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Mathematical Sciences

, 7:13

# Some expressions for the Riesz angle of weighted Orlicz sequence spaces

• Boyan Zlatanov
Open Access
Original research

## Abstract

We obtain an expression for computation of the Riesz angle in weighted Orlicz sequence spaces. We use this expression to find some estimates of the Riesz angle for large classes of weighted Orlicz sequence spaces.

## Keywords

Banach Space Sequence Space Banach Lattice Orlicz Space Weight Sequence
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

## Introduction

In order to generalize the technique in [1] for c0 to a larger class of Banach lattices, Borwein and Sims introduced in [2] the notion of a weakly orthogonal Banach lattice and Riesz angle a(X). A Banach lattice is weakly orthogonal if limn∥ |x n |∧|x| ∥ = 0 for all x ∈ X, whenever is a weakly null sequence, where |x| ∧ |y| = min(|x|, |y|). The Riesz angle a(X) of a Banach lattice (X, ∥ · ∥) is

where |x| ∨ |y| = max(|x|, |y|). Clearly 1 ≤ a(X) ≤ 2. If there exists a weakly orthogonal Banach lattice Y such that d(X, Y).a(Y) < 2, where d(X, Y) is the Banach-Mazur distance between the Banach spaces X and Y and a(Y) is the Riesz angle of Y, then X has the weak fixed point property [2]. The coefficient R(X) for a Banach space X is defined, and a connection between a(X) and R(X) is found in [3]. If there exists a Banach space Y with weak Opial condition, such that d(X,Y).R(Y) < 2, then X has the fixed point property [3]. An N-dimensional Riesz angle for a Banach lattice is introduced and studied in [4]. A fixed point theorem is proved involving the N-dimensional Riesz angle of the space [4].

The Riesz angle is an important geometric coefficient in Banach lattices [5, 6, 7]; therefore, the finding of formulas for its calculation or estimation is an interesting problem. Estimations of the Riesz angles in Orlicz function spaces are found in [8]. Some estimations of the Riesz angles in Orlicz sequence spaces equipped with Luxemburg norm and Orlicz norm were found in [9]. Later, a formula for computing the Riesz angle in Orlicz spaces is obtained in [10]. We refine the technique in [10] to obtain a formula for the Riesz angle in a wide class of weighted Orlicz sequence spaces. We apply this formula to find the Riesz angle in some classes of weighted Orlicz sequence spaces. An open problem is to find formula for the computation of the N-dimensional Riesz angle, defined recently in [4], in Orlicz spaces.

## Methods

We use the standard Banach space terminology from [11]. Let X be a real Banach space, S X be the unit sphere of X. Let 0 stand for the space of all real sequences, i.e. , is the set of natural numbers and is the set of real numbers.

### Definition 1

A Banach space (X, ∥ · ∥) is said to be a Köthe sequence space if X is a subspace of 0 such that
1. (i)

If x ∈  0, y ∈ X and |x i | ≤ |y i |for all then x ∈ X andx∥ ≤ ∥y

2. (ii)

There exists an element x ∈ X such that x i  > 0 for all .

We recall that M is an Orlicz function if M is even, convex, M(0) = 0, M(t) > 0 for t > 0. The Orlicz function M(t) is said to have the property Δ2 if there exists a constant c such that M(2t) ≤ c M(t) for every . A weight sequence is a sequence of positive reals. Following [12], we say that is from the class Λ if there exists a subsequence such that and . A weighted Orlicz sequence space M (w) generated by an Orlicz function M and a weight sequence w is the set of all sequences x ∈ 0, such that the inequality holds for some λ < .

It is well known that the space M (w) is a Banach space if endowed with Luxemburg’s norm ∥x M  =  or with Amemiya’s norm k>0}.

We will write M (w, ∥ · ∥ M ) and M (w, ||| · ||| M ) for the weighted Orlicz sequence spaces equipped with Luxemburg’s and Amemiya’s norms, respectively. Luxemburg’s and Amemiya’s norms are connected by the inequalities:

We will write M (w), when the statement holds for the weighted Orlicz sequence space equipped with both norms - Luxemburg and Amemiya. The space M (w), endowed with Luxemburg’s or Amemiya’s norm is a Köthe sequence space. In [13], Ruiz proved that the weighted Orlicz sequence spaces M (w) are all mutually isomorphic for the weight sequences . Sharp estimates are found in [14] for the cotype of M (w), which depends only on the generating Orlicz function, when the weight sequence verifies the condition . It is proved in [15] that M (w), endowed with Luxemburg’s or Amemiya’s norm, has weak uniform normal structure iff M ∈ Δ2 at zero, when weight sequence verifies the condition . Weighted Orlicz sequence spaces were investigated for example in [16, 17, 18]. Let us mention that if the weight sequence is from the class Λ, then a lot of the properties of the space M (w) depend only on the generating Orlicz function M[14, 15], which is in contrast with the results when w ∉ Λ[12, 13, 15]. All these inspired us to find the Riesz angle in a wide class of weighted Orlicz sequence spaces.

