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Mathematical Sciences

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Some expressions for the Riesz angle of weighted Orlicz sequence spaces

  • Boyan ZlatanovEmail author
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 { x n } n = 1 Open image in new window is a weakly null sequence, where |x| ∧ |y| = min(|x|, |y|). The Riesz angle a(X) of a Banach lattice (X, ∥ · ∥) is
a ( X ) = sup { ( | x | | y | ) : x 1 , y 1 } , Open image in new window

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. x = { x i } i = 1 0 Open image in new window, N Open image in new window is the set of natural numbers and R Open image in new window 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 i N Open image in new window then x ∈ X andx∥ ≤ ∥y

     
  2. (ii)

    There exists an element x ∈ X such that x i  > 0 for all i N Open image in new window.

     

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 t R Open image in new window. A weight sequence w = { w i } i = 1 Open image in new window is a sequence of positive reals. Following [12], we say that w = { w i } i = 1 Open image in new window is from the class Λ if there exists a subsequence w = { w i k } k = 1 Open image in new window such that lim k w i k = 0 Open image in new window and k = 1 w i k = Open image in new window. 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 M ~ ( x / λ ) = i = 1 w i M ( x i / λ ) < Open image in new window 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  =  inf r > 0 : i = 1 w i M ( x i / r ) 1 Open image in new window or with Amemiya’s norm | | | x | | | M = inf 1 k 1 + i = 1 w i M ( k x i ) : Open image in new windowk>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:
· M | | | · | | | M 2 · M . Open image in new window

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 w = { w n } n = 1 Λ Open image in new window. 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 w = { w n } n = 1 Λ Open image in new window. 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 w = { w n } n = 1 Λ Open image in new window. 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 w = { w i } i = 1 Λ Open image in new window. Then, the Riesz angle of X = ( M (w), ∥ · ∥) can be expressed as follows:
a ( X ) = sup k x : M ~ w x k x = 1 2 , x S M ( w ) . Open image in new window

Theorem 2

Let M be an Orlicz function with the Δ2-condition and w = { w i } i = 1 Λ Open image in new window. Then, the Riesz angle of X = ( M (w), ||| · |||) can be expressed as follows:
d ( X ) a ( X ) 3 2 d ( X ) , Open image in new window
where
d ( X ) = sup | | | x | | | = 1 inf k > 1 d x , k : M ~ w kx d x , k = k - 1 2 . Open image in new window

Lemma 1

Let w = { w i } i = 1 Λ Open image in new window and v = { v i } i = 1 Λ Open image in new window be an arbitrary subsequence of w. Then, there exist sequences of naturals { m i ( s ) } i = 1 Open image in new window, { k i ( s ) } i = 1 Open image in new window, s N Open image in new window, such that
1 m 1 ( 1 ) k 1 ( 1 ) Open image in new window
k n - 1 ( 1 ) < m 1 ( n ) , m i ( s ) k i ( s ) , k i ( s ) < m i + 1 ( s - 1 ) Open image in new window
for n , i , s N , n 2 , i + s = n + 1 , Open image in new window
and for every i N Open image in new window, there holds the equality
s = 1 j = m i ( s ) k i ( s ) v j = w i . Open image in new window

Proof

By v ∈ Λ, it follows that there is a subsequence { v i j } j = 1 Open image in new window, such that lim j v i j = 0 Open image in new window and j = 1 v i j = Open image in new window. For the simplicity of the notations, let the subsequence { v i j } j = 1 Open image in new window denoted by { v j } j = 1 Open image in new window. We will prove the Lemma by induction on n:

