L p +L and L p L are not isomorphic for all 1 p <, p 2

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1 L p +L and L p L are not isomorphic for all 1 p <, p 2 Sergei V. Astashkin and Lech Maligranda arxiv: v1 [math.fa] 6 Apr 217 Abstract We prove the result stated in the title. It comes as a consequence of the fact that the space L p L, 1 p <, p 2, does not contain a complemented subspace isomorphic to L p. In particular, as a subproduct, we show that L p L contains a complemented subspace isomorphic to l 2 if and only if p = 2. 1 Preliminaries and main result Isomorphic classification of symmetric spaces is an important problem related to the study of symmetric structures in arbitrary Banach spaces. This research was initiated in the seminal work of Johnson, Maurey, Schechtman and Tzafriri [9]. Somewhat later it was extended by Kalton to lattice structures [1]. In particular, in [9] (see also [12, Section 2.f] it was shown that the space L 2 L p for 2 p < (resp. L 2 +L p for 1 < p 2 is isomorphic to L p. A detailed investigation of various properties of separable sums and intersections of L p -spaces (i.e., with p < was undertaken by Dilworth in the papers [5] and [6]. In contrast to that, we focus here on the problem if the nonseparable spaces L p +L and L p L, 1 p <, are isomorphic or not. In this paper we use the standard notation from the theory of symmetric spaces (cf. [3], [11] and [12]. For 1 p < the space L p + L consists of all sums of p-integrable and bounded measurable functions on (, with the norm defined by x Lp+L := ( inf u Lp + v L. x(t=u(t+v(t,u L p,v L The L p L consists of all bounded p-integrable functions on (, with the norm x Lp L := max { { } ( } 1/p,esssup x Lp, x L = max x(t dt p x(t. Both L p +L and L p L for all 1 p < are non-separable Banach spaces (cf. [11, p. 79] for p = 1. The norm in L p +L satisfies the following sharp estimates t> ( 1 1/p ( 1 1/p x (t p dt x Lp+L 2 1 1/p x (t dt p (1 Research partially supported by the Ministry of Education and Science of the Russian Federation. 21 Mathematics Subject Classification: 46E3, 46B2, 46B42 Key words and phrases: symmetric spaces, isomorphic spaces, complemented subspaces 1

2 (cf. [4, p. 19], [13, p. 176] and with details in [14, Theorem 1] see also [3, pp ] and [11, p. 78], where we can find a proof of (1 in the case when p = 1, that is, x L1 +L = 1 x (tdt. Here, x (t denotes the decreasing rearrangement of x(u, that is, x (t = inf{τ > : m({u > : x(u > τ} < t} (if E R is a measurable set, then m(e is its Lebesgue measure. Note that every measurable function and its decreasing rearrangement are equimeasurable, that is, m({u > : x(u > τ} = m({t > : x (t > τ} for all τ >. Denote by L and (L p +L,1 p <, the closure of L 1 L in L and in L p +L, respectively. Clearly, (L p +L = L p +L. Note that and L p +L = {x L p +L : x (t as t } (2 (L 1 +L = L 1 L, i.e., L 1 L,1 p <, is a dual space (cf. [11, pp. 79-8] and [3, pp ]. Also, L p L and L p +L,1 < p <, are dual spaces because where 1/p+1/q = 1. (L q +L 1 = L p L and (L q L 1 = L p +L, Now, we state the main result of this paper. THEOREM 1. For every 1 p <,p 2, the spaces L p + L and L p L are not isomorphic. Clearly, thespace L p +L containsthecomplemented subspace (L p +L [,1] isomorphic to L p [,1] for every 1 p <. As a bounded projection we can take the operator Px := xχ [,1] because ( 1 1/p ( 1 1/p Px Lp = xχ [,1] Lp = x(t p dt x (t dt p x Lp+L. In the next two sections we show that L p L for p [1,2 (2, does not contain a complemented subspace isomorphic to L p, which gives our claim. At the same time, note that L p L, 1 p <, contains a subspace isomorphic to L and hence a subspace isomorphic to L p. The spaces L p +L and L are not isomorphic since L p +L contains a complemented subspace isomorphic to L p and L is a prime space (this follows from the Lindenstrauss and Pe lczyński results see [1, Theorems and 4.3.1]. Similarly, the spaces L p L and L are not isomorphic because of L p L contains a complemented subspace isomorphic to l p (take, for instance, the span of the sequence {χ [n 1,n } in L p L. If {x n } is a sequence from a Banach space X, by [x n] we denote its closed linear span in X. As usual, the Rademacher functions on [,1] are defined as follows: r k (t = sign(sin2 k πt, k N,t [,1]. 2

