The New Class Lz,P,E of S− Type Operators

Home , Compact operator

AIMS Mathematics, 4(3): 779–791. DOI:10.3934/math.2019.3.779 Received: 09 April 2019 Accepted: 02 June 2019 http://www.aimspress.com/journal/Math Published: 05 July 2019

Research article

The new class Lz,p,E of s− type operators

Pınar Zengin Alp and Emrah Evren Kara∗

Department of Mathematics, Duzce¨ University, Konuralp, Duzce, Turkey

* Correspondence: Email: [email protected]. Abstract: The purpose of this study is to introduce the class of s-type Z u, v; lp (E) operators, which we denote by Lz,p,E (X, Y), we prove that this class is an operator ideal and quasi-Banach operator ideal by a quasi-norm defined on this class. Then we define classes using other examples of s-number sequences. We conclude by investigating which of these classes are injective, surjective or symmetric. Keywords: block sequence space; operator ideal; s-numbers; quasi-norm Mathematics Subject Classification: 47B06, 47B37, 47L20.

1. Introduction

In this study, the set of all natural numbers is represented by N and the set of all nonnegative real numbers is represented by R+ . If the dimension of the range space of a bounded linear operator is finite, it is called a finite rank operator [1]. Throughout this study, X and Y denote real or complex Banach spaces. The space of all bounded linear operators from X to Y is denoted by B (X, Y) and the space of all bounded linear operators from an arbitrary Banach space to another arbitrary Banach space is denoted by B. The theory of operator ideals is a very important field in functional analysis. The theory of normed operator ideals first appeared in 1950’s in [2]. In functional analysis, many operator ideals are constructed via different scalar sequence spaces. An s- number sequence is one of the most important examples of this. The definition of s- numbers goes back to E. Schmidt [3], who used this concept in the theory of non-selfadjoint integral equations. In Banach spaces there are many different possibilities of defining some equivalents of s- numbers, namely Kolmogorov numbers, Gelfand numbers, approximation numbers, and several others. In the following years, Pietsch give the notion of s- number sequence to combine all s- numbers in one definition [4–6]. A map

S : K → (sr(K)) 780 which assigns a non-negative scalar sequence to each operator is called an s-number sequence if for all Banach spaces X, Y, X0 and Y0 the following conditions are satisﬁed:

(i) kKk = s1 (K) ≥ s2 (K) ≥ ... ≥ 0, for every K ∈ B (X, Y) , (ii) sp+r−1 (L + K) ≤ sp (L) + sr (K) for every L, K ∈ B (X, Y) and p, r ∈ N, (iii) sr (MLK) ≤ kMk sr (L) kKk for all M ∈ B (Y, Y0) , L ∈ B (X, Y) and K ∈ B (X0, X) , where X0, Y0 are arbitrary Banach spaces, (iv) If rank (K) ≤ r, then sr (K) = 0, n (v) sn−1 (In) = 1, where In is the identity map of n-dimensional Hilbert space l2 to itself [7].

sr (K) denotes the r − th s−number of the operator K. Approximation numbers are frequently used examples of s-number sequence which is deﬁned by Pietsch. ar (K), the r-th approximation number of a bounded linear operator is deﬁned as

ar (K) = inf { kK − Ak : A ∈ B (X, Y) , rank (A) < r } , where K ∈ B (X, Y) and r ∈ N [4]. Let K ∈ B (X, Y) and r ∈ N. The other examples of s-number 0 sequences are given in the following, namely Gel f and number (cr (K)), Kolmogorov number (dr (K)), Weyl number (xr (K)), Chang number (yr (K)) , Hilbert number (hr (K)) , etc. For the deﬁnitions of these sequences we refer to [1]. In the sequel there are some properties of s−number sequences. When any isometric embedding J ∈ B (Y, Y0) is given and an s-number sequence s = (sr) satisﬁes sr (K) = sr (JK) for all K ∈ B (X, Y) the s-number sequence is called injective [8, p.90].