(i) Let n = 1. We can choose m 1 ( 1 ) , k 1 ( 1 ) Open image in new window so that
m 1 ( 1 ) k 1 ( 1 ) Open image in new window
and
w 1 - w 1 2 j = m 1 ( 1 ) k 1 ( 1 ) v j < w 1 . Open image in new window
Let us use the notation
f ( i , p ) = s = 1 p j = m i s k i s v j . Open image in new window
(ii) Let n = 2. We will show that we can choose m i ( s ) , k i ( s ) Open image in new window, i + s = 3, so that
k 1 ( 1 ) < m 1 ( 2 ) k 1 ( 2 ) < m 2 ( 1 ) k 2 ( 1 ) Open image in new window
and
w i - w i 2 3 - i s = 1 3 - i j = m i ( s ) k i ( s ) v j < w i , for i = 1 , 2 . Open image in new window
Indeed, let us choose first m 1 ( 2 ) , k 1 ( 2 ) N Open image in new window: k 1 ( 1 ) < m 1 ( 2 ) k 1 ( 2 ) Open image in new window, such that
w 1 - w 1 2 2 f ( 1 , 1 ) + j = m 1 ( 2 ) k 1 ( 2 ) v j < w 1 . Open image in new window
Then, we choose m 2 ( 1 ) , k 2 ( 1 ) N Open image in new window: k 1 ( 2 ) < m 2 ( 1 ) k 2 ( 1 ) Open image in new window, such that
w 2 - w 2 2 j = m 2 ( 1 ) k 2 ( 1 ) v j < w 2 . Open image in new window
Suppose that for n = p, we have chosen { m i ( s ) } i = 1 Open image in new window, { k i ( s ) } i = 1 Open image in new window, i + s = p + 1 with the properties:
k p - 1 ( 1 ) < m 1 ( p ) , m i ( s ) k i ( s ) , k i ( s ) < m i + 1 ( s - 1 ) Open image in new window
for i + s = p + 1 and
w i - w i 2 p + 1 - i s = 1 p + 1 - i j = m i ( s ) k i ( s ) v j < w i , for i p. Open image in new window
(iv) Let n = p + 1. We will show that we can choose { m i ( s ) } i = 1 Open image in new window, { k i ( s ) } i = 1 Open image in new window, i + s = p + 2, so that
k p ( 1 ) < m 1 ( p + 1 ) , m i ( s ) k i ( s ) , k i ( s ) < m i + 1 ( s - 1 ) Open image in new window
for i + s = p + 2 and
w i - w i 2 p + 1 - i s = 1 p + 1 - i j = m i ( s ) k i ( s ) v j < w i , Open image in new window
(1)

for i ≤ p + 1.

Indeed, let us choose first m 1 ( p + 1 ) , k 1 ( p + 1 ) N Open image in new window: k p ( 1 ) < Open image in new window m 1 ( p + 1 ) k 1 ( p + 1 ) Open image in new window, such that
w 1 - w 1 2 p + 1 f ( 1 , p ) + j = m 1 ( p + 1 ) k 1 ( p + 1 ) v j < w 1 . Open image in new window
Then, we choose m 2 ( p ) , k 2 ( p ) N Open image in new window: k 1 ( p + 1 ) < m 2 ( p ) k 2 ( p ) Open image in new window, such that
w 2 - w 2 2 p f ( 2 , p - 1 ) + j = m 2 ( p ) k 2 ( p ) v j < w 2 . Open image in new window
If for i0 ≤ p - 1 we have chosen m i 0 ( p - i 0 + 2 ) , k i 0 ( p - i 0 + 2 ) N Open image in new window to satisfy the inequalities k i 0 ( p - i 0 + 1 ) < m i 0 ( p - i 0 + 2 ) k i 0 ( p - i 0 + 2 ) Open image in new window, such that
w i 0 - w i 0 2 p - i 0 + 2 f ( i 0 , p - i 0 + 1 ) + j = m i 0 ( p - i 0 + 2 ) k i 0 ( p - i 0 + 2 ) v j < w i 0 , Open image in new window
then for i0 + 1 ≤ p, we choose m i 0 + 1 ( p - i 0 + 1 ) , k i 0 + 1 ( p - i 0 + 1 ) N Open image in new window: k i 0 + 1 ( p - i 0 ) < m i 0 + 1 ( p - i 0 + 1 ) k i 0 + 1 ( p - i 0 + 1 ) Open image in new window, such that
w i 0 + 1 - w i 0 + 1 2 p - i 0 + 1 f ( i 0 + 1 , p - i 0 ) + j = m i 0 + 1 ( p - i 0 + 1 ) k i 0 + 1 ( p - i 0 + 1 ) v j < w i 0 + 1 . Open image in new window
On the last step for i = p + 1, we choose m p + 1 ( 1 ) , k p + 1 ( 1 ) N Open image in new window: k p ( 2 ) < m p + 1 ( 1 ) k p + 1 ( 1 ) Open image in new window such that
w p + 1 - w p + 1 2 j = m p + 1 ( 1 ) k p + 1 ( 1 ) v j = f ( p + 1 , 1 ) < w p + 1 . Open image in new window

By (1), it follows that lim n s = 1 n j = m i ( s ) k i ( s ) v j = w i Open image in new window holds for every i N Open image in new window. □

Theorem 3

[19] Let the iterated series n = 1 s = 1 a n s Open image in new window be given. If the series n = 1 s = 1 | a n s | Open image in new window is convergent, then for any permutations Π , σ : N N Open image in new window the series n , s a Π ( n ) σ ( s ) Open image in new window is convergent and n , s a Π ( n ) σ ( s ) = n = 1 s = 1 a n s Open image in new window.

