Kadec–Klee properties of Orlicz–Lorentz sequence spaces equipped with the Orlicz norm

The necessary and sufficient conditions for both the Kadec–Klee property as well as the Kadec–Klee property with respect to the coordinatewise convergence in Orlicz–Lorentz sequence spaces equipped with the Orlicz norm and generated by arbitrary Orlicz functions as well as any non-increasing weight sequences are given. Moreover, for their subspaces of elements with an order continuous norm the full characterization of the Kadec–Klee property with respect to the coordinatewise convergence is presented. Some tools useful in the proofs of the main results are also provided.


Introduction
Throughout this paper R, R + and N denote the sets of reals, nonnegative reals and natural numbers, respectively. By S(X ) and B(X ) denote the unit sphere and the unit ball of a Banach space X = (X , · ), respectively.
The triple (N, 2 N , m) stands for the counting measure space while 0 = 0 (m) denotes the space of all real sequences x : N → R. For every x = (x(i)) ∞ i=1 ∈ 0 , let supp(x) := {i ∈ N : x(i) = 0} and |x|(i) := |x(i)| for all i ∈ N, that is, |x| denotes the absolute value of x.
For any Orlicz function we define its complementary (in the sense of Young) function by the formula Let ω : N → R + be a non-increasing sequence, called a weight sequence. In the whole paper we will assume that ω (1) > 0.
In most of the previous papers one studied Orlicz-Lorentz spaces equipped with the Luxemburg norm It is easy to show that if ∞ i=1 ω (i) < ∞ or a ϕ > 0, then λ ,ω = l ∞ with equivalent norms. Otherwise, we have λ ,ω → c 0 .
In order to answer this question, let us define for any x ∈ λ ,ω \{0} the following constants where p is the right-hand side derivative of . Define the function f on (0, ∞), The function f is continuous on the interval (0, λ ∞ ) and left-continuous at λ ∞ provided that λ ∞ < ∞. We also have 0 < k * ≤ k * * ≤ λ ∞ ≤ ∞. Moreover, as it was shown in [8], the function f is strictly decreasing on the interval (0, k * ), strictly increasing on the interval (k * * , λ ∞ ) if k * * < λ ∞ as well as constant and equal to x O ,ω on the interval [k * , k * * ] if k * < k * * . Therefore, assuming for any x ∈ λ ,ω \{0} that [8]). In the further part of this paper we will need to use the following Remark 1 Note that if b < ∞, then for any x ∈ λ ,ω \{0}, we have that k * * ≤ λ ∞ ≤ b x * (1) . In the case of b = ∞ and lim u→∞ (u) u = lim u→∞ 1 u u 0 p(s)ds = lim u→∞ p(u) = ∞, we have also that k * * < ∞ for any x ∈ λ ,ω \{0}. Indeed, if λ ∞ = ∞, then there exists k x > 0 such that ( p(k x x * (1)))ω(1) > 1 and k * * ≤ k x . Now, assume that lim u→∞ ω(i) > 1, then k * * < ∞. In order to do it, let n x ∈ N be such that In the case when m(supp x) < ∞, we take n x = m(supp x) and in the case when m(supp x) = ∞, we can find n x ∈ N, for which inequality (4) holds. Since ω(i) = 1 and p(u) < B for any u > 0, we get that k * = ∞, which means K (x) = ∅.
Let (E, ≤, · E ) be a Banach sequence lattice. An element x ∈ E is said to be order continuous if for any sequence (x n ) in E + (the positive cone of E) such that x n ≤ |x| for any n ∈ N and x n → 0 coordinatewisely, there holds x n E → 0 as n → ∞. The subspace E a of all order continuous elements in E is an order ideal in E. Space E is called order continuous if E a = E.
Since the Luxemburg and the Orlicz norms are equivalent, the subspace of order continuous elements for both these norms is the same (as a subset of the elements of the space λ ,ω ). We will denote it as (λ ,ω ) a . The following theorem holds true: This theorem is the consequence of the fact that the space (λ ,ω , · ,ω ) (with the Luxemburg norm) is order continuous if and only if ∞ i=1 ω(i) = ∞ and ∈ δ 2 (cf. [18,Theorem 2.4]) as well as the equivalence of the Luxemburg and the Orlicz norms.
A Banach space X = (X , · ) is said to have the Kadec-Klee property or property H (resp. the Kadec-Klee property with respect to the coordinatewise convergence or property H c ) if for any sequence (x n ) ∞ n=1 in X and x ∈ X such that lim n→∞ x n = x , we have x n − x → 0 provided x n → x weakly as n → ∞ (resp. provided x n (i) → x(i) coordinatewisely). Note that this property was originally considered by Radon [29] and next by Riesz [30,31], where it was proved that L p spaces (1 < p < ∞) have property H , while L 1 [0, 1] has not. The newer results concerning the Kadec-Klee properties a reader can find for example in [3,5,6,9,10,17,32].