3 2 L 1 L does not contain a complemented subspace isomorphic to L 1 OurproofofTheorem1inthecasep=1willbebasedonanapplicationoftheHagler-Stegall theorem proved in [8] (see Theorem 1. To state it we need the following definition. The space ( ln lp, 1 p <, is the Banach space of all sequences {c n k }, (c n k n ln, n = 1,2,..., such that ( {c n k } := 1/p ( 1/p (c n k n p l = max 1 k n cn k p <. THEOREM 2 (Hagler-Stegall. Let X be a Banach space. Then its dual X contains a complemented subspace isomorphic to L 1 if and only if X contains a subspace isomorphic to ( ln l 1. Note that (L 1 +L [,1] = L 1 [,1], and hence L 1 +L contains a complemented copy of L 1 [,1], and so of l 1. Moreover, its subspace { } c k χ [k 1,k] : c k as k is isomorphic to c and so, by Sobczyk theorem (cf. [1, Theorem 2.5.8], is complemented in the separable space L 1 +L. Therefore, the latter space contains uniformly complemented copies of l,n n N. However, we have THEOREM 3. The space L 1 +L does not contain any subspace isomorphic to the space ( ln l1. Proof. On thecontrary, assume that L 1 +L contains a subspace isomorphic to ( ln l 1. Let x n k,n N,k = 1,2,...,n, form the sequence from L 1+L equivalent to the unit vector basis of ( ln l1. This means that there is a constant C > such that for all a n k R C 1 max,2,...,n an k n a n k xn k C L1 +L max,2,...,n an k. In particular, for any n N, every subset A {1,2,...,n} and all ε k = ±1,k A, we have ε k x n k = L1 +L k A 1 (3 ( (sds ε k xk n C, (4 and for all 1 k(n n,n N the sequence {x n k(n } is equivalent in L 1+L to the unit vector basis of l 1, i.e., for all a n R C 1 a n 1 k A ( (sds a n x n k(n C a n. (5 3

4 Moreover, we can assume that x n k L 1 +L = 1 for all n N,k = 1,2,...,n, i.e., 1 (x n k (sds = 1,n N,k = 1,2,...,n. (6 Firstly, we show that for every δ > there is M = M(δ N such that for all n N and any E (, with m(e 1 we have card{k = 1,2,...,n: x n k(s ds δ} M. (7 Indeed, assuming the contrary, for some δ > we can find n i,e i (,,m(e i 1,i = 1,2,..., such that card{k = 1,2,...,n i : x n i k (s ds δ }. E i Denoting A i := {k = 1,2,...,n i : E i x n i k (s ds δ }, for all ε k = ±1 we have 1 ( ε k x n i (sds k = ε k x ni k ε k x n i ds. k (s (8 L1 +L k A i k A E i i k A i Moreover, by the Fubini theorem, Khintchine s inequality in L 1 (cf. [12, pp. 5-51] or [17] and Minkowski inequality, we obtain 1 1 dsdt r k (tx ni k (s dtds = k (s E i k A E i i 1 2 E i E r k (tx ni k A i ( 1/2ds x ni k (s 2 k A i 1 2 ( k A i ( E i x n i k (s ds 2 1/2 δ 2 cardai. Therefore, for each i N there are signs ε k (i,k A i such that ε k (ix n i ds k (s δ cardai. k A i 2 E i Combining this with (8 we obtain that ε k (ix n i δ cardai,i = 1,2,... L1 +L k A i 2 k Since carda i as i, the latter inequality contradicts (4. Thus, (7 is proved. Now, we claim that for all δ > and n N { card k = 1,2,...,n: there is F [, such that m(f 1 M+1 and F xn k (s ds δ } M, (9 4