Lemma 2

Let M be an Orlicz function with the Δ2-condition and w = { w i } i = 1 Λ Open image in new window. Let M (w) be equipped with Luxemburg’s or Amemiya’s norm. Then,(1) For every x ∈  M (w), such that M ~ w ( λx ) < Open image in new window, for every λ > 0, there are y, z ∈  M (w), such that |y| ∧ |z| = 0, M ~ w ( λy ) = M ~ w ( λz ) = M ~ w ( λx ) Open image in new window, for any λ > 0(2) For every x S ( M ( w ) , · ) Open image in new window, there are y, z, such that |y| ∧ |z| = 0, y , z S ( M ( w ) , · ) Open image in new window(3) For every x S ( M ( w ) , | | | · | | | ) Open image in new window, there are y, z, such that |y| ∧ |z| = 0, y , z S ( M ( w ) , | | | · | | | ) Open image in new window.

Proof

(1) Let x = { x n } n = 1 M ( w ) Open image in new window be arbitrarily chosen. By M ∈ Δ 2, it follows that M ~ w ( λx ) < Open image in new window for every λ > 0. By w ∈ Λ, it follows that we can choose two subsequences v = { v i } i = 1 Open image in new window, u = { u i } i = 1 Open image in new window of w, such that v, u ∈ Λ, v ∩ u = , v ∪ u = w.

By Lemma 1, there are sequences of naturals { m i ( s ) } i = 1 Open image in new window, { k i ( s ) } i = 1 Open image in new window, { α i ( s ) } i = 1 Open image in new window, { β i ( s ) } i = 1 Open image in new window, s N Open image in new window, such that
1 m 1 ( 1 ) k 1 ( 1 ) , 1 α 1 ( 1 ) β 1 ( 1 ) Open image in new window
k n - 1 ( 1 ) < m 1 ( n ) , m i ( s ) k i ( s ) , k i ( s ) < m i + 1 ( s - 1 ) Open image in new window
for n , i , s N , n 2 , i + s = n + 1 Open image in new window
β n - 1 ( 1 ) < α 1 ( n ) , α i ( s ) β i ( s ) β i ( s ) < α i + 1 ( s - 1 ) Open image in new window
for n , i , s N , n 2 , i + s = n + 1 Open image in new window
and there hold the equalities
s = 1 j = m i ( s ) k i ( s ) v j = w i = s = 1 j = α i ( s ) β i ( s ) u j , Open image in new window

for every i N Open image in new window.

Put
y n = s = 1 j = m n ( s ) k n ( s ) x n e j , z n = s = 1 j = α n ( s ) β n ( s ) x n e j Open image in new window

and y = n = 1 y n Open image in new window, z = n = 1 z n Open image in new window. We will show that M ~ w ( λy ) = M ~ w ( λz ) = M ~ w ( λx ) Open image in new window, for any λ > 0.

Let λ > 0 and put a n s ( λ ) = M ( λ x n ) j = m n ( s ) k n ( s ) v j Open image in new window for n , s N Open image in new window. Let us consider the infinite matrix
a 1 1 ( λ ) a 1 2 ( λ ) a 1 3 ( λ ) a 1 s ( λ ) a 2 1 ( λ ) a 2 2 ( λ ) a 2 3 ( λ ) a 2 s ( λ ) ...... ...... ...... .. ..... .. a n 1 ( λ ) a n 2 ( λ ) a n 3 ( λ ) a n s ( λ ) ...... ...... ...... .. ..... .. Open image in new window
For every n N Open image in new window, the equality s = 1 a n s ( λ ) = M ( λ x n ) s = 1 j = m n ( s ) k n ( s ) v j = w n M ( λ x n ) Open image in new window holds and thus n = 1 s = 1 a n s ( λ ) = n = 1 w n M ( λ x n ) < Open image in new window for every λ > 0. By a n s ( λ ) 0 Open image in new window for every n , s N Open image in new window and Theorem 3, it follows that for any two permutations Π , σ : N N Open image in new window, the series n , s a Π ( n ) σ ( s ) ( λ ) Open image in new window is convergent and there hold the equalities
n , s a Π ( n ) σ ( s ) ( λ ) = n = 1 s = 1 a n s ( λ ) = M ~ w ( λx ) . Open image in new window
Consequently, there exist two permutations Π , σ : N N Open image in new window, such that we can write the chain of equalities
M ~ w ( λy ) = p = 2 n = 1 p - 1 a n p - n ( λ ) = p = 2 n = 1 p - 1 j = m n ( s ) k n ( s ) v j M ( λ x n ) = n , s a Π ( n ) σ ( s ) ( λ ) = n = 1 s = 1 a n s ( λ ) = M ~ w ( λx ) . Open image in new window
(2)
Similarly, if we put b n s ( λ ) = M ( λ x n ) j = α n ( s ) β n ( s ) v j Open image in new window for n , s N Open image in new window, we get the chain of equalities
M ~ w ( λz ) = p = 2 n = 1 p - 1 b n p - n ( λ ) = p = 2 n = 1 p - 1 j = α n ( s ) β n ( s ) u j M ( λ x n ) = n , s b Π ( n ) σ ( s ) ( λ ) = n = 1 s = 1 b n s ( λ ) = M ~ w ( λx ) . Open image in new window
(3)

Para>(2) If λ = 1, then by M ~ w ( y ) = M ~ w ( z ) = M ~ w ( x ) = 1 Open image in new window we get that if x S ( M ( w ) , · ) Open image in new window then y , z S ( M ( w ) , · ) Open image in new window.