Proposition 1. [8, p.90–94] The number sequences (cr (K)) and (xr (K)) are injective.

When any quotient map S ∈ B (X0, X) is given and an s-number sequence s = (sr) satisﬁes sr (K) = sr (KS) for all K ∈ B (X, Y) the s-number sequence is called surjective [8, p.95].

Proposition 2. [8, p.95] The number sequences (dr (K)) and (yr (K)) are surjective. Proposition 3. [8, p.115] Let K ∈ B (X, Y). Then the following inequalities hold:

i)h r (K) ≤ xr (K) ≤ cr (K) ≤ ar (K) and ii)h r (K) ≤ yr (K) ≤ dr (K) ≤ ar (K) .

Lemma 1. [5] Let S, K ∈ B (X, Y) , then |sr (K) − sr (S )| ≤ kK − S k for r = 1, 2,.... Let ω be the space of all real valued sequences. Any vector subspace of ω is called a sequence space. In [9] the space Z u, v; lp is deﬁned by Malkowsky and Savas¸as follows:

 ∞ n p   X X  Z u, v; l = x ∈ ω : u v x < ∞ p  n k k   n=1 k=1  where 1 < p < ∞ and u = (un) and v = (vn) are positive real numbers.

The Cesaro sequence space cesp is deﬁned as ( [10, 11, 19])

 ∞  n p   X 1 X   ces = x = (x ) ∈ ω :  |x | < ∞ , 1 < p < ∞. p  k n k    n=1 k=1 

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∞ P p If an operator K ∈ B (X, Y) satisfies (an (K)) < ∞ for 0 < p < ∞, K is defined as an lp type n=1 operator in [4] by Pietsch. Afterwards ces-p type operators which is a new class obtained via Cesaro sequence space is introduced by Constantin [12]. Later on Tita in [14] proved that the class of lp type operators and ces-p type operators coincide. (s) In [15], ςp , the class of s−type Z u, v; lp operators is given. For more information about sequence spaces and operator ideals we refer to [1, 13, 16, 18, 20]. Let X0 , the dual of X, be the set of continuous linear functionals on X. The map x∗ ⊗ y : X → Y is defined by (x∗ ⊗ y)(x) = x∗ (x) y where x ∈ X, x∗ ∈ X0 and y ∈ Y. A subcollection = of B is said to be an operator ideal if for each component = (X, Y) = = ∩ B (X, Y) the following conditions hold:

(i) if x∗ ∈ X0 , y ∈ Y, then x∗ ⊗ y ∈ = (X, Y) , (ii) if L, K ∈ = (X, Y) , then L + K ∈ = (X, Y) , (iii) if L ∈ = (X, Y) , K ∈ B (X0, X) and M ∈ B (Y, Y0) , then MLK ∈ = (X0, Y0) [6]. Let = be an operator ideal and ρ : = → R+ be a function on =. Then, if the following conditions hold:

(i) if x∗ ∈ X0 , y ∈ Y, then ρ (x∗ ⊗ y) = kx∗k kyk ; (ii) if ∃C ≥ 1 such that ρ (L + K) ≤ C ρ (L) + ρ (K) ; (iii) if L ∈ = (X, Y) , K ∈ B (X0, X) and M ∈ B (Y, Y0) , then ρ (MLK) ≤ kMk ρ (L) kKk,

ρ is said to be a quasi-norm on the operator ideal = [6]. For special case C = 1, ρ is a norm on the operator ideal =. If ρ is a quasi-norm on an operator ideal =, it is denoted by =, ρ. Also if every component = (X, Y) is complete with respect to the quasi-norm ρ, =, ρ is called a quasi-Banach operator ideal. Let =, ρ be a quasi-normed operator ideal and J ∈ B (Y, Y0) be a isometric embedding. If for every operator K ∈ B (X, Y) and JK ∈ = (X, Y0) we have K ∈ = (X, Y) and ρ (JK) = ρ (K), =, ρ is called an injective quasi-normed operator ideal. Furthermore, let =, ρ be a quasi-normed operator ideal and S ∈ B (X0, X) be a quotient map. If for every operator K ∈ B (X, Y) and KS ∈ = (X0, Y) we have K ∈ = (X, Y) and ρ (KS) = ρ (K), =, ρ is called an surjective quasi-normed operator ideal [6]. Let K0 be the dual of K. An s− number sequence is called symmetric (respectively, completely 0 0 symmetric) if for all K ∈ B, sr (K) ≥ sr(K ) (respectively,sr (K) = sr(K )) [6].