5 where M depending on δ is taken from (7. Indeed, otherwise, we can find δ >,n N and I {1,2,...,n }, cardi = M + 1,M = M(δ, such that for every k I there is F k (, with m(f k 1 M +1 and F k x n k (s ds δ. Setting E = k I F k, we see that m(e k I m(f k 1. Moreover, by the definition of I and E, card{k = 1,2,...,n : x n k (s ds δ } cardi > M, E which is impossible because of (7. Now, we construct a special sequence of pairwise disjoint functions, which is equivalent in L 1 +L to the unit vector basis in l 1. By (7, for arbitrary δ 1 > there is M 1 = M 1 (δ 1 N such that for all n N card{k = 1,2,...,n: 1 x n 1 (s ds δ 1 } M 1. Therefore, taking n 1 > 2M 1, we can find = 1,2,...,n 1 satisfying 1 x n 1 (s ds < δ 1 and, by (9, such that from F (, with m(f 1 M 1 +1 x n 1 (s ds < δ 1. F it follows that Moreover, recalling (2 we have (x n 1 (t as t. Therefore, since x n 1 L 1 +L and any measurable function is equimeasurable with its decreasing rearrangement, there exists m 1 N such that x n 1 χ [m1, L1 +L δ 1. Then, setting y 1 := x n 1 χ [1,m1 ], we have x n 1 y 1 L1 +L 2δ 1. Next, by (7, for arbitrary δ 2 > there is M 2 = M 2 (δ 2 N such that for all n N and j = 1,2,...,m 1 card{k = 1,2,...,n: j j 1 x n k (s ds δ 2} M 2. Let n 2 N be such that n 2 > M 2 m 1 +M 2 +M 1. Then, by the preceding inequality and (9, there is 1 k 2 n 2 such that for all j = 1,2,...,m 1 we have and from F (, with m(f 1 M i,i = 1,2, it follows that +1 x n 2 k 2 (s ds δ i. j j 1 x n 2 k 2 (s ds δ 2, (1 F 5

6 Note that (1 implies m 1 x n 2 k 2 (s ds m 1 δ 2, whence x n 2 k 2 χ [,m1 ] L1 +L m 1 δ 2. As above, by (2, there is m 2 > m 1 such that x n 2 k 2 χ [m2, L1 +L m 1 δ 2. Thus, putting y 2 := x n 2 k 2 χ [m1,m 2 ], we have x n 2 k 2 y 2 L1 +L 2m 1 δ 2. Continuing this process, for any δ 3 >, by (7, we can find M 3 N such that for all n N and j = 1,2,...,m 2 it holds card{k = 1,2,...,n: j j 1 x n k (s ds δ 3} M 3. So, again, applying (9 and taking n 3 > m 2 M 3 +M 1 +M 2 +M 3 we find 1 k 3 n 3 such that and j j 1 x n 3 k 3 (s ds δ 3, j = 1,2,...,m 2, F x n 3 k 3 (s ds δ i, whenever m(f 1 M i +1,i = 1,2,3. This implies that m 2 x n 3 k 3 (s ds m 2 δ 3, and so x n 3 k 3 χ [,m2 ] L1 +L m 2 δ 3. Choosing m 3 > m 2 so that x n 3 k 3 χ [m3, L1 +L m 2 δ 3 and setting y 3 := x n 3 k 3 χ [m2,m 3 ], we obtain x n 3 k 3 y 3 L1 +L 2m 2 δ 3. As a result, we get the increasing sequences n i,m i,k i of natural numbers, 1 k i n i,i = 1,2,... and the sequence {y i } of pairwise disjoint functions from L 1 +L such that x n i k i y i L1 +L 2m i 1 δ i, where m := 1. Noting that the sequence of positive reals {δ i } i=1 can be chosen in such a way that the numbers m i 1 δ i would be arbitrarily small, we can assume, by the principle of small perturbations (cf. [1, Theorem 1.3.1] and by inequalities (5, that {y i } is equivalent in L 1 +L to the unit vector basis of l 1. Moreover, by construction, for all j = 1,2,... and i = 1,2,3,...,j we have 1 y j (s ds δ i whenever m(f M i +1. (11 F Let 1 l < m be arbitrary. Since y i,i = 1,2,... are disjoint functions, then m y i = L1 +L i=l 1 ( m (sds m y i = i=l i=l E i y i (s ds, (12 6

