- Email: [email protected]
Essential norm of substitution vector-valued integral operator on L1(Σ) space
JID:YJMAA
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.1 (1-6)
J. Math. Anal. Appl. ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Essential norm of substitution vector-valued integral operator on L1 (Σ) space Z. Moayyerizadeh Department of Mathematical Sciences, University of Lorestan, 5166617766, Khorramabad, Iran
a r t i c l e
i n f o
Article history: Received 17 October 2016 Available online xxxx Submitted by J.A. Ball
a b s t r a c t In this paper, we determine essential norm of a substitution vector-valued integral operator Tuϕ on L1 (Σ) space using conditional expectation operators. © 2016 Elsevier Inc. All rights reserved.
Keywords: Integral operator Conditional expectation Essential norm
1. Introduction and preliminaries Let (X, Σ, μ) be a σ-finite measure space and ϕ : X → X be a non-singular measurable transformation; i.e. μ ◦ ϕ−1 μ. Here the non-singularity of ϕ guarantees that the operator f → f ◦ ϕ is well defined as a mapping on L0 (Σ) where L0 (Σ) denotes the linear space of all equivalence classes of Σ-measurable functions on X. Let h0 = dμ ◦ ϕ−1 /dμ be the Radon–Nikodym derivative. We also assume that h0 is almost everywhere finite-valued, or equivalently ϕ−1 (Σ) ⊆ Σ is a sub-σ-finite algebra see [6]. All comparisons between two functions or two sets are to be interpreted as holding up to a μ-null set. For a sub-σ-finite algebra A ⊆ Σ, the conditional expectation operator associated with A is the mapping f → E A f , defined for all non-negative f as well as for all f ∈ Lp (Σ), 1 ≤ p ≤ ∞, where E A f , by Radon–Nikodym Theorem, is the unique A-measurable function satisfying
f dμ = A
E A f dμ,
∀A ∈ A.
A
We recall that E A : L2 (Σ) → L2 (A) is an orthogonal projection. For more details on the properties of E A −1 see [3,4]. Throughout this paper, we assume that A = ϕ−1 (Σ) and E ϕ (Σ) = E. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jmaa.2016.11.063 0022-247X/© 2016 Elsevier Inc. All rights reserved.
JID:YJMAA 2
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.2 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
Recall that an atom of the measure μ is an element A ∈ Σ with μ(A) > 0, such that for each B ∈ Σ, if B ⊂ A then either μ(B) = 0 or μ(B) = μ(A). A measure with no atoms is called non-atomic. We can easily check the following well-known facts (see [8]): (a) Every σ-finite measure space (X, Σ, μ) can be partitioned uniquely as X = (∪n∈N An ) ∪ B,
(1.1)
where {An }n∈N ⊆ Σ is a countable collection of pairwise disjoint atoms and B, being disjoint from each An , is non-atomic. (b) Let E be a non-atomic set with μ(E) > 0. Then there exists a sequence of positive disjoint Σ-measurable subsets of E, {En }n∈N such that μ(En ) > 0 for each n ∈ N and limn→∞ μ(En ) = 0. For a given complex Hilbert space H, let u : X → H be a mapping. We say that u is weakly measurable if for each h ∈ H the mapping x → u(x), h of X to C is measurable. We will denote this map by u, h. Let Lp (X) be the class of all measurable mappings f : X → C such that f pp = X |f (x)|p dμ < ∞ for p ≥ 1. Let ϕ : X → X be a non-singular measurable transformation and let u : X → H be a weakly measurable function. Then the pair (u, ϕ) induces a substitution vector-valued integral operator Tuϕ : Lp (X) → H defined by
Tuϕ f, h =
u, hf ◦ ϕdμ,
h ∈ H, f ∈ Lp (X).