(3) If x S ( M ( w ) , | | | · | | | ) Open image in new window, then by (1) it follows that 1 λ ( 1 + M ~ w ( λy ) ) = 1 λ ( 1 + M ~ w ( λz ) ) = 1 λ ( 1 + M ~ w ( λx ) ) Open image in new window 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 w = { w i } i = 1 Λ Open image in new window. Then, for every x S ( M ( w ) , | | | · | | | ) Open image in new window and k > 1, there exists a unique dx,k > 1, such that M ~ w kx d x , k = k - 1 2 Open image in new window.

Proof

Let x S ( M ( w ) , | | | · | | | ) Open image in new window and k > 1 be arbitrarily chosen and fixed, then we define the function S ( d ) : [ 1 , + ) R Open image in new window by S ( d ) = M ~ w kx d = i = 1 w i M k x i d Open image in new window. By the inequality M k x i d M ( k x i ) Open image in new window and the convergence of the series i = 1 w i M ( k x i ) Open image in new window, it follows that i = 1 w i M k x i d Open image in new window 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
f ( 1 ) = M ~ w ( kx ) | | | kx | | | M - 1 = k - 1 > k - 1 2 Open image in new window
and
lim d + f ( d ) = lim d + M ~ w kx d = 0 < k - 1 2 , Open image in new window

we get that there is a unique dx,k > 0, such that f ( d x , k ) = M ~ w kx d x , k = k - 1 2 Open image in new window. □

Lemma 4

([10]) For a Köthe sequences space (X, ∥ · ∥), the Riesz angle a(X) can be expressed as
a ( X ) = sup { ( | x | | y | ) : x , y S X , | x | | y | = 0 } , Open image in new window

where |x| ∧ |y| = min{|x|, |y|}.

Proof of Theorem 1 (1)

Let x = { x n } n = 1 S ( M ( w ) , · ) Open image in new window be arbitrarily chosen. By w = { w i } i = 1 Λ Open image in new window, it follows that we can choose two subsequences v = { v i } i = 1 Open image in new window, u = { u i } i = 1 Open image in new window of w, such that v, u ∈ Λ, v ∩ u = , v ∪ u = w.

It follows from Lemma 1 that there exist sequences of naturals { m i ( s ) } i = 1 Open image in new window, { k i ( s ) } i = 1 Open image in new window, { α i ( s ) } i = 1 Open image in new window, { β i ( s ) } i = 1 Open image in new window, s N Open image in new window, such that
1 m 1 ( 1 ) k 1 ( 1 ) , 1 α 1 ( 1 ) β 1 ( 1 ) Open image in new window
k n - 1 ( 1 ) < m 1 ( n ) , m i ( s ) k i ( s ) , k i ( s ) < m i + 1 ( s - 1 ) Open image in new window
for n , i , s N , n 2 , i + s = n + 1 , Open image in new window
β n - 1 ( 1 ) < α 1 ( n ) , α i ( s ) β i ( s ) , β i ( s ) < α i + 1 ( s - 1 ) Open image in new window
for n , i , s N , n 2 , i + s = n + 1 Open image in new window
and there holds the equalities
s = 1 j = m i ( s ) k i ( s ) v j = w i = s = 1 j = α i ( s ) β i ( s ) u j Open image in new window

for every i N Open image in new window.

We can put
y n = s = 1 j = m n ( s ) k n ( s ) x n e j , z n = s = 1 j = α n ( s ) β n ( s ) x n e j Open image in new window

and y = n = 1 y n Open image in new window, z = n = 1 z n Open image in new window.

Using Lemma 2, we get that y , z S ( M ( w ) , · ) Open image in new window and M ~ w ( λx ) = M ~ w ( λy ) = M ~ w ( λz ) Open image in new window for any λ R Open image in new window.

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:
1 = M ~ w y + z y + z = M ~ w y y + z + M ~ w z y + z = 2 M ~ w x ( | y | | z | ) . Open image in new window
Consequently, it follows that for every x S ( M ( w ) , · ) Open image in new window, there exists k x  = ∥(|y| ∨ |z|)∥, such that M ~ w x k x = 1 2 Open image in new window. By Lemma 4, we get the inequality
a ( X ) sup k x : M ~ w x k x = 1 2 , x S X , Open image in new window

where X = ( M (w), ∥ · ∥).

