlicz and Lorentz spaces, are uniformly isomorphic to subspaces of L1. The approach we choose is based on combinatorial methods, which are of independent interest. As powerful as these averaging techniques are, as technical they can typically be, and only few experts are really familiar with them. In order to present the main result of this article, let us briefly introduce some notation. For details, we refer the reader to Section 2 or the standard literature on Banach space geometry [17, 18, 24, 33]. We shall denote by dBM the Banach-Mazur distance, which is a measure for the geometric similarity or difference of two isomorphic n spaces. We define an Orlicz-Lorentz space ℓM,a with non-increasing weight sequence a a . . . a 0 and Orlicz function M to be Rn with the norm 1 ≥ 2 ≥ ≥ n ≥ n ∗ aixi x M,a = inf ρ> 0 : M 1 , k k ( ρ ! ≤ ) Xi=1 ∗ n where (xi )i=1 is the non-increasing rearrangement of the coordinates of the vector x. The standard unit vector basis is a symmetric basis for this space. We shall write Sn for the symmetric group of all permutations of the set 1,...,n . Lastly, we denote by n!322n { } L1 the space
n!322n n!322n n L = x(π,σ,ε,δ) R : ε, δ 1, 1 ,π,σ Sn 1 π,σ,ε,δ ∈ ∈ {− } ∈ with the norm
1 3 2n x = x(π,σ,ε,δ) , x =(x(π,σ,ε,δ)) Rn! 2 . k k1 n!322n | | π,σ,ε,δ ∈ π,σ,ε,δX The following theorem shows that for any p (1, 2), every Orlicz function M ∈ for which M(t)/tp−ǫ is decreasing (ǫ > 0), and all weight sequences a for which the Orlicz-Lorentz norm satisfies two Hardy-type inequalities, the corresponding sequence of Orlicz-Lorentz spaces ℓn , n N embeds uniformly into L . M,a ∈ 1
Theorem 1.1. Let n N and 1
2 n!322n where C1,C2 (0, ) are absolute constants. Then there is a subspace Yn of L1 with ∈ ∞ n dim Yn = n such that dBM(ℓM,a,Yn) D, where D (0, ) is a constant depending only on p. ≤ ∈ ∞
The parameter p appears in Theorem 1.1, because in our proof we need to pass through an ℓp space to embed into L1 and this seems to be a technical matter. It can be seen as a kind of convexification argument. A condition similar to M(t)/tp−ǫ is decreasing has already been used in [27]. This condidtion is satisfied, for instance, if M is an r-concave Orlicz function with r 2 Preliminaries We briefly present the notions and some background material used throughout this text. We split this part into several smaller subsections. Basic notions from Banach spaces theory Let X and Y be isomorphic Banach spaces. We say that they are C-isomorphic if there is an isomorphism T : X Y with T T −1 C. We define the Banach-Mazur distance of X and Y by → k kk k ≤ d (X,Y ) = inf T T −1 : T L(X,Y ) isomorphism . BM k kk k ∈ n o Let (Xn)n be a sequence of n-dimensional normed spaces and let Z be another normed space. If there exists a constant C (0, ), such that for all n N there exists a normed space Y Z with dim(Y )=∈n and∞ d (X ,Y ) C, then∈ we say that (X ) n ⊆ n BM n n ≤ n n embeds uniformly into Z or in short: Xn embeds into Z. For a detailed introduction to the concept of Banach-Mazur distances, we refer the reader to [33]. 