Auxiliary results
Remembering Remark 1, we can formulate the following Note that (B) < ∞ and (u) = ∞ for any u > B. Hence, for any y ∈ λ ,ω such that I ,ω (y) ≤ 1, we get that y(i) ≤ B for any i ∈ N. Therefore, for the same y, we obtain Hence it follows that .
Let us note the following observation. It is known that if a sequence ( is uniformly convergent to x * . However, the above implication is not true in general, if the assumption about uniform convergence of (x n ) ∞ n=1 to x is replaced by the assumption that (x n ) ∞ n=1 is convergent to x coordinatewisely. This fact can be easily visualized by the following example. Let us take the sequence (x n ) ∞ n=1 such that x n (n) = 1 and x n (i) = 0 for i = n, where n ∈ N. Then (x n ) ∞ n=1 converges to zero coordinatewisely but x * n (1) does not converge to zero. However, Lemma 2 holds. This Lemma is a kind of generalization of Lemma 3.1 from [10]. We will present its proof below just for the completeness of this presentation as well as for the convenience of a reader.

Lemma 2
Let vanishes only at zero and ω(i) > 0 for any i ∈ N. Then, for any x ∈ c 0 and any sequence (x n ) ∞ n=1 from c 0 such that x n → x coordinatewisely and Assume that x and a sequence (x n ) ∞ n=1 satisfy the assumptions of the lemma and that, passing to a subsequence if necessary, there exists i 0 ∈ N such that |x * n (i 0 ) − x * (i 0 )| > ε for some ε > 0 and all n ∈ N. We will consider now three cases (passing to a subsequence if necessary).
for the same n. Note that the set A = {i ∈ N : |x(i)| > ε 2 } is finite because x ∈ c 0 . Let us denote m 0 = m(A), y = xχ A and y n = x n χ A for each n ∈ N. Since y n → y uniformly, we get that y * n tends to y * coordinatewisely (so uniformly as well). (5), there holds that y * n (i 0 ) ≥ x * n (i 0 ) + ε 2 for n ≥ n 0 . On the other hand, y n ≤ x n for any n ∈ N, whence y * n ≤ x * n for any n ∈ N, which arrive us to a contradiction.
for any n ≥ n 0 . Without loss of generality, we can also assume that for n ≥ n 0 . Since z n ≤ x n , we get that z * n ≤ x * n and, consequently, applying (6) and (8), we get for n ≥ n 0 , where the last inequality comes from (7), a contradiction.
, similarly as before, we will get a contradiction.

Remark 2
In Lemma 2 (also in Lemma 3) the assumptions that vanishes only at zero and ω(i) > 0 for any i ∈ N can not be omitted. Indeed, assume firstly that a > 0 and define x = a e 1 and x n = a e 1 + a e n for n ≥ 2. Then I ,ω (x) = I ,ω (x n ) = 0 for n ≥ 2, x n → x coordinatewisely and x * n (2) = a > 0 = x * (2) for n ≥ 2.
Now, assume that a = 0 and there exists j ∈ N such that w( j) > w( j + 1) = 0. Let u > 0 be such that (u) < ∞ and let us define Next lemma can be proved analogously as Lemma 3.2. from [10].

Lemma 3
Let vanishes only at zero and ω(i) > 0 for any i ∈ N. Then (x n −x) * → 0 uniformly as n → ∞ for any x ∈ c 0 and any sequence (x n ) ∞ n=1 from c 0 such that x n → x coordinatewisely and I ,ω (x n ) → I ,ω (x) = α < ∞ as n → ∞.