## Results and discussion

### Theorem 1

Let M be an Orlicz function with the Δ2-condition and . Then, the Riesz angle of X = ( M (w), ∥ · ∥) can be expressed as follows:

### Theorem 2

Let M be an Orlicz function with the Δ2-condition and . Then, the Riesz angle of X = ( M (w), ||| · |||) can be expressed as follows:
where

### Lemma 1

Let and be an arbitrary subsequence of w. Then, there exist sequences of naturals , , , such that
and for every , there holds the equality

### Proof

By v ∈ Λ, it follows that there is a subsequence , such that and . For the simplicity of the notations, let the subsequence denoted by . We will prove the Lemma by induction on n:

(i) Let n = 1. We can choose so that
and
Let us use the notation
(ii) Let n = 2. We will show that we can choose , i + s = 3, so that
and
Indeed, let us choose first : , such that
Then, we choose : , such that
Suppose that for n = p, we have chosen , , i + s = p + 1 with the properties:
for i + s = p + 1 and
(iv) Let n = p + 1. We will show that we can choose , , i + s = p + 2, so that
for i + s = p + 2 and
(1)

for i ≤ p + 1.

Indeed, let us choose first : , such that
Then, we choose : , such that
If for i0 ≤ p - 1 we have chosen to satisfy the inequalities , such that
then for i0 + 1 ≤ p, we choose : , such that
On the last step for i = p + 1, we choose : such that

By (1), it follows that holds for every . □

### Theorem 3

[19] Let the iterated series be given. If the series is convergent, then for any permutations the series is convergent and .

### Lemma 2

Let M be an Orlicz function with the Δ2-condition and . Let M (w) be equipped with Luxemburg’s or Amemiya’s norm. Then,(1) For every x ∈  M (w), such that , for every λ > 0, there are y, z ∈  M (w), such that |y| ∧ |z| = 0, , for any λ > 0(2) For every , there are y, z, such that |y| ∧ |z| = 0, (3) For every , there are y, z, such that |y| ∧ |z| = 0, .

### Proof

(1) Let be arbitrarily chosen. By M ∈ Δ 2, it follows that for every λ > 0. By w ∈ Λ, it follows that we can choose two subsequences , of w, such that v, u ∈ Λ, v ∩ u = , v ∪ u = w.

By Lemma 1, there are sequences of naturals , , , , , such that
and there hold the equalities

for every .

Put

and , . We will show that , for any λ > 0.

Let λ > 0 and put for . Let us consider the infinite matrix
For every , the equality holds and thus for every λ > 0. By for every and Theorem 3, it follows that for any two permutations , the series is convergent and there hold the equalities
Consequently, there exist two permutations , such that we can write the chain of equalities
(2)
Similarly, if we put for , we get the chain of equalities
(3)

Para>(2) If λ = 1, then by we get that if then .

(3) If , then by (1) it follows that for any λ > 0. Therefore, if |||x||| = 1, then |||y||| = |||z||| = |||x||| = 1.

### Lemma 3

Let M be an Orlicz function with the Δ2-condition and . Then, for every and k > 1, there exists a unique dx,k > 1, such that .

### Proof

Let and k > 1 be arbitrarily chosen and fixed, then we define the function by . By the inequality and the convergence of the series , it follows that is uniformly convergent on [1, + ). Thus, S is a continuous function. It is easy to see that S is a strictly decreasing function on [1, + ). Therefore, by the inequalities
and

we get that there is a unique dx,k > 0, such that . □

### Lemma 4

([10]) For a Köthe sequences space (X, ∥ · ∥), the Riesz angle a(X) can be expressed as

where |x| ∧ |y| = min{|x|, |y|}.

### Proof of Theorem 1 (1)

Let be arbitrarily chosen. By , it follows that we can choose two subsequences , of w, such that v, u ∈ Λ, v ∩ u = , v ∪ u = w.

It follows from Lemma 1 that there exist sequences of naturals , , , , , such that
and there holds the equalities

for every .

We can put

and , .

Using Lemma 2, we get that and for any .