X
It is easy to see that Tuϕ is well defined and linear. Moreover for each f ∈ Lp (X), sup | Tuϕ f, h| ≤ sup Tuϕ f h = Tuϕ f
h∈H1
h∈H1
T ϕf = | Tuϕ f, uϕ | ≤ sup | Tuϕ f, h|, Tu f h∈H1 where H1 is the closed unit ball of H. Hence Tuϕ f = sup | Tuϕ f, h|, for each f ∈ Lp (X). Some fundamental h∈H1
properties of this operator on L2 (X) space are studied by the author et al. in [2]. Definition 1.1. Let u : X → H be a weakly measurable function. We say that (u, ϕ, H) has absolute property, if for each f ∈ Lp (X), there exists hf ∈ H1 such that suph∈H1 X | u, h||f ◦ ϕ|dμ = X | u, hf ||f ◦ ϕ|dμ, and u, hf = ei(− arg f ◦ϕ+θf ) | u, hf |, for a constant θf . It is easy to see that some of obtained results in [2] for Tuϕ : L2 (X) → H are valid for substitution vector-valued integral operator Tuϕ on Lp (X) space for p ≥ 1 such as the following proposition. Proposition 1.2 ([2]). Assume that (u, ϕ, H) has the absolute property. Then sup | h∈H1
u, hf ◦ ϕdμ| = sup
X
h∈H1
| u, h||f ◦ ϕ|dμ. X
Throughout of this paper we assume that (u, ϕ, H) has the absolute property. The aim of this paper is to carry some of the results obtained for the composition operator and weighted composition operators in [1,6,7] to a substitution vector-valued integral operator on L1 (X) space. In this note, we will determine under certain conditions the essential norm of Tuϕ on L1 (X) space. Then, we present an example to illustrate our result.
JID:YJMAA
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.3 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
3
2. Main results In this section we give sufficient condition for the boundedness of Tuϕ on L1 (X). Also by using conditional expectation operator we characterize compactness Tuϕ on L1 (X). Next, we determine the essential norm Tuϕ on L1 (X) space. Definition 2.1. Let u : X → H be a weakly measurable function. We say that u is a semi-weakly bounded function if for some M > 0, u, h∞ ≤ M h,
for each h ∈ H.
Theorem 2.2. Let u : X → H be a weakly measurable function. If u is a semi-weakly bounded function and h0 ∈ L∞ (Σ), then Tuϕ : L1 (X) → H is bounded. Proof. Let f ∈ L1 (X). By Holder’s inequality and change of variable formula we have Tuϕ f = sup | h∈H1
u, h(f ◦ ϕ)dμ| X
u, h∞ |f ◦ ϕ|dμ
≤ sup h∈H1
X
≤ sup (M h) h∈H1
h0 |f |dμ X
≤ M h0 ∞ f 1 . This shows that Tuϕ is bounded. 2 Theorem 2.3. The substitution vector-valued integral operator Tuϕ is a compact operator on L1 (X) if and only if for any > 0 the set K := {x ∈ X : sup h0 E| u, h| ◦ ϕ−1 ≥ } h∈H1
consists of finitely many atoms. Proof. Suppose that Tuϕ is compact but for some > 0 the set K either contains a non-atomic subset or has infinitely many atoms. Then there exist δ > 0 and h1 ∈ H1 such that C := {x ∈ K : h0 E| u, h1 | ◦ ϕ−1 ≥ δ} has positive measure. We may also assume μ(C) < ∞. Since C ⊆ K , we can find a sequence of pairwise χAn disjoint measurable subsets An ∈ Σ of positive measure satisfying An ⊆ C for all n ∈ N. Put fn := μ(A , n) 1 1 note that for each n ∈ N, fn ∈ L (X). Also {fn }n is a bounded sequence in L (X). Then we observe that for any n, m ∈ N, Tuϕ fm − Tuϕ fn = sup
h∈H1
≥ X
| u, h||fm − fn | ◦ ϕdμ X
h0 E(| u, h1 |) ◦ ϕ−1 |fm − fn |dμ
≥ An ∪Am
h0 E(| u, h1 |) ◦ ϕ−1 |fm − fn |dμ
JID:YJMAA
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.4 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
4
= An
h0 E(| u, h1 |) ◦ ϕ−1
+
1 dμ μ(A(n))
h0 E(| u, h1 |) ◦ ϕ−1