Remark 3
Note that in Lemma 3 assumptions x ∈ c 0 and x n ∈ c 0 for any N are essential. Indeed, for any fixed u > 0 let us define x(i) = u for i ∈ N, x n (n) = 0 and x n (i) = u for n ∈ N and i = n. Obviously, x ∈ ∞ \c 0 , x n ∈ ∞ \c 0 for n ∈ N and x n → x coordinatewisely. We also have that The proof of the lemma below can be found in [8].

Lemma 4
Let (x n ) be a sequence of elements of the space λ ,ω . Then the following properties hold.

Main results
Before we present the main results of our paper, we will recall what so far has been known about the property of the Kadec-Klee type in Orlicz, Lorentz and Orlicz-Lorentz spaces. In the case of Orlicz spaces the suitable criteria for both the Orlicz as well as the Luxemburg norms was given in [4,9,26]. Note that in the case of nonatomic measure in place of the Kadec-Klee property with respect to the coordinatewise convergence we consider the Kadec-Klee property for the local convergence in measure. Analogous results for Lorentz spaces can be found in [5]. Whereas, in the case of Orlicz-Lorentz spaces the research connected to the properties of the Kadec-Klee type were led only for the Luxemburg norm (see [3,16,17] and also [10]). In particular, we get that the Orlicz-Lorentz sequence space λ ,ω equipped with the Luxemburg norm · ,ω has the Kadec-Klee property with respect to the coordinatewise convergence if and only if ∈ δ 2

Theorem 2
The following conditions are equivalent: ,ω ) has the Kadec-Klee property with respect to the coordinatewise convergence.
,ω ) has the Kadec-Klee property. Proof It is obvious that (2) implies (3). Since the Kadec-Klee property implies an order continuity of the space (λ ,ω , . O ,ω ) (see [9, Proposition 2.1]) and the last means that ∈ δ 2 and ∞ i=1 ω(i) = ∞ (see Theorem 1), we obtain that (3) implies (1). Now, we will show (1) implies (2). Case 1. At the beginning let us assume that b < ∞ or b = ∞ and lim u→∞ ,ω ) for n ∈ N and x n → x coordinatewisely. By the assumptions of (see Remark 1), for any n ∈ N we can find k n > 1 such that 1 = x n O ,ω = 1 k n 1 + I ,ω (k n x n ) . First, we will show the boundedness of the sequence (k n ) ∞ n=1 . Define . By virtue of the rearrangement definition, there exist natural numbers i 1 , i 2 , . . . , i m 1 such that for j = 1, 2, . . . , m 1 and n ≥ n 0 . Hence x * n (i) ≥ b for i = 1, 2, . . . , m 1 and n ≥ n 0 . Therefore, if b < ∞, then we have k n ≤ b /b for any n ≥ n 0 . Let now b = ∞ and lim u→∞ (u) u = ∞. If lim n→∞ k n = ∞ (passing to a subsequence if necessary), then by the assumption that lim u→∞ beginning from some natural n 1 ≥ n 0 , a contradiction. So (k n ) ∞ n=1 is bounded. Now, we will show that any cluster point k 0 of the sequence (k n ) ∞ n=1 belongs to K (x), that is, From the properties of the Orlicz norm In order to get the opposite inequality, let us define the sequence (m l ) in the following way: m 1 is given by formula (9) and In the case of m(supp x) = p < ∞, we can find l x ≤ p such that m l x = p. For any l the set N l = {i ∈ N : |x(i)| ≥ x * (m l )} is finite, more precisely, m(N l ) = m l . Let lim s→∞ k n(s) = k 0 . Since k n(s) x n(s) → k 0 x coordinatewisely, then k n(s) x n(s) χ N l → k 0 xχ N l uniformly for any fixed l, whence k n(s) x * n(s) χ N l → k 0 x * χ N l coordinatewisely for any fixed l and, consequently, as s → ∞. Hence, for any fixed l, there holds From the arbitrariness of l ∈ N, we get This inequality together with inequality (11) give us equality (10). Let now (x n(s) ) be an arbitrary subsequence of the sequence (x n ). Without loss of generality, passing to a subsequence if necessary, we can assume that sequence (k n(s) ) is convergent to some k 0 . Thus k n(s) x n(s) → k 0 x coordinatewisely and k 0 ∈ K (x), whence and, consequently, lim s→∞ I ,ω (k n(s) x n(s) ) = I ,ω (k 0 x).