By the choice of the subsequences v, u ⊂ w, we have that |y| ∧ |z| = 0 and ∥(|y| ∨ |z|)∥ = ∥y + z∥. Therefore, we can write the chain of equalities:
Consequently, it follows that for every , there exists k x  = ∥(|y| ∨ |z|)∥, such that . By Lemma 4, we get the inequality

where X = ( M (w), ∥ · ∥).

On the other hand, let us put

It follows from Lemma 4 that for every ε > 0, there are , |x| ∧ |y| = 0, such that ∥(|x|∨|y|)∥ > a( M (w)) - ε.

Since

we get the inequality ∥(|x| ∨ |y|)∥ ≤ d, which implies a( M (w)) ≤ d + ε. By the arbitrariness of ε > 0, we obtain that d ≥ a( M (w)).

### Proof of Theorem 2

Let us denote
(4)

For any ε > 0, there exist and k > 1, such that dx,k ≥ d - ε.

By , it follows that we can choose two subsequences , of w, such that v, u ∈ Λ, v ∩ u = , v ∪ u = w.

It follows from Lemma 1 that there exist sequences of naturals , , , , , such that
and there hold the equalities

for every .

We can put

and , .

Using Lemma 2, we see that , and for any . Let us put
and
Therefore, by the chain of inequalities
and

we get that |||y + z||| ≥ d - ε. By the arbitrariness of ε > 0 and Lemma 4, we obtain the inequality a(( M (w), ||| · |||)) ≥ d.

On the other hand, for any ε > 0, there are , |x| ∧ |y| = 0, such that there holds the inequality |||(|x| ∨ |y|)||| > a(( M (w), ||| · |||)) - ε. It follows from (4) and Lemma 3 that for every ε > 0, there are k, h > 1, such that dx,k < d + ε, dy,h < d + ε, where dx,k and dy,h are the solutions of the equations and , respectively. WLOG, we may assume that 1 < h ≤ k. By the chain of inequalities

we obtain the inequality and hence . Since ε > 0 is arbitrarily chosen, it follows that .

For the estimation of the Riesz angle in weighted Orlicz sequence spaces, we will need some well-known indices. For an Orlicz function M, we consider the index function , u ∈ (0, + ) [20]. Following [10, 21], we define the indices:
(5)

Let us mention that for an Orlicz sequence space M , only the behavior of the Orlicz function M at zero is important, and therefore, the above indices are defined only at zero in [10, 21].

### Theorem 4

Let M be an Orlicz function with the Δ2-condition and . Then,

### Proof

We will prove first that .

We chose arbitrary and put . It is easy to check the equality . For any u i  ∈ (0, + ), the inequality holds. Then,
(6)
For every , there exists k x , such that , and thus by (6), it follows the inequality . Therefore,

where X = ( M (w), ∥ · ∥).

Now, we will prove that .

For any u ∈ (0, + ), there are sequences of naturals , , such that p n  ≤ q n  < pn+1, for and .

Put . It is easy to see that and thus . By the equality
we get that for any u ∈ (0, + ), there holds the inequality
and therefore,
Thus, we have proven that
(7)

The proof of the inequality follows directly by equality (7) and the inequality . □

### Theorem 5

Let M be an Orlicz function with the Δ2-condition and . Then,

### Proof

(i) We will prove first that
Let us choose arbitrary . By M ∈ Δ2 and the equality

it follows that there exists k0 > 0, such that the equality holds [20].

We put . Then, similarly to (6), we can write the inequality:
and consequently we get that , where dx,k is the solution of the equation , k > 1. Thus, for any k > 1, there holds the inequality
(8)
By the arbitrary choice of and (8), we obtain that
From the inequalities ∥ · ∥ ≤ ||| · ||| ≤ 2 ∥ · ∥, it follows that for any with |x| ∧ |y| = 0, there holds the inequality
(9)

where X = ( M (w), ||| · |||).

Therefore, we get that
(ii) By (9), we obtain the inequality

### Definition 2

We say that the Orlicz function M satisfies the ∇2 condition if there exists l > 1, such that , for every x ∈ [0, + ), and we denote this by M ∈ ∇2.

The function M-1 is a concave function and thus . According to [20] (p. 22), M ∈ ∇2 iff and , i.e.
(10)

### Corollary 1

Let M be an Orlicz function with M ∈ Δ2 and w ∈ Λ is a weight sequence. Then,
1. (a)

M ∉ ∇2 iff a(( M (w), ∥ · ∥)) = 2

2. (b)

M ∈ ∇2 iff a(( M (w), ∥ · ∥)) < 2.