0 Lemma 2. [6] The approximation numbers are symmetric, i.e., ar(K ) ≤ ar(K) for K ∈ B . Lemma 3. [6] Let K ∈ B. Then

0 0 cr(K) = dr(K ) and cr(K ) ≤ dr(K).

0 In addition , if K is a compact operator then cr(K ) = dr(K) .

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Lemma 4. [8] Let K ∈ B. Then

0 0 xr(K) = yr(K ) and yr(K ) = xr(K) . The dual of an operator ideal = is denoted by =0 and it is defined as [6] 0 n 0 0 o = (X, Y) = K ∈ B (X, Y) : K0 ∈ = Y , X . An operator ideal = is called symmetric if = ⊂ =0 and is called completely symmetric if = = =0 [6]. Let E = (En) be a partition of finite subsets of the positive integers which satisfies

max En < min En+1 + for n ∈ N . In [21] Foroutannia deﬁned the sequence space lp (E) by  p   X∞ X    lp (E) = x = (xn) ∈ ω : x j < ∞ , (1 ≤ p < ∞)   n=1 j∈En with the seminorm k|·|kp,E , which deﬁned as:

p 1  ∞  p X X  k|x|k =  x  . p,E  j  n=1 j∈En

For example, if En = {3n − 2, 3n − 1, 3n} for all n, then x = (xn) ∈ lp (E) if and only if ∞ P p |x3n−2 + x3n−1 + x3n| < ∞. It is obvious that k|·|kp,E is not a norm, since we have k|x|kp,E = 0 while n=1 + x = (−1, 1, 0, 0,...) and En = {3n − 2, 3n − 1, 3n} for all n. For the particular case En = {n} for n ∈ N we get lp (E) = lp and k|x|kp,E = kxkp . For more information about block sequence spaces, we refer the reader to [17, 22–25].

2. Results

Let u = (u ) and v = (v ) be positive real number sequences. In this section, by replacing l with n n p lp (E) we get the sequence space Z u, v; lp (E) deﬁned as follows:  p   X∞ Xn X    Z u, v; lp (E) = x ∈ ω : un v j x j < ∞ .   n=1 k=1 j∈Ek An operator K ∈ B (X, Y) is in the class of s-type Z u, v; lp (E) if

∞  n p X  X X  u v s (K) < ∞, (1 < p < ∞) .  n j j  n=1 k=1 j∈Ek The class of all s-type Z u, v; lp (E) operators is denoted by Lz,p,E (X, Y) . (s) In particular case if En = {n} for n = 1, 2,..., then the class Lz,p,E (X, Y) reduces to the class ςp . Conditions used in Theorem 1 hold throughout the remainder of the paper.

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∞ P p Theorem 1. Fix 1 < p < ∞. If (un) < ∞ and M > 0 is such that v2k−1 + v2k ≤ Mvk, M > 0 for all n=1 k ∈ N, then Lz,p,E is an operator ideal. ∗ 0 ∗ ∗ Proof. Let x ∈ X and y ∈ Y. Since the rank of the operator x ⊗ y is one, sn (x ⊗ y) = 0 for n ≥ 2. By using this fact

∞  n p ∞ X  X X  X u v s (x∗ ⊗ y) = (u )p (v s (x∗ ⊗ y))p  n j j  n 1 1 n=1 k=1 j∈Ek n=1 ∞ X p p ∗ p = (un) (v1) kx ⊗ yk n=1 ∞ X p p ∗ p p = (un) (v1) kx k kyk n=1 < ∞.