, k n(s(t)) x n(s(t)) − k 0 x O
,ω → 0. Since k n(s(t)) → k 0 , we obtain Hence, by (12) and (13), we can find t 0 ∈ N such that for any t ≥ t 0 , we have (k n(s(t)) x n(s(t)) − k 0 x) * (1) ≤ u 0 and k n( by the formula we get again that z ∈ 1 and • (k n(s(t)) x n(s(t)) − k 0 x) * · ω ≤ z for t ≥ t 0 . Proceeding analogously as above we obtain x n(s(t)) − x O ,ω → 0. Finally, applying the double extract convergence theorem, we get ,ω ) for n ∈ N and x n → x coordinatewisely. First, assume additionally that (B) for i ∈ N x and n ≥ n 0 . Hence, I ,ω ( p(k 1 x * n )) > 1 for k 1 = 2u x x * ( j x )+x * ( j x +1) and n ≥ n 0 and, consequently, k * * n := k * * (x n ) ≤ k 1 for the same n. Next, proceeding analogously as in Case 1, we obtain that x n → x in norm.
We proceed analogously in the case when (B) ω(i) = 1 and there exists u 0 > 0 such that p(u) = B for all u ≥ u 0 . Indeed, note that in this case (B) > 0 and m(supp x) < ∞ by virtue of the assumption that ∞ i=1 ω(i) = ∞. Taking this time j x = m(supp x) and using again the fact that x n → x coordinatewisely, we can find n 0 ∈ N such that |x n (i)| ≥ 1 2 x * ( j x ) for any i ∈ supp x and n ≥ n 0 . Hence, all n ≥ n 0 , which means that for any n ≥ n 0 there is k n ≤ k 1 such that x n O ,ω = 1 k n (1 + I ,ω (k n x n )). Next, proceeding again in the same way as in Case 1, we can finish this part of the proof. Finally, assume that (B) ω(i) = 1 and p(u) < B for any u > 0, which means that K (x) = ∅, or equivalently, for any k > 0 (see Remark 1). Let (x n(s) ) be an arbitrary subsequence of the sequence (x n ). We will consider two subcases. Subcase 1. Assume, passing to a subsequence if necessary, that there exists s 0 ∈ N such that (B) m(supp x n(s) ) i=1 ω(i) ≤ 1 for any s ≥ s 0 . By virtue of Lemma 1, we obtain for all s ≥ s 0 . Hence, by Lemma 2 (for 1 (u) = |u|, we have lim s→∞ I 1 ,ω (x n(s) ) = Applying again the fact that 1 has the Kadec-Klee property with respect to coordinatewise convergence, we obtain Hence, by virtue of Lemma 4, p. 97 from [20], there exist y ∈ 1 + and a subsequence x * n(s(t)) of the sequence x * n(s) such that Hence, similarly to Case 1, we get