On the other hand, let us put
D = sup k x : M ~ w x k x = 1 2 , x S ( M ( w ) , · ) . Open image in new window

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

Since
M ~ w ( | x | | y | ) d = M ~ w x d + M ~ w y d 1 2 + 1 2 = 1 , Open image in new window

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
d = sup | | | x | | | = 1 inf k > 1 d x , k : M ~ w kx d x , k = k - 1 2 . Open image in new window
(4)

For any ε > 0, there exist x = { x n } n = 1 S ( M ( w ) , | | | · | | | ) Open image in new window and k > 1, such that dx,k ≥ d - ε.

By w = { w i } i = 1 Λ Open image in new window, it follows that we can choose two subsequences v = { v i } i = 1 Open image in new window, u = { u i } i = 1 Open image in new window of w, such that v, u ∈ Λ, v ∩ u = , v ∪ u = w.

It follows from Lemma 1 that there exist sequences of naturals { m i ( s ) } i = 1 Open image in new window, { k i ( s ) } i = 1 Open image in new window, { α i ( s ) } i = 1 Open image in new window, { β i ( s ) } i = 1 Open image in new window, s N Open image in new window, such that
1 m 1 ( 1 ) k 1 ( 1 ) , 1 α 1 ( 1 ) β 1 ( 1 ) Open image in new window
k n - 1 ( 1 ) < m 1 ( n ) , m i ( s ) k i ( s ) , k i ( s ) < m i + 1 ( s - 1 ) Open image in new window
for n , i , s N , n 2 , i + s = n + 1 , Open image in new window
β n - 1 ( 1 ) < α 1 ( n ) , α i ( s ) β i ( s ) , β i ( s ) < α i + 1 ( s - 1 ) Open image in new window
for n , i , s N , n 2 , i + s = n + 1 , Open image in new window
and there hold the equalities
s = 1 j = m i ( s ) k i ( s ) v j = w i = s = 1 j = α i ( s ) β i ( s ) u j Open image in new window

for every i N Open image in new window.

We can put
y n = s = 1 j = m n ( s ) k n ( s ) x n e j , z n = s = 1 j = α n ( s ) β n ( s ) x n e j Open image in new window

and y = n = 1 y n Open image in new window, z = n = 1 z n Open image in new window.

Using Lemma 2, we see that y , z S ( M ( w ) , | | | · | | | ) Open image in new window, and M ~ w ( λx ) = M ~ w ( λy ) = M ~ w ( λz ) Open image in new window for any λ R Open image in new window. Let us put
D 1 = inf 0 < k 1 1 k 1 + M ~ w k ( y + z ) d - ε Open image in new window
and
D 2 = inf k > 1 1 k 1 + M ~ w k ( y + z ) d - ε Open image in new window
Therefore, by the chain of inequalities
D 2 = inf k > 1 1 k 1 + M ~ w k ( y + z ) d - ε = inf k > 1 1 k 1 + 2 M ~ w kx d - ε inf k > 1 1 k 1 + 2 M ~ w kx d x , k = inf k > 1 1 k 1 + 2 k - 1 2 = 1 Open image in new window
and
y + z d - ε = min { D 1 , D 2 } min 1 , D 2 = 1 , Open image in new window

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 S ( M ( w ) , | | | · | | | ) Open image in new window, |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 M ~ w kx d x , k = k - 1 2 Open image in new window and M ~ w hy d y , h = h - 1 2 Open image in new window, respectively. WLOG, we may assume that 1 < h ≤ k. By the chain of inequalities
( | x | | y | ) d + ε 1 h 1 + M ~ w h ( | x | | y | ) d + ε = 1 h 1 + M ~ w hx d + ε + M ~ w hy d + ε = 1 h + 1 h M ~ w hx d h , x + 1 h M ~ w hy d k , y 1 h + 1 h M ~ w hx d h , x + 1 k M ~ w ky d k , y = 1 h + 1 h h - 1 2 + 1 k k - 1 2 = 1 + 1 2 h - 1 2 k < 3 2 , Open image in new window

we obtain the inequality | | | ( | x | | y | ) | | | 3 2 d + 3 2 ε Open image in new window and hence a ( M ( w ) ) < 3 2 d + 5 ε 2 Open image in new window. Since ε > 0 is arbitrarily chosen, it follows that a ( M ( w ) ) 3 2 d Open image in new window.