3 n Let X be a Banach space with basis ei i=1. We call the basis C-symmetric if there { } n R exists a constant C (0, ) such that for all signs εi = 1, all sequences (ai)i=1 in , and all permutations∈π ∞S ± ∈ n n n aiei C εiaπ(i)ei . ≤ i=1 X i=1 X X X We shall use the asymptotic notation a b to express that there exist two positive absolute constants c ,c such that c a b≈ c a and similarly use a . b or a & b. 1 2 1 ≤ ≤ 2 If the constants depend on a parameter α, we indicate this by a α b, a .α b, or ≈ −1 a &α b. We shall use the notations Ave and Ave to denote the averages (n!) S π ε π∈ n −n ∗ and 2 ε∈{−1,1}n , respectively. For a parameter p, we shall denote by p its conjugateP 1 1 for whichP the relation p + p∗ = 1 is satisfied. Orlicz spaces A convex function M : [0, ) [0, ) with M(0) = 0 and M(t) > 0 for t> 0 is called ∞ → ∞ an Orlicz function. An Orlicz function (as we define it) is bijective and continuous on [0, ). We assume that an Orlicz function satisfies ∞ M(t) M(t) lim = 0 and lim = , t→0 t t→∞ t ∞ n Rn which is typically called an N-function. The Orlicz space ℓM is defined as the space equipped with the Luxemburg norm n xi x M = inf ρ> 0 : M | | 1 . k k ( ρ ! ≤ ) Xi=1 An Orlicz function M is called r-concave if t M(t1/r) is a concave function. Given an Orlicz function M, we define its conjugate function7→ M ∗ by the Legendre-Transform, i.e., M ∗(x) = sup xt M(t) . t∈[0,∞) − ∗ ∗∗ 1 p Again, M is an Orlicz function and M = M. For instance, taking M(t)= p t , p 1, ∗ 1 p∗ 1 1 ≥ the conjugate function is given by M (t) = p∗ t with p∗ + p = 1. Notice also that n ∗ the norm of the dual space (ℓ ) is equivalent to ∗ . Moreover, one has the duality M k·kM relation t M −1(t)(M ∗)−1(t) 2t ≤ ≤ for all t 0 (see, e.g., [5]). We say that two Orlicz functions M and N are equivalent ≥ if there are positive constants a and b such that for all t 0 ≥ M(at) N(t) M(bt) ≤ ≤ which is equivalent to aN −1(t) M −1(t) bN −1(t). ≤ ≤ If two Orlicz functions are equivalent so are their norms. Notice that it is enough for the functions M and N to be equivalent in a neighborhood of 0 for the corresponding sequence spaces ℓM and ℓN to coincide [17]. For a detailed and thorough introduction to the theory of Orlicz spaces, we refer the reader to [11], [22], or [28]. 4 Lorentz spaces Let 1 p< and a a . . . a 0. We define the Lorentz space dn(a, p) to be ≤ ∞ 1 ≥ 2 ≥ ≥ n ≥ Rn equipped with the norm n 1/p ∗p x dn(a,p) = aixi , k k ! Xi=1 ∗ n n where (xi )i=1 is the non-increasing rearrangement of ( xi )i=1. For ai 1, we simply n |p/q|−1 ≡ obtain the space ℓp and for the special choice ai = i , 1 p Orlicz-Lorentz spaces Glueing together Orlicz and Lorentz spaces in an ℓM -fashion, we define an Orlicz- Lorentz space ℓn with weight sequence a a . . . a 0 and Orlicz function M,a 1 ≥ 2 ≥ ≥ n ≥ M to be Rn with the norm n ∗ aixi x M,a = inf ρ> 0 : M 1 . k k ( ρ ! ≤ ) Xi=1 The standard unit vector basis in Rn is a 1-symmetric basis for these spaces and turns them into symmetric Banach spaces. For the choice M(t) = tp, we obtain the Lorentz n n n space d (a, p), and for ai 1 the space ℓM,a is simply the Orlicz space ℓM . These