Consequently, defining
Since, by virtue of Lemma 3, (x n(s(t)) − x) * → 0 uniformly as t → ∞, applying again the Lebesgue dominated convergence theorem, we get that lim t→∞ Hence, it is enough to show that, starting with some t 0 , there holds whence we finally get Obviously, formula (16) is satisfied if (B) = 0. Now, assume that (B) > 0. Since ∞ i=1 ω(i) = ∞, we get that m(supp x) < ∞. Hence, and by the assumption that x n is coordinatewise convergent to x, we obtain that supp x ⊂ supp x n(s(t)) starting with some t 0 and, without loss of generality, we can assume that s(t 0 ) ≥ s 0 . Consequently, ω(i) ≤ 1 for all t ≥ t 0 , whence from Lemma 1 formula (16) arrives for the same t. Subcase 2. Assume, starting with some s 0 ∈ N, that (B) Since x n(s) ∈ c 0 , then for any s ∈ N there exists i s such that b n(s) = |x n(s) (i s )|. We will show indirectly that lim s→∞ b n(s) = 0. In order to do it suppose, passing to a subsequence if necessary, that there is δ > 0 such that b n(s) ≥ δ starting with some s 1 ≥ s 0 . Without loss of generality we can assume that δ ≤ 1 2 x * (i x ). Moreover, since x n(s) → x coordinatewisely, we can assume that |x n(s) (i)| ≥ 1 2 x * (i x ) for any i ∈ supp x and s ≥ s 1 . If , then by virtue of Lemma 1, for any s ≥ s 1 there holds Since lim s→∞ (x n(s) χ supp x ) * (i) = x * (i) for any i ∈ supp x, we have whence we conclude that there is s 2 ≥ s 1 such that for any s ≥ s 2 . Hence, for any s ≥ s 2 , a contradiction to (17).
i=1 ω(i) > 1. Defining k 1 = u x δ , we have that I ( p(k 1 (x n(s) χ supp x + δe i s ) * )) > 1 for all s ≥ s 1 . Therefore, for each s ≥ s 1 there is k s ∈ [1, k 1 ] such that where the inequality (δ) ≤ (k s δ) k s is a consequence of the fact that function (u) u is non-decreasing.
On the other hand, by (15), Hence, by the fact that function f x n(s) χ supp x (see formula (2)) is strictly decreasing, there is s 2 ≥ s 1 such that for all s ≥ s 2 , we have a contradiction to (18). Therefore, lim s→∞ b n(s) = 0 holds. Let s 1 ≥ s 0 be such that b n(s) < 1 2 x * (i x ) for s ≥ s 1 and |x n(s) (i)| > 1 2 x * (i x ) for any i ∈ supp x and s ≥ s 1 . Hence, Indeed, in the opposite case, there exists a subsequence (k n(s(t)) ) such that lim t→∞ k n(s(t)) = k < ∞. Then, by inequality (15), we get a contradiction. We also have for any s ≥ s 1 that Defining Consequently, for the same s. Note that for any s ≥ s 1 there holds Hence, by virtue of and equality (19), we get that lim s→∞ x n(s) Summarizing both subcases and applying the double extract convergence theorem, we conclude that lim n→∞ x − x n O ,ω = 0, which finishes the proof.
Finally, let us present a criterion for the subspace of order continuous elements (λ ,ω ) a of the space (λ ,ω , · O ,ω ) to possess the Kadec-Klee property with respect to the coordinatewise convergence.
,ω ) has the Kadec-Klee property with respect to the coordinatewise convergence if and only if satisfies the i=1 ω(i) = ∞ and ∈ δ 2 , then by virtue of Theorem 1, we obtain that (λ ,ω ) a = λ ,ω , whence by Theorem 2, we get that ((λ ,ω ) a , · O ,ω ) has the Kadec-Klee property with respect to the coordinatewise convergence. Now, we will show the necessity of the δ 2 -condition. Assume that / ∈ δ 2 . Let j ∈ N be such that K (x j ) = ∅ where x j = (1, 1, . and, in consequence, K (x j ) = ∅ (see Remark 1). Let h ∈ K (x j ). First, assume that a > 0. For n ≥ j + 1 let us define Note that x j and x j,n for n ≥ j + 1 belong to (λ ,ω ) a . Since 0 ≤ x j ≤ x j,n for ,ω for the same n. On the other hand Assume right now that a = 0. Since / ∈ δ 2 , we find a decreasing to zero sequence (u n ) ∞ n=1 such that u 1 h < 1, (2u n ) > 2 n+1 (u n ) and (u n )ω(1) ≤ 1 2 n+1 . Let j n , where n ∈ N, be such that for n ∈ N. Obviously, x j and x j,n for n ∈ N belong to (λ ,ω ) a and x j,n → x j coordinatewisely. Applying the orthogonal subadditivity of the modular, we get for n ∈ N, whence h(x j,n − x j ) ,ω ≥ 1 2 for the same n, where · ,ω is the Luxemburg norm. Therefore, x j,n − x j O ,ω ≥ x j,n − x j ,ω ≥ 1 2h for n ∈ N and we conclude again that ((λ ,ω ) a , · O ,ω ) does not possess the Kadec-Klee property with respect to the coordinatewise convergence.
Coming to the proof of Subcase 2, note that from condition (B) m(supp(x n(s) )) i=1 ω(i) > 1, we conclude that (B) > 0. Hence and from condition (B) m(supp(x)) i=1 ω(i) ≤ 1, we get that m(supp(x)) < ∞. Further, the proof can be led and finished similarly as this for Theorem 2.
Necessity. First, suppose that there is j ∈ N such that ω( j) > ω( j + 1) = 0. Define e i and x j,n = j i=1 e i + e n for n ≥ j + 1. Then, x j and x j,n for n ≥ j + 1 belong to (λ ,ω ) a , x j,n → x j coordinatewisely and in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.