∗ Therefore x ⊗ y ∈ Lz,p,E(X, Y). Let L, K ∈ Lz,p,E(X, Y). Then

∞  n p ∞  n p X  X X  X  X X  u v s (L) < ∞, u v s (K) < ∞.  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek

To show that L + K ∈ Lz,p,E(X, Y), let us begin with

n n   X X X X X  v s (L + K) ≤  v s (L + K) + v s (L + K) j j  2 j−1 2 j−1 2 j 2 j  k=1 j∈Ek k=1 j∈Ek j∈Ek Xn X ≤ v2 j−1 + v2 j s2 j−1 (L + K)

k=1 j∈Ek Xn X ≤ M v j s j (L) + s j (K)

k=1 j∈Ek n   X X X  ≤ M  v s (L) + v s (K) .  j j j j  k=1 j∈Ek j∈Ek By using Minkowski inequality we get;

1  ∞  n  p p X  X X    u  v s (L + K)    n  j j   n=1 k=1 j∈Ek 1  ∞  n  p p X  X X X   ≤ M  u  v s (L) + v s (K)    n  j j j j   n=1 k=1 j∈Ek j∈Ek   p 1   p 1   ∞ n p ∞ n p  X  X X   X  X X    ≤ M  un v j s j (L)  +  un v j s j (K)   < ∞.          n=1 k=1 j∈Ek n=1 k=1 j∈Ek

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Hence L + K ∈ Lz,p,E(X, Y). Let M ∈ B(Y, Y0), L ∈ Lz,p,E(X, Y) and K ∈ B(X0, X). Then,

∞  n p ∞  n p X  X X  X  X X  u v s (MLK) ≤ u kMk kKk v s (L)  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek  ∞  n p X  X X   ≤ kMkp kKkp  u v s (L)  < ∞.   n j j   n=1 k=1 j∈Ek

So MLK ∈ Lz,p,E(X0,Y0). Therefore Lz,p,E(X, Y) is an operator ideal.

!p! 1 P∞ Pn P p un v j s j (K) n=1 k=1 j∈Ek Theorem 2. kKkz,p,E = 1 is a quasi-norm on the operator ideal Lz,p,E. ∞ ! p P p (un) v1 n=1

∗ 0 ∗ ∗ Proof. Let x ∈ X and y ∈ Y. Since the rank of the operator x ⊗ y is one , sn (x ⊗ y) = 0 for n ≥ 2. Then

p 1 1 ∞ n ! ! p ∞ ! ! p P P P ∗ ⊗ P p p ∗ p un v j s j (x y) (un) v1 kx ⊗ yk n=1 k=1 j∈Ek n=1 1 = 1 ∞ ! p ∞ ! p P p P p (un) v1 (un) v1 n=1 n=1 = kx∗ ⊗ yk = kx∗k kyk .

∗ ∗ Therefore kx ⊗ ykz,p,E = kx k kyk . Let L, K ∈ Lz,p,E(X, Y). Then

Xn X Xn X X v j s j (L + K) ≤ v2 j−1 s2 j−1 (L + K) + v2 j s2 j (L + K)

k=1 j∈Ek k=1 j∈Ek j∈En Xn X ≤ v2 j−1 + v2 j s2 j−1 (L + K)

k=1 j∈Ek Xn X ≤ M v j s j (L) + s j (K) .

k=1 j∈Ek

By using Minkowski inequality we get;

1 1  ∞  n p p  ∞  n p p X  X X   X  X X    u v s (L + K)  ≤  Mu v s (L) + s (K)     n j j     n j j j   n=1 k=1 j∈Ek n=1 k=1 j∈Ek

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 !p! 1   ∞ n p   P u P P v s (L)   n j j   n=1 k=1 j∈Ek  ≤ M  1  .  ∞ n !p! p   P P P   + un v j s j (K)  n=1 k=1 j∈Ek

Hence kL + Kkz,p,E ≤ M kS kz,p,E + kKkz,p,E .