1 dμ μ(A(m))
Am
≥ 2δ This implies that the sequence {Tuϕ fn }n does not contain a convergent subsequence, but this shows that Tuϕ is not compact and this is a contradiction. Conversely, by Theorem 2.10 in [2], it is easy to see that if for any > 0 the set K := {x ∈ X : suph∈H1 h0 E| u, h| ◦ ϕ−1 ≥ } consists of finitely many atoms then Tuϕ is a compact operator on L1 (X). 2 Let B be a Banach space and K be the set of all compact operators on B, the essential norm of T means the distance from T to K in the operator norm, namely T e = inf{T − S : S ∈ K}. Clearly, T is compact if and only if T e = 0. As is seen [5], the essential norm plays an interesting role in the compact problem of concrete operators. We are concerned with the case that T is a substitution vector-valued integral operator Tuϕ on L1 (X). Theorem 2.4. Let Tuϕ be a bounded operator on L1 (X). Then Tuϕ e = inf{r > 0 : Gr consists of finitely many atoms}
(2.1)
where Gr = {x ∈ X : suph∈H1 h0 E| u, h| ◦ ϕ−1 ≥ r}. Proof. Denote inf{r > 0 : Gr consists of finitely many atoms} by α. If α = 0, then by Theorem 2.3 is nothing to prove and so we assume that α > 0. Take > 0 arbitrary. The definition of α implies that E = Gα− 2 either contains a non-atomic subset or has infinitely many atoms. If E contains a non-atomic n) subset, then there are measurable sets En , n ∈ N, such that En+1 ⊆ En ⊆ E, μ(En+1 ) = μ(E 2n . Define χEn fn = μ(En ) . Then fn 1 = 1 for all n ∈ N. We claim that fn → 0 weakly. For this we show that X fn g → 0 for all g ∈ L∞ (Σ). Let F ⊆ E with 0 < μ(F ) < ∞ and g = χF . Hence
|
fn χF dμ| ≤
X
F ∩En
μ(F ∩ En ) 1 1 dμ = ≤ n → 0, μ(En) μ(En ) 2
as
n → ∞.
Since simple functions are dense in L∞ (Σ), so fn is proved to converge to 0 weakly. Now assume that E consists of infinitely many atoms. Let {En }∞ n=1 be disjoints atoms in E. Again put fn as above. It is easy to see that for F ⊆ E with 0 < μ(E) < ∞ we have μ(F ∩ En ) = 0, for sufficiently large n. Hence in both cases f g → 0. Now, take a compact operator T on L1 (Σ) such that Tuϕ − T < Tuϕ e + . The definition of X n E = Gα− 2 implies that there exists h ∈ H1 such as h1 such that h0 E| u, h1 | ◦ ϕ−1 > α − on E. Then for all n ∈ N and h1 ∈ H1 , we have Tuϕ e > Tuϕ − T − ≥ Tuϕ fn − T fn − ≥ Tuϕ fn − T fn − = sup h0 E| u, h| ◦ ϕ−1 |fn |dμ − T fn − h∈H1
X
JID:YJMAA
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.5 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
5
h0 E| u, h1 | ◦ ϕ−1 |fn |dμ − T fn −
≥ X
h0 E| u, h1 | ◦ ϕ−1
≥
1 dμ − T fn − μ(En )
En
≥ (α − ) − T fn − . Since a compact operator maps weakly convergence sequences into norm convergence ones, it follows that T fn → 0. Therefore Tuϕ e ≥ α − 2. Since was arbitrary, we obtain Tuϕ e ≥ α. For the opposite inequality, take arbitrary. Put A = Gα+ and v = χA u. The definition of α implies that A consists of finitely many atoms. So we can write A = {A1 , A2 , ..., Am }, where A1 , A2 , ..., Am are distinct. We show that Tvϕ is a finite rank operator. To show that Tvϕ (L1 (X)) is finite-dimensional, we only need to prove that the set V := {h ∈ Tvϕ (L1 (X)) : h ≤ 1} is compact in H. Let {Tvϕ fn }n be an arbitrary sequence in V . Then we have Tvϕ fn = sup
h∈H1
A
h0 E(| u, h|) ◦ ϕ−1 |fn |dμ
h0 E(| u, h|) ◦ ϕ−1 |fn |dμ
= sup h∈H1 ∪m i=1 Ai
=
m
sup h0 E(| u, h|) ◦ ϕ−1 (Ai )μ(Ai )|fn (Ai )|.
i=1 h∈H1
Put βi = suph∈H1 h0 E(| u, h|) ◦ ϕ−1 (Ai )μ(Ai ). Then: Tuϕ fn =
m
βi |fn (Ai )|.