### Proof

(a) Let M ∉ ∇2. Then from (10), it follows that or , and thus by Theorem 4, it follows that a(( M (w), ∥ · ∥)) ≥ 2. Therefore by the inequalities 1 ≤ a(( M (w), ∥ · ∥)) ≤ 2, it follows that a( M (w), ∥ · ∥) = 2.

Let a(( M (w), ∥ · ∥)) = 2. There are three cases: , or there exists t0 ∈ (0, + ), such that .

Let or holds; then by (10), it follows that M ∉ ∇2.

There exists t0 ∈ (0, + ), such that . Then, we can write the equality , and consequently by the concavity of the function M-1, it follows that the points (0, 0), (t0, M-1(t0)) and (2t0, M-1(2t0)) lie on a line. Thus, the function M-1 is linear on the segment [0, 2t0] and therefore . Therefore by (10), it follows that M ∉ ∇2.

(b) The proof follows directly from (a). Indeed, let M∈∇2 holds, but a(( M (w), ∥ · ∥)) < 2 does not hold. Then a(( M (w), ∥ · ∥)) = 2 and by (a), it follows that M ∉ ∇2 which is a contradiction.

Let a(( M (w), ∥ · ∥)) < 2 holds, but M ∈ ∇2 does not hold. Then M ∉ ∇2 and by (a), it follows that a(( M (w), ∥ · ∥)) = 2 which is a contradiction.

For the next Corollary, we will need the indices form [22].

We put , t ∈ (0, + ), where p is the right derivative of M. Let us define
The above indices are connected by the formulas ([11], p.149; [20], p.27).
and

where N is the complementary function to M.

The inequalities
(11)
(12)

hold [23]. Let us mention that inequalities (11) are proven in [23]. The proof of inequalities (12) is similar. We are sure that inequalities (12) are proven somewhere. Just for completeness, we will prove (12) using the technique from [23].

If , then clearly . Assume that . For any ε > 0, there exists t0 > 0 such that for every t ∈ [t0, + ). Then for any t0 ≤ t1 < t2 < + , we have
We put t1 = M-1(u) and t2 = M-1(2u). Thus for any u ∈ [M(t0), + ), there holds the inequality

By the arbitrary choice of ε > 0, it follows the proof of the right side of the inequality (12). The proof of the left side is similar.

If limt→0+F M (t) exists, we denote it by and if limt→+F M (t) exists we denote it by . We put .

### Corollary 2

Let M be an Orlicz function with the Δ2-condition and w ∈ Λ is a weight sequence. Then,
1. (a)

If F M is an increasing function on (0, + ), then

2. (b)

If F M is a decreasing function on (0, + ), then

3. (c)

If there is t 0 ∈ (0, + ), such that F M is increasing on (0, M -1(t 0)) and decreasing on (M -1(t 0), + ), then

### Proof

1. (a)
If F M is an increasing function in (0, +), exists and is increasing in (0, +) [21]. Then from (11), we get

and therefore .
1. (b)
If F M is a decreasing function in (0, +), then exists, and is decreasing in (0, +) [21]. Then by (11), we obtain

and therefore .
1. (c)

If F M is an increasing function in (0, M -1(t 0)), then exists, and is increasing in (0, t 0 / 2). If F M is a decreasing function in (M -1(t 0), +), then exists, and is decreasing in (t 0 / 2, +) [21].

From (11), it follows that , and hence . By Theorem 4, we get that .

## Conclusions

### Example 1

Let M1(t) = 2|t| p  + |t|2p, p ∈ [1, +). Then, for t ∈ [0, +). The function is an increasing function and . By Corollary 2, we get that .

### Example 2

Let , p ∈ [1, +). Then, for t ∈ [0, +). The function is an increasing function and . By Corollary 2, we get that .

### Example 3

Let M3(t) = |t| p logr(1 + |t|), p ∈ [1, +), r ∈ (0, +). Then,

The function is a decreasing function and . By Corollary 2, we get that .

### Example 4

Let q ≥ 2 and p ∈ [q - 1, 2q - 1]. We define the function
The function M4 is an Orlicz function. Then,

where and . The index function is increasing on [0,1] and is decreasing on [0, +), and . By Corollary 2, we get that .

### Example 5

Let M5(t) = (1 + |t|) log(1 + |t|) - |t|. Then is a decreasing function on (0, +) and . Thus, . By Corollary 1 it follows that M6 ∉ ∇2.

## Notes

### Acknowledgements

This research is partially supported by Plovdiv University ‘Paisii Hilendarski’, NPD, Project NI11-FMI-004.

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## Copyright information

© Zlatanov; licensee Springer. 2013

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## Authors and Affiliations

1. 1.Faculty of Mathematics and InformaticsUniversity of PlovdivPlovdivBulgaria