Let M ∈ B(Y, Y0), L ∈ Lz,p,E(X, Y) and K ∈ B(X0, X)

1 1  ∞  n p p  ∞  n p p X  X X   X  X X    u v s (MLK)  ≤  u kMk kKk v s (L)    n j j     n j j   n=1 k=1 j∈Ek n=1 k=1 j∈Ek 1  ∞  n p p X  X X   ≤ kMk kKk  u v s (L)    n j j   n=1 k=1 j∈Ek < ∞

kMLKkz,p,E ≤ kMk kKk kLkz,p,E .

Therefore kKkz,p,E is a quasi-norm on Lz,p,E. h i Theorem 3. Let 1 < p < ∞. Lz,p,E(X, Y), kKkz,p,E is a quasi-Banach operator ideal. Proof. Let X, Y be any two Banach spaces and 1 ≤ p < ∞. The following inequality holds

!p! 1 P∞ Pn P p un v j s j (K) n=1 k=1 j∈Ek kKkz,p,E = 1 ≥ kKk ∞ ! p P p (un) v1 n=1 for K ∈ Lz,p,E(X, Y). Let (Km) be Cauchy in Lz,p,E(X, Y). Then for every ε > 0 there exists n0 ∈ N such that

kKm − Klkz,p,E < ε (2.1) for all m, l ≥ n0. It follows that k − k ≤ k − k Km Kl Km Kl z,p,E < ε.

Then (Km) is a Cauchy sequence in B (X, Y) . B (X, Y) is a Banach space since Y is a Banach space. Therefore kKm − Kk → 0 as m → ∞ for some K ∈ B (X, Y) . Now we show that kKm − Kkz,p,E → 0 as m → ∞ for K ∈ Lz,p,E (X, Y) . The operators Kl − Km, K − Km are in the class B (X, Y) for Km, Kl, K ∈ B (X, Y) .

|sn (Kl − Km) − sn (K − Km)| ≤ kKl − Km − (K − Km)k = kKl − Kk .

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Since Kl → K as l → ∞ we obtain

sn (Kl − Km) → sn (K − Km) (2.2)

It follows from (2.1) that the statement

!p! 1 P∞ Pn P p un v j s j (Km − Kl) n=1 k=1 j∈Ek kKm − Klkz,p,E = 1 < ε ∞ ! p P p (un) v1 n=1 holds for all m, l ≥ n0. We obtain from (2.2) that

!p! 1 P∞ Pn P p un v j s j (Km − K) n=1 k=1 j∈Ek 1 ≤ ε. ∞ ! p P p (un) v1 n=1

Hence we have

kKm − Kkz,p,E < ε for all m ≥ n0.

Finally we show that K ∈ Lz,p,E (X, Y) .

∞  n p ∞  n n p X  X X  X  X X X X  u v s (K) ≤ u v s (K) + u v s (K)  n j j   n 2 j−1 2 j−1 n 2 j 2 j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek k=1 j∈Ek ∞  n p X  X X  ≤ u v + v s (K − K + K )  n 2 j−1 2 j 2 j−1 m m  n=1 k=1 j∈Ek ∞  n p X  X X  ≤ M u v s (K − K ) + s (K )   n j j m j m  n=1 k=1 j∈Ek

By using Minkowski inequality; since Km ∈ Lz,p,E (X, Y) for all m and kKm − Kkz,p,E → 0 as m → ∞, we have

1  ∞  n p p X  X X   M  u v s (K − K ) + s (K )     n j j m j m   n=1 k=1 j∈Ek  !p! 1   ∞ n p   P u P P v s (K − K )   n j j m   n=1 k=1 j∈Ek  ≤ M  1  < ∞  ∞ n !p! p   P P P   + un v j s j (Km)  n=1 k=1 j∈Ek which means K ∈ Lz,p,E (X, Y) .