(2.2)
i=1
Since Tvϕ fn ≤ 1 for all n ∈ N, then |fn (Ai )| ≤ β1i for each 1 ≤ i ≤ m and each n ∈ N. Hence, by Bolzano–Weierstrass Theorem, there exists a subsequence of natural numbers {nj }j∈N such that for each fixed 1 ≤ i ≤ m, the sequence {fnj (Ai )}j∈N converges. Assume that limj→∞ fnj (Ai ) = ξi and let m m ϕ ϕ f := i=1 ξi χAi . Now, from (2.2) we have Tv f = i=1 βi |ξi | ≤ 1. Hence, Tv f ∈ V and so we get that Tvϕ fnj
−
Tvϕ f
=
m
βi |fnj (Ai ) − f (Ai )| =
i=1
m
βi |fnj (Ai ) − ξi | → 0
as j → ∞.
i=1
It follows that V is compact in H. Therefore we deduce that Tvϕ (L1 (X)) is finite-dimensional. Notice that Tvϕ is a compact operator, then we get that ϕ ϕ Tuϕ − Tvϕ = Tu−v = sup Tu−v f f 1 ≤1
= sup
sup
f 1 ≤1 h∈H1
h0 E| (1 − χK )u, h| ◦ ϕ−1 |f |dμ
X
= sup
sup
f 1 ≤1 h∈H1 X\K
h0 E| u, h| ◦ ϕ−1 |f |dμ
JID:YJMAA
AID:20916 /FLA
6
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.6 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
≤ (α + ) sup
f 1 ≤1 X\K
|f |dμ
≤α+ Since was arbitrary, we obtain Tuϕ e ≤ α.
2
Example 2.5. Let X = [−1, 0] ∪ {1, 2, 3, 4, 5, 6}, Σ be the Lebesgue subsets of X. Let μ be the Lebesgue measure on [−1, 0] and μ({n}) = 21n , if n = 1, 2, 3, 4, 5, 6. Define ϕ : X → X as: ϕ = 2χ{1} + 3χ{2,3} + 4χ{4,5,} + 5χ{6} + xχ[−1,0] . Put H = R2 . Also let u : X → R be defined by u(x) = 1. Then it is easy to see that there exists M > 0 such that for each h ∈ R2 , | h, u(x)∞ = h∞ ≤ M h. Thus by using Theorem 2.2, we deduce that Tuϕ : L1 (X) → R2 is a bounded operator. It is easy to check that ϕ is a non-singular measurable transformation and 3 1 h0 = 2χ{2} + 3χ{3} + χ{4} + χ{5} + χ[−1,0] . 2 2 On the other hand we have suph∈R1 h0 E| u, h| ◦ ϕ−1 = h0 , where R21 is the closed unit ball of R. It is clear that (u, ϕ, R2 ) has the absolute property. Now by using Theorem 2.4, we get that Tuϕ e = 32 . References [1] J.T. Chan, A note on compact weighted composition operators on Lp (μ), Acta Sci. Math. 56 (1992) 165–168. [2] H. Emamalipour, M.R. Jabbarzadeh, Z. Moayyerizadeh, A substitution vector-valued integral operator, J. Math. Anal. Appl. 431 (2015) 812–821. [3] J. Herron, Weighted conditional expectation operators, Oper. Matrices 5 (2011) 107–118. [4] M.M. Rao, Conditional Measure and Applications, Marcel Dekker, New York, 1993. [5] J.H. Shapiro, The essential norm of a composition operator, Ann. of Math. 125 (1987) 375–404. [6] R.K. Singh, J.S. Manhas, Composition Operators on Function Spaces, North-Holl., Amsterdam, 1993. [7] H. Takagi, T. Miura, S.E. Takahasi, Essential norms and stability constants of weighted composition operators on C(X), Bull. Korean Math. Soc. 40 (2003) 583–591. [8] A.C. Zaanen, Integration, 2nd ed., North-Holl., Amsterdam, 1967.