7 where E i are disjoint sets from (, such that m i=l m(e i 1. Clearly, for a fixed l we have 1 k (m := card{i N : l i m and m(e i > M l +1 } M l +1. Hence, by (11, (12 and (6, m y i k (m+(m l k (mδ l. L1 +L So, assuming that m (M l +1/δ l +l, we obtain i=l m y i 2δ l (m l. L1 +L i=l Since δ l as l, the latter inequality contradicts the fact that {y i } is equivalent in L 1 +L to the unit vector basis of l 1. The proof is complete. Remar. Since the space L p, 1 p <, is of Rademacher cotype max(p,2, the result of Theorem 3 can be generalized as follows: For every 1 p < the space L p +L does not contain any subspace isomorphic to the space ( ln lp. Proof of Theorem 1 for p = 1. By the Hagler-Stegall theorem 2, Theorem 3 and the fact that L 1 L = (L 1 +L, we obtain that (in contrast to L 1 +L the space L 1 L does not contain a complemented subspace isomorphic to L 1 [,1], which gives our claim. There is a natural question (cf. also [2, p. 28] if the space L 1 +L is isomorphic to a dual space? Our guess is that not, but we don t have a proof. Of course, the answer not would imply immediately the result of Theorem 1 for p = 1. Problem 1. Is the space L 1 +L isomorphic to a dual space? 3 L p L for p 2 does not contain a complemented subspace isomorphic to L p The well-known Raynaud s result (cf. [15, Theorem 4] presents the conditions under which a separable symmetric space (on [, 1] or on (, contains a complemented subspace isomorphic to l 2. The following theorem can be regarded as its extension to a special class of nonseparable spaces. THEOREM 4. Let 1 p <. Then the space L p L contains a complemented subspace isomorphic to the space l 2 if and only if p = 2. Proof. If p = 2, then clearly the sequence {χ [n 1,n } is equivalent in L 2 L to the unit vector basis of l 2 and spans a complemented subspace. Let us prove necessity. On the contrary, let {x n } L p L be a sequence equivalent in L p L to the unit vector basis of l 2 so that [x n ] is a complemented subspace of L p L. 7

8 Firstly, let us show that there is not a > such that for all c k R,k = 1,2,... c k x k χ [,a] L1 Indeed, the latter equivalence implies L1 c k x k χ [,a] c k x k. Lp L c k x k (c k l2. Lp L [,a] Since L p L [,a] = L [,a], we see that the sequence {x n χ [,a] } spans in both spaces L 1 [,a] and L [,a] the same infinite-dimensional space. However, by the well-known Grothendieck s theorem (cf. [7, Theorem 1]; see also [16, p. 117] it is impossible. As a result, we can find a sequence {f n } [x k ], f n Lp L = 1,n = 1,2,..., such that for every a > a f n (t dt as n. Hence, f m n (convergence in Lebesgue measure m on any interval [,a]. Since [x k ] spans l 2, then passing to a subsequence if it is necessary (and keeping the same notation, we can assume that f n weakly in L p L. Therefore, combining the Bessaga-Pe lczyński Selection Principle (cf. [1, Theorem 1.3.1] and the principle of small perturbations (cf. [1, Theorem 1.3.1], we can select a further subsequence, which is equivalent to the sequence {x k } in L p L (and so to the unit vector basis in l 2 and which spans a complemented subspace in L p L. Let it be denoted still by {f n }. Now, we will select a special subsequence from {f n }, which is equivalent to a sequence of functions whose supports intersect only over some subset of (, with Lebesgue measure at most 1. Let {ε n } be an arbitrary (by now decreasing sequence of positive reals, ε 1 < 1. Since f m n on [,1], there is n 1 N such that m({t [,1]: f n1 (t > ε 1 } < ε 1. (13 Moreover, the fact that f n1 χ (m, Lp as m allows us to find m 1 N, for which Clearly, from (14 it follows that Denoting f n1 χ [m1, Lp ε 2 2. (14 m({t [m 1, : f n1 (t > ε 2 } ε 2. (15 A 1 := {t [,1]: f n1 (t > ε 1 }, B 1 := {t [m 1, : f n1 (t > ε 2 } and g 1 := f n1 (χ A1 +χ B 1 +χ [1,m1 ], from (13, (14 and (15 we have f n1 g 1 Lp L ε 1 +max(ε 2,ε 2 2 2ε 1. 8