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Deﬁnition 1. Let µ = (µi (K)) be one of the sequences s = (sn (K)) , c = (cn (K)) , d = (dn (K)) , (µ) x = (xn (K)) , y = (yn (K)) and h = (hn (K)) . Then the space Lz,p,E generated via µ = (µi (K)) is deﬁned as

  p   X∞ Xn X  (µ)     L (X, Y) = K ∈ B (X, Y) : un v jµ j (K) < ∞, (1 < p < ∞) . z,p,E     n=1 k=1 j∈Ek (µ) And the corresponding norm kKkz,p,E for each class is deﬁned as

!p! 1 P∞ Pn P p un v jµ j (K) n=1 k=1 j∈E k k(µ) k K z,p,E = 1 . ∞ ! p P p (un) v1 n=1

(a) (a) Proposition 4. The inclusion Lz,p,E ⊆ Lz,q,E holds for 1 < p ≤ q < ∞. (a) (a) Proof. Since lp ⊆ lq for 1 < p ≤ q < ∞ we have Lz,p,E ⊆ Lz,q,E.

h (s) (s) i Theorem 4. Let 1 < p < ∞. The quasi-Banach operator ideal Lz,p,E, kKkz,p,E is injective, if the sequence sn(K) is injective.

Proof. Let 1 < p < ∞ and K ∈ B (X, Y) and J ∈ B (Y, Y0) be any isometric embedding. Suppose that (s) JK ∈ Lz,p,E (X, Y0) . Then ∞  n p X  X X  u v s (JK) < ∞  n j j  n=1 k=1 j∈Ek

Since s = (sn) is injective, we have

sn (K) = sn (JK) for all K ∈ B (X, Y) , n = 1, 2,.... (2.3) Hence we get ∞  n p ∞  n p X  X X  X  X X  u v s (K) = u v s (JK) < ∞  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek (s) Thus K ∈ Lz,p,E (X, Y) and we have from (2.3)

!p! 1 P∞ Pn P p un v j s j (JK) n=1 k=1 j∈E kJ k(s) k K z,p,E = 1 ∞ ! p P p (un) v1 n=1 !p! 1 P∞ Pn P p un v j s j (K) n=1 k=1 j∈E k k k(s) = 1 = K z,p,E ∞ ! p P p (un) v1 n=1

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h (s) (s) i Hence the operator ideal Lz,p,E, kKkz,p,E is injective.

Conclusion 1. [8, p.90–94] Since the number sequences (cn(K)) and (xn(K)) are injective , the quasi- h (c) (c) i h (x) (x) i Banach operator ideals Lz,p,E, kKkz,p,E and Lz,p,E, kKkz,p,E are injective.

h (s) (s) i Theorem 5. Let 1 < p < ∞. The quasi-Banach operator ideal Lz,p,E, kKkz,p,E is surjective, if the sequence (sn(K)) is surjective.

Proof. Let 1 < p < ∞ and K ∈ B (X, Y) and S ∈ B (X0, X) be any quotient map. Suppose that (s) KS ∈ Lz,p,E (X0, Y). Then ∞  n p X  X X  u v s (KS) < ∞.  n j j  n=1 k=1 j∈Ek

Since s = (sn) is surjective, we have

sn (K) = sn (KS) for all K ∈ B (X, Y) , n = 1, 2,.... (2.4)

Hence we get ∞  n p ∞  n p X  X X  X  X X  u v s (K) = u v s (KS) < ∞.  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek (s) Thus K ∈ Lz,p,E (X, Y) and we have from (2.4)

!p! 1 P∞ Pn P p un v j s j (KS) n=1 k=1 j∈E k Sk(s) k K z,p,E = 1 ∞ ! p P p (un) v1 n=1 !p! 1 P∞ Pn P p un v j s j (K) n=1 k=1 j∈E k k k(s) = 1 = K z,p,E . ∞ ! p P p (un) v1 n=1

h (s) (s) i Hence the operator ideal Lz,p,E, kKkz,p,E is surjective.