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.1 (1-6)
J. Math. Anal. Appl. ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Essential norm of substitution vector-valued integral operator on L1 (Σ) space Z. Moayyerizadeh Department of Mathematical Sciences, University of Lorestan, 5166617766, Khorramabad, Iran
a r t i c l e
i n f o
Article history: Received 17 October 2016 Available online xxxx Submitted by J.A. Ball
a b s t r a c t In this paper, we determine essential norm of a substitution vector-valued integral operator Tuϕ on L1 (Σ) space using conditional expectation operators. © 2016 Elsevier Inc. All rights reserved.
Keywords: Integral operator Conditional expectation Essential norm
1. Introduction and preliminaries Let (X, Σ, μ) be a σ-finite measure space and ϕ : X → X be a non-singular measurable transformation; i.e. μ ◦ ϕ−1 μ. Here the non-singularity of ϕ guarantees that the operator f → f ◦ ϕ is well defined as a mapping on L0 (Σ) where L0 (Σ) denotes the linear space of all equivalence classes of Σ-measurable functions on X. Let h0 = dμ ◦ ϕ−1 /dμ be the Radon–Nikodym derivative. We also assume that h0 is almost everywhere finite-valued, or equivalently ϕ−1 (Σ) ⊆ Σ is a sub-σ-finite algebra see [6]. All comparisons between two functions or two sets are to be interpreted as holding up to a μ-null set. For a sub-σ-finite algebra A ⊆ Σ, the conditional expectation operator associated with A is the mapping f → E A f , defined for all non-negative f as well as for all f ∈ Lp (Σ), 1 ≤ p ≤ ∞, where E A f , by Radon–Nikodym Theorem, is the unique A-measurable function satisfying
f dμ = A
E A f dμ,
∀A ∈ A.
A
We recall that E A : L2 (Σ) → L2 (A) is an orthogonal projection. For more details on the properties of E A −1 see [3,4]. Throughout this paper, we assume that A = ϕ−1 (Σ) and E ϕ (Σ) = E. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jmaa.2016.11.063 0022-247X/© 2016 Elsevier Inc. All rights reserved.
JID:YJMAA 2
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.2 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
Recall that an atom of the measure μ is an element A ∈ Σ with μ(A) > 0, such that for each B ∈ Σ, if B ⊂ A then either μ(B) = 0 or μ(B) = μ(A). A measure with no atoms is called non-atomic. We can easily check the following well-known facts (see [8]): (a) Every σ-finite measure space (X, Σ, μ) can be partitioned uniquely as X = (∪n∈N An ) ∪ B,
(1.1)
where {An }n∈N ⊆ Σ is a countable collection of pairwise disjoint atoms and B, being disjoint from each An , is non-atomic. (b) Let E be a non-atomic set with μ(E) > 0. Then there exists a sequence of positive disjoint Σ-measurable subsets of E, {En }n∈N such that μ(En ) > 0 for each n ∈ N and limn→∞ μ(En ) = 0. For a given complex Hilbert space H, let u : X → H be a mapping. We say that u is weakly measurable if for each h ∈ H the mapping x → u(x), h of X to C is measurable. We will denote this map by u, h. Let Lp (X) be the class of all measurable mappings f : X → C such that f pp = X |f (x)|p dμ < ∞ for p ≥ 1. Let ϕ : X → X be a non-singular measurable transformation and let u : X → H be a weakly measurable function. Then the pair (u, ϕ) induces a substitution vector-valued integral operator Tuϕ : Lp (X) → H defined by
Tuϕ f, h =
u, hf ◦ ϕdμ,
h ∈ H, f ∈ Lp (X).