9 Further, since f n m on [,m 1 ], there exists n 2 > n 1,n 2 N such that m({t [,m 1 ]: f n2 (t > ε 2 m 1 } < ε 2. (16 Again, using the fact that f n2 χ (m, Lp as m, we can choose m 2 > m 1 in such a way that f n2 χ [m2, Lp ε 2 3. (17 and also From (17, obviously, it follows that Setting m(b 1 1 < ε 3, where B 1 1 := B 1 [m 2,. (18 m({t [m 2, : f n2 (t > ε 3 } ε 3. (19 A 2 := {t [,m 1 ]: f n2 (t > ε 2 m 1/p 1 }, B 2 := {t [m 2, : f n2 (t > ε 3 } and g 2 := f n2 (χ A2 +χ B 2 +χ [m1,m 2 ], by (16, (17 and (19, we get f n2 g 2 Lp L max(ε 2 m 1/p 1,ε 2 +max(ε 3,ε 2 3 < 2ε 2. Let s do one more step. Since f n m on [,m 2 ], there exists n 3 > n 2,n 3 N such that m({t [,m 2 ]: f n3 (t > ε 3 m 1/p 2 } < ε 3. (2 As above, we can choose m 3 > m 2 with the properties f n3 χ [m3, Lp ε 2 4, (21 and From (21 we infer that m(b 2 1 < ε 4, where B 2 1 := B 1 [m 3,, (22 m(b 1 2 < ε 4, where B 1 2 := B 2 [m 3,. (23 m({t [m 3, : f n3 (t > ε 4 } ε 4. (24 Finally, putting A 3 := {t [,m 2 ]: f n3 (t > ε 3 m 1/p 2 }, B 3 := {t [m 3, : f n3 (t > ε 4 } and g 3 := f n3 (χ A3 +χ B 3 +χ [m2,m 3 ], 9

10 by (2, (21 and (24, we have f n3 g 3 Lp L max(ε 3 m 1/p 2,ε 3 +max(ε 4,ε 2 4 < 2ε 3. Continuing in the same way, we get the increasing sequences of natural numbers {n k },{m k }, the sequences of sets {A k },{Bi k } i=,k = 1,2,... and the sequence of functions g k := f nk (χ Ak +χ B k +χ [mk 1,m k ], (where m = 1, satisfying the properties (see (18, (22 and (23 and m(a k ε k,k = 1,2,..., (25 m(b i k ε k+i+1,k = 1,2,...,i =,1,2,..., (26 f nk g k Lp L 2ε k,k = 1,2,... In particular, by the last inequality, choosing sufficiently small ε k, k = 1,2,..., and applying once more the principle of small perturbations [1, Theorem 1.3.1], we may assume that the sequence {g k } is equivalent to {f nk } (and so to the unit vector basis of l 2 and the subspace [g k ] is complemented in L p L. Thus, for some C > and all (c k l 2, i=1 C 1 (c k l2 i=2 c k g k C (c k l2. (27 Lp L Now, denote C 1 := A i B1,C 2 := A i B1 B 2,C 3 := A i B1 1 B 2 B 3,......,C j := i=j i=3 A i B j 2 1 B j B 1 j 2 B j 1 B j,... Setting C := j=1 C j and applying (25 and (26, we have m(c m(c j j=1 ( ε i +jε j 1 (28 j=1 whenever ε k,k = 1,2,..., are sufficiently small. Putting ( ( D 1 = [1,m 1 ]\ A i,d 2 = [m 1,m 2 ]\ A i B1, i=2 ( D 3 = [m 2,m 3 ]\ A i B1 1 B2,..., D j = [m j 1,m j ]\ i=4 ( i=j+1 i=j i=3 A i B j 2 1 B j B 1 j 2 B j 1 1,...,