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 G M ( u ) = M - 1 ( u ) M - 1 ( 2 u ) Open image in new window, u ∈ (0, + ) [20]. Following [10, 21], we define the indices:
α M 0 = liminf u 0 G M ( u ) , α M + = liminf u G M ( u ) , α M 0 , + = min { α M 0 , α M + } , α ~ M = inf M - 1 ( u ) M - 1 ( 2 u ) : u ( 0 , + ) . Open image in new window
(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 w = { w i } i = 1 Λ Open image in new window. Then,
1 α M 0 , a ( ( M ( w ) , · ) ) = 1 α ~ M . Open image in new window

Proof

We will prove first that a ( ( M ( w ) , · ) ) = 1 α ~ M Open image in new window.

We chose arbitrary x = { x i } i = 1 S ( M ( w ) , · ) Open image in new window and put u i = 1 2 M ( x i ) Open image in new window. It is easy to check the equality M x i . G M M ( x i ) 2 = 1 2 M ( x i ) Open image in new window. For any u i  ∈ (0, + ), the inequality α ~ M G M ( u i ) Open image in new window holds. Then,
M ~ w ( α ~ M x ) = i = 1 w i M ( α ~ M x i ) i = 1 w i M ( x i G M ( u i ) ) = i = 1 w i M x i G M M ( x i ) 2 = i = 1 w i M ( x i ) 2 = 1 2 . Open image in new window
(6)
For every x = { x i } i = 1 S ( M ( w ) , · ) Open image in new window, there exists k x , such that M ~ w x k x = 1 2 Open image in new window, and thus by (6), it follows the inequality k x 1 α ~ M Open image in new window. Therefore,
a ( X ) = sup k x : M ~ w x k x = 1 2 , x S X 1 α ~ M , Open image in new window

where X = ( M (w), ∥ · ∥).

Now, we will prove that a ( ( M ( w ) , · ) ) 1 α ~ M Open image in new window.

For any u ∈ (0, + ), there are sequences of naturals { p n } n = 1 Open image in new window, { q n } n = 1 Open image in new window, such that p n  ≤ q n  < pn+1, for n N Open image in new window and n = 1 i = p n q n w i = 1 u Open image in new window.

Put x = n = 1 i = p n q n M - 1 ( u ) e i Open image in new window. It is easy to see that M ~ w ( x ) = 1 Open image in new window and thus x S ( M ( w ) , · ) Open image in new window. By the equality
M ~ w x M - 1 ( u ) M - 1 ( u / 2 ) = 1 2 , Open image in new window
we get that for any u ∈ (0, + ), there holds the inequality
M - 1 ( u ) M - 1 ( u / 2 ) a ( ( M ( w ) , · ) ) , Open image in new window
and therefore,
1 α ~ M = sup M - 1 ( u ) M - 1 ( u / 2 ) : u ( 0 , + ) a ( M ( w ) ) . Open image in new window
Thus, we have proven that
a ( ( M ( w ) , · ) ) = 1 α ~ M . Open image in new window
(7)

The proof of the inequality 1 α M 0 , a ( ( M ( w ) , · ) ) Open image in new window follows directly by equality (7) and the inequality α M 0 , α ~ M Open image in new window. □

Theorem 5

Let M be an Orlicz function with the Δ2-condition and w = { w i } i = 1 Λ Open image in new window. Then,
1 α M 0 , a ( ( M ( w ) , | | | · | | | ) ) 3 2 α ~ M . Open image in new window

Proof

(i) We will prove first that
a ( ( M ( w ) , | | | · | | | ) ) 3 2 α ~ M . Open image in new window
Let us choose arbitrary x = { x i } i = 1 S ( M ( w ) , | | | · | | | ) Open image in new window. By M ∈ Δ2 and the equality
1 = | | | x | | | = inf 1 k 1 + M ~ w ( kx ) : k > 0 , Open image in new window

it follows that there exists k0 > 0, such that the equality M ~ w ( k 0 x ) = k 0 - 1 Open image in new window holds [20].

We put u i = 1 2 M ( k 0 x i ) Open image in new window. Then, similarly to (6), we can write the inequality:
M ~ w ( k 0 α ~ M x ) = i = 1 w i M ( k 0 α ~ M x i ) i = 1 w i M ( k 0 x i G M ( u i ) ) = i = 1 w i M k 0 x i G M M ( k 0 x i ) 2 = i = 1 w i M ( k 0 x i ) 2 = k 0 - 1 2 Open image in new window
and consequently we get that d x , k 0 < 1 α ~ M Open image in new window, where dx,k is the solution of the equation M ~ w kx d x , k = k - 1 2 Open image in new window, k > 1. Thus, for any k > 1, there holds the inequality
inf { d x , k : k > 1 } < d x , k 0 < 1 α ~ . Open image in new window
(8)
By the arbitrary choice of x = { x i } i = 1 S ( M ( w ) , | | | · | | | ) Open image in new window and (8), we obtain that
a ( ( M ( w ) , | | | · | | | ) ) 3 2 sup | | | x | | | = 1 inf { d x , k : k > 0 } 3 2 α ~ M . Open image in new window
From the inequalities ∥ · ∥ ≤ ||| · ||| ≤ 2 ∥ · ∥, it follows that for any x , y S ( M ( w ) , | | | · | | | ) Open image in new window with |x| ∧ |y| = 0, there holds the inequality
a ( X ) = sup { | | | ( | x | | y | ) | | | : | | | x | | | , | | | y | | | = 1 } sup { ( | x | | y | ) : x , y 1 } = a ( ( M ( w ) , · ) ) , Open image in new window
(9)

where X = ( M (w), ||| · |||).