X
It is easy to see that Tuϕ is well defined and linear. Moreover for each f ∈ Lp (X), sup | Tuϕ f, h| ≤ sup Tuϕ f h = Tuϕ f
h∈H1
h∈H1
T ϕf = | Tuϕ f, uϕ | ≤ sup | Tuϕ f, h|, Tu f h∈H1 where H1 is the closed unit ball of H. Hence Tuϕ f = sup | Tuϕ f, h|, for each f ∈ Lp (X). Some fundamental h∈H1
properties of this operator on L2 (X) space are studied by the author et al. in [2]. Definition 1.1. Let u : X → H be a weakly measurable function. We say that (u, ϕ, H) has absolute property, if for each f ∈ Lp (X), there exists hf ∈ H1 such that suph∈H1 X | u, h||f ◦ ϕ|dμ = X | u, hf ||f ◦ ϕ|dμ, and u, hf = ei(− arg f ◦ϕ+θf ) | u, hf |, for a constant θf . It is easy to see that some of obtained results in [2] for Tuϕ : L2 (X) → H are valid for substitution vector-valued integral operator Tuϕ on Lp (X) space for p ≥ 1 such as the following proposition. Proposition 1.2 ([2]). Assume that (u, ϕ, H) has the absolute property. Then sup | h∈H1
u, hf ◦ ϕdμ| = sup
X
h∈H1
| u, h||f ◦ ϕ|dμ. X
Throughout of this paper we assume that (u, ϕ, H) has the absolute property. The aim of this paper is to carry some of the results obtained for the composition operator and weighted composition operators in [1,6,7] to a substitution vector-valued integral operator on L1 (X) space. In this note, we will determine under certain conditions the essential norm of Tuϕ on L1 (X) space. Then, we present an example to illustrate our result.
JID:YJMAA
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.3 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
3
2. Main results In this section we give sufficient condition for the boundedness of Tuϕ on L1 (X). Also by using conditional expectation operator we characterize compactness Tuϕ on L1 (X). Next, we determine the essential norm Tuϕ on L1 (X) space. Definition 2.1. Let u : X → H be a weakly measurable function. We say that u is a semi-weakly bounded function if for some M > 0, u, h∞ ≤ M h,
for each h ∈ H.
Theorem 2.2. Let u : X → H be a weakly measurable function. If u is a semi-weakly bounded function and h0 ∈ L∞ (Σ), then Tuϕ : L1 (X) → H is bounded. Proof. Let f ∈ L1 (X). By Holder’s inequality and change of variable formula we have Tuϕ f = sup | h∈H1
u, h(f ◦ ϕ)dμ| X
u, h∞ |f ◦ ϕ|dμ
≤ sup h∈H1
X
≤ sup (M h) h∈H1
h0 |f |dμ X
≤ M h0 ∞ f 1 . This shows that Tuϕ is bounded. 2 Theorem 2.3. The substitution vector-valued integral operator Tuϕ is a compact operator on L1 (X) if and only if for any > 0 the set K := {x ∈ X : sup h0 E| u, h| ◦ ϕ−1 ≥ } h∈H1
consists of finitely many atoms. Proof. Suppose that Tuϕ is compact but for some > 0 the set K either contains a non-atomic subset or has infinitely many atoms. Then there exist δ > 0 and h1 ∈ H1 such that C := {x ∈ K : h0 E| u, h1 | ◦ ϕ−1 ≥ δ} has positive measure. We may also assume μ(C) < ∞. Since C ⊆ K , we can find a sequence of pairwise χAn disjoint measurable subsets An ∈ Σ of positive measure satisfying An ⊆ C for all n ∈ N. Put fn := μ(A , n) 1 1 note that for each n ∈ N, fn ∈ L (X). Also {fn }n is a bounded sequence in L (X). Then we observe that for any n, m ∈ N, Tuϕ fm − Tuϕ fn = sup
h∈H1
≥ X
| u, h||fm − fn | ◦ ϕdμ X
h0 E(| u, h1 |) ◦ ϕ−1 |fm − fn |dμ
≥ An ∪Am
h0 E(| u, h1 |) ◦ ϕ−1 |fm − fn |dμ
JID:YJMAA
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.4 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
4
= An
h0 E(| u, h1 |) ◦ ϕ−1
+
1 dμ μ(A(n))
h0 E(| u, h1 |) ◦ ϕ−1
1 dμ μ(A(m))