spaces have a rich structure and≡ we refer the reader to, for instance, the work of Montgomery- Smith [21]. Combinatorial results As we have already mentioned, our proof is based on a combinatorial approach. We will need the following deep result obtained by the author and Carsten Schütt in their joined work [27, Lemma 2.6]. n Lemma 2.1. Let n N, 1 p ∗ ∗ 1/r 1/p ℓ 1/p 1/r n ∗−1 ℓ ℓ 1 p ℓ 1 r N ai + ai . n! ≈ n! n | | ! n ! n | | Xi=1 i=Xℓ+1 For all such Orlicz functions and all x Rn, we have ∈ 1/p n p/r r c1(r, p) x N Ave xiaπ(i) c2(r, p) x N , k k ≤ π | | ! ≤ k k Xi=1 where c (r, p),c (r, p) (0, ) are constants depending only on r and p. 1 2 ∈ ∞ 5 For the better understanding of the combinatorial methods involved, let us show how an ℓp-norm is generated by averaging over permutations. We will use this result frequently throughout this text and therefore include a proof. n n 1/p Corollary 2.2. Let n N and 1 ∗ 1/r 1 m m 1/r 1 n 1 m a + a r a , n i n n | i| ≥ n i Xi=1 i=Xm+1 Xi=1 where m m 1/p m m 1 1 n 1 −1/p 1 −1/p ai = = ∗ i = ∗ 1+ i . n n i n1/p n1/p ! Xi=1 Xi=1 Xi=1 Xi=2 A standard integral estimate shows that ∗ 1 m n 1/p 2 m 1/p . n i ≤ 1 1 n Xi=1 − p Furthermore, a similar argument shows that, for all m 2, ≥ ∗ m 1 m1/p i−1/p . ≥ 1 1 2 Xi=1 − p Hence, for all m n ∗ ≤ 1 m n 1/p 1 m 1/p , n i ≥ 2(1 1 ) n Xi=1 − p since the case m = 1 is obvious. Therefore, we have for all m n ≤ ∗ ∗ 1 m 1/p 1 m n 1/p 2 m 1/p 2(1 1 ) n ≤ n i ≤ 1 1 n − p Xi=1 − p and hence ∗ 1/r ∗ 1 m m 1/r 1 n 1 m 1/p a + a r . n i n n | i| ≥ 2(1 1 ) n Xi=1 i=Xm+1 − p In addition, we have 1/2 1/2 1/2 n n m 1 2 √m −2/p a = ∗ i . n n | i| n1/p i=Xm+1 i=Xm+1 6 Again, a simple integral estimate yields n 1 i−2/p m−2/p+1 n−2/p+1 , ≤ 2 1 − i=Xm+1 p − and so we have 1/2 1/2 1/2 n 1 1 i−2/p m−2/p+1 = m−1/p+1/2. ≤ 2 1 2 1 i=Xm+1 p − p − Therefore, 1/2 1/2 1/2 n 1/p∗ √m −2/p √m 1 −1/p+1/2 1 m ∗ i ∗ m = . n1/p ≤ n1/p 2 1 2 1 n i=Xm+1 p − p − Hence, for all m n ≤ 1/2 ∗ 1/2 ∗ 1 m m 1/2 1 n 2 m 1/p 1 m 1/p a + a 2 + n i n n | i| ≤ 1 1 n 2 1 n Xi=1 i=Xk+1 − p p − ∗ 2 1 m 1/p = + , 1 1 2 1 n − p p − q which concludes the proof. Another combinatorial tool that we shall use is the following result of Schütt taken from [31, Lemma 2.3]. Lemma 2.3. Let n N and 1 p 2. Then, for all x Rn and any k =1....,n, ∈ ≤ ≤ ∈ 1/p p/2 k 1/p n 1/2 2 1 ∗ p 1 ∗ 2 Ave xπ(i) x + x , π i i n ! ≈ k i=1 | | ! k i=k+1 | | ! i≤ k X X X ∗ where xi , i = 1,...,n denotes the non-increasing rearrangement of the numbers xi , i =1,...,n. | | Readers familiar with interpolation theory might recognize expressions of this type. Indeed, a result of this flavor has also been obtained via real interpolation techniques by Lechner, Passenbrunner, and Prochno in [14]. 3 Combinatorial elements of the proof Before we come to the embedding of Orlicz-Lorentz spaces into L1, we require a new combinatorial averaging device. We prove that an abstract Orlicz-Lorentz norm can be generated by permutation-averages of an ℓ2-norm of suitably chosen vectors. Recall that to any vector d Rn whose coordinates are non-increasing there corresponds an Orlicz function as shown∈ in Lemma 2.1. 