Therefore, we get that
a ( ( M ( w ) , | | | · | | | ) ) a ( ( M ( w ) , · ) ) = 1 α ~ M . Open image in new window
(ii) By (9), we obtain the inequality
a ( ( M ( w ) , | | | · | | | ) ) a ( ( M ( w ) , · ) ) 1 α M 0 , . Open image in new window

Definition 2

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

The function M-1 is a concave function and thus M - 1 ( t ) M - 1 ( 2 t ) 1 2 Open image in new window. According to [20] (p. 22), M ∈ ∇2 iff liminf t + M - 1 ( t ) M - 1 ( 2 t ) > 1 2 Open image in new window and liminf t 0 M - 1 ( t ) M - 1 ( 2 t ) > 1 2 Open image in new window, i.e.
M 2 α M 0 > 1 / 2 and α M + > 1 / 2 . Open image in new window
(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 α M 0 = 1 / 2 Open image in new window or α M + = 1 / 2 Open image in new window, 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: α M 0 = 1 / 2 Open image in new window, α M + = 1 / 2 Open image in new window or there exists t0 ∈ (0, + ), such that M - 1 ( t 0 ) M - 1 ( 2 t 0 ) = 1 / 2 Open image in new window.

Let α M 0 = 1 / 2 Open image in new window or α M + = 1 / 2 Open image in new window holds; then by (10), it follows that M ∉ ∇2.

There exists t0 ∈ (0, + ), such that M - 1 ( t 0 ) M - 1 ( 2 t 0 ) = 1 / 2 Open image in new window. Then, we can write the equality M - 1 ( 2 t 0 ) - M - 1 ( t 0 ) 2 t 0 - t 0 = M - 1 ( t 0 ) t 0 Open image in new window, 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 α M 0 = 1 / 2 Open image in new window. 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 F M ( t ) = tp ( t ) M ( t ) Open image in new window, t ∈ (0, + ), where p is the right derivative of M. Let us define
A M 0 = liminf t 0 + F M ( t ) , B M 0 = limsup t 0 + F M ( t ) Open image in new window
A M + = liminf t + F M ( t ) , B M + = limsup t + F M ( t ) Open image in new window
The above indices are connected by the formulas ([11], p.149; [20], p.27).
1 A M 0 + 1 B N 0 = 1 A N 0 + 1 B M 0 = 1 Open image in new window
and
1 A M + + 1 B N + = 1 A N + + 1 B M + = 1 , Open image in new window

where N is the complementary function to M.

The inequalities
2 - 1 / A M 0 α M 0 2 - 1 / B M 0 Open image in new window
(11)
2 - 1 / A M + α M + 2 - 1 / B M + Open image in new window
(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 B M + = Open image in new window, then clearly α M + = liminf u M - 1 ( u ) M - 1 ( 2 u ) 1 = 2 - 1 / B M + Open image in new window. Assume that B M + < Open image in new window. For any ε > 0, there exists t0 > 0 such that tp ( t ) M ( t ) = F M ( t ) < B M + + ε Open image in new window for every t ∈ [t0, + ). Then for any t0 ≤ t1 < t2 < + , we have
log M ( t 2 ) M ( t 1 ) = t 1 t 2 p ( t ) M ( t ) dt t 1 t 2 B M + + ε t dt = log t 2 t 1 B M + + ε . Open image in new window
We put t1 = M-1(u) and t2 = M-1(2u). Thus for any u ∈ [M(t0), + ), there holds the inequality
M - 1 ( u ) M - 1 ( 2 u ) 2 - 1 / B M + + ε . Open image in new window

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 C M 0 Open image in new window and if limt→+F M (t) exists we denote it by C M + Open image in new window. We put C M 0 , + = min { C M 0 , C M + } Open image in new window.