Am
≥ 2δ This implies that the sequence {Tuϕ fn }n does not contain a convergent subsequence, but this shows that Tuϕ is not compact and this is a contradiction. Conversely, by Theorem 2.10 in [2], it is easy to see that if for any > 0 the set K := {x ∈ X : suph∈H1 h0 E| u, h| ◦ ϕ−1 ≥ } consists of finitely many atoms then Tuϕ is a compact operator on L1 (X). 2 Let B be a Banach space and K be the set of all compact operators on B, the essential norm of T means the distance from T to K in the operator norm, namely T e = inf{T − S : S ∈ K}. Clearly, T is compact if and only if T e = 0. As is seen [5], the essential norm plays an interesting role in the compact problem of concrete operators. We are concerned with the case that T is a substitution vector-valued integral operator Tuϕ on L1 (X). Theorem 2.4. Let Tuϕ be a bounded operator on L1 (X). Then Tuϕ e = inf{r > 0 : Gr consists of finitely many atoms}
(2.1)
where Gr = {x ∈ X : suph∈H1 h0 E| u, h| ◦ ϕ−1 ≥ r}. Proof. Denote inf{r > 0 : Gr consists of finitely many atoms} by α. If α = 0, then by Theorem 2.3 is nothing to prove and so we assume that α > 0. Take > 0 arbitrary. The definition of α implies that E = Gα− 2 either contains a non-atomic subset or has infinitely many atoms. If E contains a non-atomic n) subset, then there are measurable sets En , n ∈ N, such that En+1 ⊆ En ⊆ E, μ(En+1 ) = μ(E 2n . Define χEn fn = μ(En ) . Then fn 1 = 1 for all n ∈ N. We claim that fn → 0 weakly. For this we show that X fn g → 0 for all g ∈ L∞ (Σ). Let F ⊆ E with 0 < μ(F ) < ∞ and g = χF . Hence
|
fn χF dμ| ≤
X
F ∩En
μ(F ∩ En ) 1 1 dμ = ≤ n → 0, μ(En) μ(En ) 2
as
n → ∞.
Since simple functions are dense in L∞ (Σ), so fn is proved to converge to 0 weakly. Now assume that E consists of infinitely many atoms. Let {En }∞ n=1 be disjoints atoms in E. Again put fn as above. It is easy to see that for F ⊆ E with 0 < μ(E) < ∞ we have μ(F ∩ En ) = 0, for sufficiently large n. Hence in both cases f g → 0. Now, take a compact operator T on L1 (Σ) such that Tuϕ − T < Tuϕ e + . The definition of X n E = Gα− 2 implies that there exists h ∈ H1 such as h1 such that h0 E| u, h1 | ◦ ϕ−1 > α − on E. Then for all n ∈ N and h1 ∈ H1 , we have Tuϕ e > Tuϕ − T − ≥ Tuϕ fn − T fn − ≥ Tuϕ fn − T fn − = sup h0 E| u, h| ◦ ϕ−1 |fn |dμ − T fn − h∈H1
X
JID:YJMAA
AID:20916 /FLA
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.5 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
5
h0 E| u, h1 | ◦ ϕ−1 |fn |dμ − T fn −
≥ X
h0 E| u, h1 | ◦ ϕ−1
≥
1 dμ − T fn − μ(En )
En
≥ (α − ) − T fn − . Since a compact operator maps weakly convergence sequences into norm convergence ones, it follows that T fn → 0. Therefore Tuϕ e ≥ α − 2. Since was arbitrary, we obtain Tuϕ e ≥ α. For the opposite inequality, take arbitrary. Put A = Gα+ and v = χA u. The definition of α implies that A consists of finitely many atoms. So we can write A = {A1 , A2 , ..., Am }, where A1 , A2 , ..., Am are distinct. We show that Tvϕ is a finite rank operator. To show that Tvϕ (L1 (X)) is finite-dimensional, we only need to prove that the set V := {h ∈ Tvϕ (L1 (X)) : h ≤ 1} is compact in H. Let {Tvϕ fn }n be an arbitrary sequence in V . Then we have Tvϕ fn = sup
h∈H1
A
h0 E(| u, h|) ◦ ϕ−1 |fn |dμ
h0 E(| u, h|) ◦ ϕ−1 |fn |dμ
= sup h∈H1 ∪m i=1 Ai
=
m
sup h0 E(| u, h|) ◦ ϕ−1 (Ai )μ(Ai )|fn (Ai )|.
i=1 h∈H1
Put βi = suph∈H1 h0 E(| u, h|) ◦ ϕ−1 (Ai )μ(Ai ). Then: Tuϕ fn =
m
βi |fn (Ai )|.