11 and recalling the definition of g k,k = 1,2,..., we infer that g k = u k +v k, where u k := g k χ Ck and v k := g k χ Dk,k = 1,2,... Note that ( C k ( D k =, whence (27 can be rewritten as follows 1 2 C 1 (c k l2 max ( c k u k, c k v k C (c k l2. (29 Lp L Lp L Moreover, the subspace [u k ] is also complemented in L p L and, by (28, we have ( m suppu k 1. (3 Now, suppose that liminf k u k Lp L =. Then passing to a subsequence (and keeping the same notation, by (29, we obtain 1 2C (c k l2 c k v k C (c k l2. (31 Lp L Since v k,k = 1,2,..., are pairwise disjoint, we have ( Lp L c k v k = max c k v k, c k v k Lp L ( ( 1/p, max c k p v k p L p sup c k v k L. (32 k N Firstly, let us assume that 1 p < 2. If limsup k v k Lp >, then selecting a further subsequence (and again keeping notation, we obtain the inequality ( 1/p, c k v k c c k Lp L p which contradicts the right-hand estimate in (31. So, lim k v k Lp =, and then from (32 for some subsequence of {v k } (we still keep notation we have c k v k C 1 sup c k, (33 Lp L k N and now the left-hand side of (31 fails. Thus, if 1 p < 2, inequality (31 does not hold. Let p > 2. Clearly, from (32 it follows that ( 1/p, c k v k C 2 c k Lp L p (34 11

12 and so the left-hand side estimate in (31 cannot be true. Thus, (31 fails for every p [1,2 (2,, and as a result we get lim inf k u k Lp L >. Now, if1 p < 2,then, asabove, lim k v k Lp =,andwecome(forsomesubsequence of {v k } to inequality (33. Clearly, c k u k csup c k, (35 Lp L k N and from (33 and (29 it follows that for some C > and all (c k l 2 we have C 1 (c k l2 c k u k C (c k l2. (36 Lp L Therefore, the subspace [u k ] is isomorphic in L p L to l 2. We show that the last claim holds also in the case p > 2. On the contrary, assume that the left-hand side of (36 fails (note that the opposite side of (36 follows from (29. In other words, assume that there is a sequence (c n k l 2,n = 1,2,..., such that (c n k l 2 = 1 for all n N and c n k u k as n. Lp L Then, by (35, we have sup k N c n k as n. Therefore, since c n k p ( sup k N c n k p 2 c n k 2 = ( sup c n k p 2, k N we have cn k p as n. Combining this together with (34, we obtain c n k v k as n, Lp L and so the left-hand estimate in (29 does not hold. This contradiction shows that (36 is valid for every p [1,2 (2,. Thus, the subspace [u k ] is complemented in L p L and isomorphic to l 2. As an immediate consequence of that, we infer that [u k ] is a complemented subspace of the space L p L (E, where E = suppu k = C k. Since by (3 m(e 1, it follows that L p L (E is isometric to L (E. As a result we come to a contradiction, because L does not contain any complemented reflexive subspace (cf. [1, Theorem 5.6.5]. Proof of Theorem 1 for p (1,2 (2,. Clearly, if 1 < p < then L p (and hence L p + L containsacomplemented copyofl 2 (forinstance, thespanoftherademacher sequence. Therefore, by applying Theorem 4, we complete the proof. 12