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 a ( M ( w ) , · ) = 2 1 / C M 0 Open image in new window

     
  2. (b)

    If F M is a decreasing function on (0, + ), then a ( M ( w ) , · ) = 2 1 / C M + Open image in new window

     
  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 a ( M ( w ) , · ) = 2 1 / C M 0 , + Open image in new window

     

Proof

  1. (a)
    If F M is an increasing function in (0, +), C M 0 = lim t 0 + F M ( t ) Open image in new window exists and G M ( u ) = M - 1 ( u ) M - 1 ( 2 u ) Open image in new window is increasing in (0, +) [21]. Then from (11), we get
    α M 0 = 2 - 1 C M 0 = lim t 0 G ( u ) = α ~ M Open image in new window
     
and therefore a ( M ( w ) , · ) = 2 1 / C M 0 Open image in new window.
  1. (b)
    If F M is a decreasing function in (0, +), then C M + = lim t + F M ( t ) Open image in new window exists, and G M ( u ) = M - 1 ( u ) M - 1 ( 2 u ) Open image in new window is decreasing in (0, +) [21]. Then by (11), we obtain
    α M + = 2 - 1 C M + = lim t + G ( u ) = α ~ M Open image in new window
     
and therefore a ( M ( w ) , · ) = 2 1 / C M + Open image in new window.
  1. (c)

    If F M is an increasing function in (0, M -1(t 0)), then C M 0 = lim t 0 + F M ( t ) Open image in new window exists, and G M ( u ) = M - 1 ( u ) M - 1 ( 2 u ) Open image in new window is increasing in (0, t 0 / 2). If F M is a decreasing function in (M -1(t 0), +), then C M + = lim t + F M ( t ) Open image in new window exists, and G M ( u ) = M - 1 ( u ) M - 1 ( 2 u ) Open image in new window is decreasing in (t 0 / 2, +) [21].

     

From (11), it follows that α M 0 = 2 - 1 C M 0 Open image in new window, α M + = 2 - 1 C M + Open image in new window and hence α ~ M = 2 - 1 C M 0 , + Open image in new window. By Theorem 4, we get that a ( M ( w ) , · ) = 2 1 / C M 0 , + Open image in new window.

Conclusions

Example 1

Let M1(t) = 2|t| p  + |t|2p, p ∈ [1, +). Then, F M 1 ( t ) = 2 p 1 - 1 t p + 2 Open image in new window for t ∈ [0, +). The function F M 1 Open image in new window is an increasing function and lim t 0 F M 1 ( t ) = p Open image in new window. By Corollary 2, we get that a ( M 1 ( w ) , · ) = 2 p Open image in new window.

Example 2

Let M 2 ( t ) = | t | p log ( 1 + | t | ) Open image in new window, p ∈ [1, +). Then, F M 2 ( t ) = p - t ( 1 + t ) log ( 1 + t ) Open image in new window for t ∈ [0, +). The function F M 2 Open image in new window is an increasing function and lim t 0 F M 2 ( t ) = p - 1 Open image in new window. By Corollary 2, we get that a ( M 2 ( w ) , · ) = 2 1 p - 1 Open image in new window.

Example 3

Let M3(t) = |t| p logr(1 + |t|), p ∈ [1, +), r ∈ (0, +). Then,
F M 3 ( t ) = p + rt ( 1 + t ) log ( 1 + t ) for t [ 0 , + ) . Open image in new window

The function F M 3 Open image in new window is a decreasing function and lim t + F M 3 ( t ) = p Open image in new window. By Corollary 2, we get that a ( M 3 ( w ) , · ) = 2 1 / p Open image in new window.

Example 4

Let q ≥ 2 and p ∈ [q - 1, 2q - 1]. We define the function
M 4 ( t ) = | t | p ( 1 + log | t | ) , | t | 1 2 q - p - 1 q | t | q + p - q + 1 q | t | 2 q , | t | 1 . Open image in new window
The function M4 is an Orlicz function. Then,
F M 4 ( t ) = p + 1 1 + log t , t 1 2 q 1 - a 2 ( a + b t q ) , t [ 0 , 1 ] , Open image in new window

where a = 2 q - p - 1 q Open image in new window and b = p - q + 1 q Open image in new window. The index function F M 4 Open image in new window is increasing on [0,1] and is decreasing on [0, +), lim t 0 F M 4 ( t ) = q Open image in new window and lim t + F M 4 ( t ) = p Open image in new window. By Corollary 2, we get that a ( M 4 ( w ) , · ) = 2 1 min { p , q } Open image in new window.

Example 5

Let M5(t) = (1 + |t|) log(1 + |t|) - |t|. Then F M 5 ( t ) = t log ( 1 + t ) ( 1 + t ) log ( 1 + t ) - t Open image in new window is a decreasing function on (0, +) and lim t + F M 5 ( t ) = 1 Open image in new window. Thus, a ( M 5 ( w ) , · ) = 2 Open image in new window. 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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© Zlatanov; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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Authors and Affiliations

  1. 1.Faculty of Mathematics and InformaticsUniversity of PlovdivPlovdivBulgaria

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