(2.2)
i=1
Since Tvϕ fn ≤ 1 for all n ∈ N, then |fn (Ai )| ≤ β1i for each 1 ≤ i ≤ m and each n ∈ N. Hence, by Bolzano–Weierstrass Theorem, there exists a subsequence of natural numbers {nj }j∈N such that for each fixed 1 ≤ i ≤ m, the sequence {fnj (Ai )}j∈N converges. Assume that limj→∞ fnj (Ai ) = ξi and let m m ϕ ϕ f := i=1 ξi χAi . Now, from (2.2) we have Tv f = i=1 βi |ξi | ≤ 1. Hence, Tv f ∈ V and so we get that Tvϕ fnj
−
Tvϕ f
=
m
βi |fnj (Ai ) − f (Ai )| =
i=1
m
βi |fnj (Ai ) − ξi | → 0
as j → ∞.
i=1
It follows that V is compact in H. Therefore we deduce that Tvϕ (L1 (X)) is finite-dimensional. Notice that Tvϕ is a compact operator, then we get that ϕ ϕ Tuϕ − Tvϕ = Tu−v = sup Tu−v f f 1 ≤1
= sup
sup
f 1 ≤1 h∈H1
h0 E| (1 − χK )u, h| ◦ ϕ−1 |f |dμ
X
= sup
sup
f 1 ≤1 h∈H1 X\K
h0 E| u, h| ◦ ϕ−1 |f |dμ
JID:YJMAA
AID:20916 /FLA
6
Doctopic: Functional Analysis
[m3L; v1.191; Prn:1/12/2016; 15:52] P.6 (1-6)
Z. Moayyerizadeh / J. Math. Anal. Appl. ••• (••••) •••–•••
≤ (α + ) sup
f 1 ≤1 X\K
|f |dμ
≤α+ Since was arbitrary, we obtain Tuϕ e ≤ α.
2
Example 2.5. Let X = [−1, 0] ∪ {1, 2, 3, 4, 5, 6}, Σ be the Lebesgue subsets of X. Let μ be the Lebesgue measure on [−1, 0] and μ({n}) = 21n , if n = 1, 2, 3, 4, 5, 6. Define ϕ : X → X as: ϕ = 2χ{1} + 3χ{2,3} + 4χ{4,5,} + 5χ{6} + xχ[−1,0] . Put H = R2 . Also let u : X → R be defined by u(x) = 1. Then it is easy to see that there exists M > 0 such that for each h ∈ R2 , | h, u(x)∞ = h∞ ≤ M h. Thus by using Theorem 2.2, we deduce that Tuϕ : L1 (X) → R2 is a bounded operator. It is easy to check that ϕ is a non-singular measurable transformation and 3 1 h0 = 2χ{2} + 3χ{3} + χ{4} + χ{5} + χ[−1,0] . 2 2 On the other hand we have suph∈R1 h0 E| u, h| ◦ ϕ−1 = h0 , where R21 is the closed unit ball of R. It is clear that (u, ϕ, R2 ) has the absolute property. Now by using Theorem 2.4, we get that Tuϕ e = 32 . References [1] J.T. Chan, A note on compact weighted composition operators on Lp (μ), Acta Sci. Math. 56 (1992) 165–168. [2] H. Emamalipour, M.R. Jabbarzadeh, Z. Moayyerizadeh, A substitution vector-valued integral operator, J. Math. Anal. Appl. 431 (2015) 812–821. [3] J. Herron, Weighted conditional expectation operators, Oper. Matrices 5 (2011) 107–118. [4] M.M. Rao, Conditional Measure and Applications, Marcel Dekker, New York, 1993. [5] J.H. Shapiro, The essential norm of a composition operator, Ann. of Math. 125 (1987) 375–404. [6] R.K. Singh, J.S. Manhas, Composition Operators on Function Spaces, North-Holl., Amsterdam, 1993. [7] H. Takagi, T. Miura, S.E. Takahasi, Essential norms and stability constants of weighted composition operators on C(X), Bull. Korean Math. Soc. 40 (2003) 583–591. [8] A.C. Zaanen, Integration, 2nd ed., North-Holl., Amsterdam, 1967.