7 n 1/p Theorem 3.1. Let 1 1/2 1/p n n n p/2 k 2 k 2 Ave xicπ(i)dσ(k)zη(k) p Ave xicπ(i)dσ(k) . π,σ,η | | ≈ π,σ | | ! i,kX=1 kX=1 Xi=1 The triangle inequality yields the estimate 1/p 1/p n n p/2 n n 1/2 p k 2 k 2 Ave xiciπ(i)dσ(k) Ave Ave xicπ(i)dσ(k) π,σ | | ! ≥ σ π | | ! k=1 i=1 k=1 i=1 X X X X 1/p n n 1 /2 p p k 2 = Ave dσ(k) Ave xicπ(i) . σ | | π | | ! k=1 i=1 X X An application of Lemma 2.3, where we simply omit the quadratic term, shows that 1/p n n 1/2 p n k p 1/p p k 2 p 1 ∗ p Ave dσ(k) Ave xicπ(i) Ave dσ(k) xi ak . σ | | π | | ! ≈ σ | | k ! k=1 i=1 k=1 i=1 X X X X By Lemma 2.1 there exists an Orlicz function Md such that n k p 1/p k n p 1 ∗ p 1 ∗ ∗ n Ave dσ(k) x a p ak x (akx ) , σ k i k k i k k=1 Md k=1 | | i=1 ! ≈ i=1 !k=1 ≥k k X X X Md where we used that k x∗ kx∗ for any k = 1,...,n. Therefore, we obtain the i=1 i ≥ k desired lower bound P 1/2 n k 2 Ave xic dσ(k)zη(k) c1(p) x , π,σ,η | π(i) | ≥ k kMd,a i,kX=1 8 where the constant c (p) (0, ) depends only on p. 1 ∈ ∞ We now proceed with the upper bound. As we have seen, since z generates the ℓp-norm, we have 1/2 1/p n n n p/2 k 2 k 2 Ave xicπ(i)dσ(k)zη(k) p Ave xicπ(i)dσ(k) . π,σ,η | | ≈ π,σ | | ! i,kX=1 kX=1 Xi=1 It follows from Jensen’s inequality that 1/p 1/p n n p/2 n n p/2 k 2 p k 2 Ave xicπ(i)dσ(k) Ave dσ(k) Ave xicπ(i) . π,σ | | ! ≤ σ | | π | | ! kX=1 Xi=1 kX=1 Xi=1 Using again Lemma 2.1, we obtain that 1/p 1/p n n n p/2 n p/2 p k 2 k 2 Ave dσ(k) Ave xicπ(i) p Ave xicπ(i) . σ | | π | | ! ≈ π | | ! kX=1 Xi=1 Xi=1 k=1 Md An application of Lemma 2.3 and the triangle inequality yields 1/p n n p/2 k 2 Ave xicπ(i) π | | ! Xi=1 k=1 Md n 1/p n 1/2 1 k 1 n . a x∗p + a x∗2 . k i k i k i=1 ! k i=k+1 X k=1 Md X k=1 Md Therefore, conditions (3) and (4) show that 1/2 n k 2 Ave xic dσ(k)zη(k) c2(p) x , π,σ,η | π(i) | ≤ k kMd,a i,kX=1 with constant c (p) (0, ) depending only on p. This completes the proof. 2 ∈ ∞ Remark 3.2. Let us note that the first Hardy-type inequality in Theorem 3.1 (and Theorem 1.1) holds, for instance, if the weight sequence a defining the Orlicz-Lorentz space does not decay too slowly, i.e., if for all k =1,...,n, n a i Ca k1−1/p, (5) i1/p ≤ k i=Xk+1 where C (0, ) is an absolute constant. Indeed, using the Hahn-Banach theorem and ∈ ∞ n that the norm of the dual space of ℓ is up to a factor 2 equivalent to M ∗ , we obtain Md k·k d n k 1/p n k 1/p 1 ∗p ak ∗p ak xi sup yk xi , n 1/p k ≈ y∈B ∗ k i=1 ! M k=1 i=1 ! X k=1 Md d X X 9 n n where B ∗ is the unit ball of the space ℓ ∗ . Since we may assume without loss of Md Md generality that y1 . . . yn and because of the Lorentz-norm estimate p 2 p,1, we obtain ≥ ≥ k·k ≤ k·k n k 1/p n k ak ∗p ak 1 ∗ sup yk xi 2 sup yk x 1/p n 1/p 1−1/p i ∗ k ∗ k i y∈BM k=1 i=1 ! ≤ y∈BM k=1 i=1 X X d X X n n 1 ∗ ak = 2 sup x yk n 1−1/p i 1/p ∗ i k y∈BM i=1 k=i d X X n n 1 ∗ ak 2 sup x yi . n 1−1/p i 1/p ∗ i k ≤ y∈BM i=1 k=i d X X Applying condition (5) and again the duality relation used before, we find that n n n 1 ∗ ak 1 ∗ 1−1/p ∗ n sup x yi C sup x yii ai (x ai) . n 1−1/p i 1/p n 1−1/p i i i=1 Md ∗ i k ∗ i y∈BM i=1 k=i ≤ y∈BM i=1 ≈k k d X X d X 4 The embedding of Orlicz-Lorentz spaces into L1 We are now prepared to present the proof of Theorem 1.1, which is based on Theorem 3.1. In fact, we need to prove two things. One is that we can choose the vector d such that it yields an Orlicz function equivalent to the one from Theorem 3.1. The second is that the average over permutations from Theorem 3.1 is equivalent to an L1-norm. The latter will be a consequence of Khintchine’s inequality. n Proof of Theorem 1.1. We consider z = (n/i)1/p and choose the sequence d such i=1 that, for all ℓ =1,...,n, ℓ 1 ℓ (M ∗)−1 = d . n n i Xi=1 Let c1,...,cn Rn be the vectors as chosen in Theorem 3.1. Then Theorem 3.1 shows that ∈ 1/2 n k 2 Ave xic dσ(k)zη(k) p x , π,σ,η | π(i) | ≈ k kMd,a i,kX=1 where M is an Orlicz function such that for all ℓ n d ≤ ∗ 1/p ℓ 1/p n ∗ −1 ℓ 1 ℓ 1 p (Md ) di + di . n! ≈ n | | n! n | | Xi=1 i=Xℓ+1 ∗ ∗ We show that Md and M are equivalent Orlicz functions. The lower bound is imme- diate, because of the choice of the sequence d and so we obtain ℓ ∗ −1 ℓ 1 ∗ −1 ℓ (Md ) & di =(M ) . n! n | | n! Xi=1 Let us proceed with the upper bound. We have ∗ 1/p 1/p n ∗ −1 ℓ ∗ −1 ℓ ℓ 1 p (Md ) . (M ) + di n ! n! n! n | | i=Xℓ+1 10 and so it remains to estimate the second term on the right-hand side. First, we observe that for all ℓ =1,...,n ∗ −1 ℓ ℓ (M ) dℓ. n! ≥ n Using the relation t M −1(t)(M ∗)−1(t) 2t that holds for all t 0, we obtain that, for all ℓ =1,...,n, ≤ ≤ ≥ (M ∗)−1(ℓ/n) 2 d . ℓ ≤ ℓ/n ≤ M −1(ℓ/n) Hence, for all ℓ =1,...,n, n p n p 1 p 2 1 di −1 n i=ℓ+1 | | ≤ n i=ℓ+1 M (i/n) X X p− ǫ Consider pǫ := p ǫ for ǫ (0,p 1) such that M(t)/t is decreasing. Then also t − ∈ − t −1 is a decreasing function. In particular, the function 7→ (M (t))pǫ t p/pǫ tp/pǫ t = −1 pǫ −1 p 7→ (M (t)) (M (t)) is decreasing. Therefore, we obtain the estimate 1 n 2p n (i/n)p/pǫ n p/pǫ 2p (ℓ/n)p/pǫ n n p/pǫ d p n | i| ≤ n (M −1(i/n))p i ≤ n (M −1(ℓ/n))p i i=Xℓ+1 i=Xℓ+1 i=Xℓ+1 2p ℓp/pǫ n ℓp/pǫ = i−p/pǫ . ℓ1−p/pǫ n (M −1(ℓ/n))p p n (M −1(ℓ/n))p i=Xℓ+1 ℓ/n = , (M −1(ℓ/n))p where in the penultimate step we used that pǫ < p. Putting everything together, we arrive at ∗ 1/p 1/p n 1/p∗ 1/p ℓ 1 p (ℓ/n) (ℓ/n) ℓ/n ∗ −1 ℓ di .p = (M ) , n! n | | M −1(ℓ/n) M −1(ℓ/n) ≤ n! i=Xℓ+1 where in the last step we used the duality relation t/M −1(t) (M ∗)−1(t) valid for all t 0. This shows that the Orlicz functions M and M are equivalent≤ and we have ≥ d 1/2 n k 2 Ave xic dσ(k)zη(k) p x . π,σ,η | π(i) | ≈ k kM,a i,kX=1 We now define our embedding into L1 as follows, n 3 2n Ψ : ℓn Ln! 2 , x x ck d z ε δ . n M,a → 1 7→ i π(i) σ(k) η(k) i k i,kX=1 π,σ,η,ε,δ It is a direct consequence of Khintchine’s inequality (see, e.g., [33, Theorem 6.1]) that Ψ (x) x . k n k1 ≈k kM,a This means that there exists a constant D (0, ) depending only on p such that N ∈ n∞!322n for any n there exists a subspace Yn of L1 with dim(Yn) = n such that d (ℓn ,Y∈) D. BM M,a n ≤ 11 Let us close this article with a comment on the special case of embeddings of Lorentz spaces into L1. Remark 4.1. 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