13 Note that if X is a symmetric space on (,, then X +L contains a complemented space isomorphic to X[,1] = {x X: suppx [,1]} since xχ [,1] X C X x X+L for x X +L, where C X max(2 χ [,1] X,1. In fact, for x X +L, using estimate (4.2 from [11, p. 91], we obtain x X+L xχ [,1] X+L inf ( xχ A X + xχ [,1]\A L A [,1] inf ( xχ 1 A X + xχ [,1]\A X 1 xχ [,1] X. A [,1] 2 χ [,1] X C X So, an inspection of the proofs of Theorems 4 and 1 (in the case when p (1,2 (2, shows that the following more general result is true. THEOREM 5. Suppose X is a separable symmetric space on (, satisfying either the upper p-estimate for p > 2 or lower q-estimate for q < 2. Then the space X L does not contain any complemented subspace isomorphic to l 2. If, in addition, the space X[,1] contains a complemented subspace isomorphic to l 2, then the spaces X L and X +L are not isomorphic. 4 Concluding remarks related to the spaces L 2 + L and L 2 L We do not know whether the spaces L 2 +L and L 2 L are isomorphic or not. Problem 2. Are the spaces L 2 +L and L 2 L isomorphic? We end up the paper with the following remarks related to the above problem. Remark 2. The predual spaces L 1 L 2 and L 1 +L 2 for L 2 +L and L 2 L, respectively, are not isomorphic. In fact, L 1 L 2 is a separable dual space since (L 2 +L = L 2 L 1 (cf. [5, Proposition 2(a]. Therefore, the space L 1 [,1] cannot be embedded in this space (cf. [1, p. 147] but L 1 +L 2 hasacomplementedsubspaceisomorphictol 1 [,1],whichcompletesourobservation. Remark 3. Either of the spaces L 2 +L and L 2 L is isomorphic to a (uncomplemented subspace of l, and hence L 2 +L is isomorphic to a subspace of L 2 L and vice versa. To see this, for instance, for L 2 +L, it is sufficient to take arbitrary dense sequence of the unit ball of the space L 1 L 2, say, {ϕ n }, and to set ( Tx := x(tϕ n (tdt for all x L 2 +L. It is easy to see that this mapping defines an isometrical embedding of L 2 +L into l. 13

14 References [1] F. Albiac, N. J. Kalton, Topics in Banach Space Theory, Springer-Verlag, New York 26. [2] S. V. Astashkin, K. Leśnik and L. Maligranda, Isomorphic structure of Cesàro and Tandori spaces, to appear. Preprint of 33 pages submitted on 1 December 215 at arxiv: [3] C. Bennett and R. Sharpley, Interpolation of Operators, Academic Press, Boston [4] J. Bergh and J. Löfström, Interpolation Spaces. An Introduction, Springer-Verlag, Berlin-New Yor976. [5] S. J. Dilworth, Intersection of Lebesgue spaces L 1 and L 2, Proc. Amer. Math. Soc. 13 (1988, no. 4, [6] S. J. Dilworth, A scale of linear spaces related to the L p scale, Illinois J. Math. 34 (199, no. 1, [7] A. Grothendieck, Sur certains sous-espaces vectoriels de L p, Canadian J. Math. 6, (1954, [8] J. Hagler and Ch. Stegall, Banach spaces whose duals contain complemented subspaces isomorphic to C([,1], J. Funct. Anal. 13 (1973, [9] W. B. Johnson, B. Maurey, G. Schechtman and L. Tzafriri, Symmetric structures in Banach spaces, Mem. Amer. Math. Soc. 19 (1979, no. 217, v+298 pp. [1] N. J. Kalton, Lattice structures on Banach spaces, Mem. Amer. Math. Soc. 13 (1993, no. 493, vi+92 pp. [11] S. G. Krein, Yu. I. Petunin and E. M. Semenov, Interpolation of Linear Operators, Amer. Math. Soc., Providence [12] J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces, II. Function Spaces, Springer- Verlag, Berlin-New Yor979. [13] L. Maligranda, The K-functional for symmetric spaces, Lecture Notes in Math. 17 (1984, [14] L. Maligranda, The K-functional for p-convexifications, Positivity 17 (213, no. 3, [15] Y. Raynaud, Complemented Hilbertian subspaces in rearrangement invariant function spaces, Illinois J. Math. 39 (1995, no. 2, [16] W. Rudin, Functional Analysis, 2nd Edition, MacGraw-Hill, New Yor991. [17] S. J. Szarek, On the best constants in the Khintchine inequality, Studia Math. 58 (1976, Department of Mathematics Samara National Research University, Moskovskoye shosse , Samara, Russia address: astash56@mail.ru Department of Mathematics, Luleå University of Technology SE Luleå, Sweden address: lech@sm.luth.se 14