Conclusion 2. [8, p.95] Since the number sequences (dn(K)) and (yn(K)) are surjective , the quasi- h (d) (d) i h (y) (y) i Banach operator ideals Lz,p,E, kKkz,p,E and Lz,p,E, kKkz,p,E are surjective. Theorem 6. Let 1 < p < ∞. Then the following inclusion relations holds: (a) (c) (x) (h) iL z,p,E ⊆ Lz,p,E ⊆ Lz,p,E ⊆ Lz,p,E (a) (d) (y) (h) iiL z,p,E ⊆ Lz,p,E ⊆ Lz,p,E ⊆ Lz,p,E. (a) Proof. Let K ∈ Lz,p,E. Then ∞  n p X  X X  u v s (K) < ∞  n j j  n=1 k=1 j∈Ek

AIMS Mathematics Volume 4, Issue 3, 779–791. 789 where 1 < p < ∞. And from Proposition 3, we have;

∞  n p ∞  n p X  X X  X  X X  u v h (K) ≤ u v x (K)  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek ∞  n p X  X X  ≤ u v c (K)  n j j  n=1 k=1 j∈Ek ∞  n p X  X X  ≤ u v a (K)  n j j  n=1 k=1 j∈Ek < ∞ and ∞  n p ∞  n p X  X X  X  X X  u v h (K) ≤ u v y (K)  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek ∞  n p X  X X  ≤ u v d (K)  n j j  n=1 k=1 j∈Ek ∞  n p X  X X  ≤ u v a (K)  n j j  n=1 k=1 j∈Ek < ∞.

So it is shown that the inclusion relations are satisﬁed. (a) (h) Theorem 7. For 1 < p < ∞,Lz,p,E is a symmetric operator ideal and Lz,p,E is a completely symmetric operator ideal. Proof. Let 1 < p < ∞. 0 (a) (a) (a) (a) Firstly, we show that Lz,p,E is symmetric in other words Lz,p,E ⊆ Lz,p,E holds. Let K ∈ Lz,p,E. Then

∞  n p X  X X  u v a (K) < ∞.  n j j  n=1 k=1 j∈Ek

0 It follows from [6, p.152] an K ≤ an (K) for K ∈ B. Hence we get

∞  n p ∞  n p X  X X  X  X X  u v a T 0 ≤ u v a (K) < ∞.  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek

0 (a) (a) Therefore K ∈ Lz,p,E . Thus Lz,p,E is symmetric. 0 (h) (h) 0 Now we prove that the equation Lz,p,E = Lz,p,E holds. It follows from [8, p.97] that hn K = hn (K) for K ∈ B. Then we can write

∞  n p ∞  n p X  X X  X  X X  u v h K0 = u v h (K) .  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek

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∞  n p ∞  n p X  X X  X  X X  u v h K0 = u v h (K) .  n j j   n j j  n=1 k=1 j∈Ek n=1 k=1 j∈Ek (h) Hence Lz,p,E is completely symmetric.

0 0 (c) (d) (d) (c) Theorem 8. Let 1 < p < ∞. The equation Lz,p,E = Lz,p,E and the inclusion relation Lz,p,E ⊆ Lz,p,E (d) 0 (c) holds. Also, for a compact operator K, K ∈ Lz,p,E if and only if K ∈ Lz,p,E .

0 0 Proof. Let 1 < p < ∞. For K ∈ B it is known from [8] that cn (K) = dn K and cn K ≤ dn (K) . 0 Also, when K is a compact operator, the equality cn K = dn (K) holds. Thus the proof is clear.

0 0 (x) (y) (y) (x) Theorem 9. Lz,p,E = Lz,p,E and Lz,p,E = Lz,p,E hold for 1 < p < ∞.

0 0 Proof. Let 1 < p < ∞. For K ∈ B we have from [8] that xn (K) = yn K and yn (K) = xn K . Thus the proof is clear.

Acknowledgments

The authors would like to thank anonymous referees for their careful corrections and valuable comments on the original version of this paper.

Conﬂict of interest

The authors declare no conﬂict of interest.

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AIMS Mathematics Volume 4, Issue 3, 779–791.