1. Introduction

Let be a σ-finite measure space. A measurable transformation T is said to be non-singular if whenever for every .

If T is non-singular, then we say that is absolutely continuous with respect to μ. Hence, by Radon-Nikodym theorem there exists a unique non-negative essentially bounded function f_T such that for .

Let f be any complex-valued measurable function. For s ≥ 0, the distribution function μ_f of f is defined as

The non-increasing rearrangement of f is defined as

The maximal (average) operator is given by

One can refer to ^[4] for the properties of these functions.

Definition 1. A positive and Lebesgue measurable function b is said to be slowly varying (s.v.) on (0, ∞) if, for each , is equivalent to a non-decreasing function and is equivalent to a non-increasing function on (0, ∞).

Given a s.v. function b on (0, ∞), we denote by the positive function defined by

For various properties of slowly varying function we can refer to ^{[4, 10]}.

For 1 < p < ∞, 1 ≤ q < ∞ and for a measurable function f on Ω, define

The Lorentz-Karamata space introduced in ^[4] is the set of all measurable functions f on Ω such that

Let be a strongly measurable function on a Banach space X. Define a function as

for all . Then the Lorentz-Karamata-Bochner space is a rearrangement invariant-Bochner space for where the norm is given as

The Lorentz-Karamata space is a Banach space and we still have the density of simple functions in it and its dual is

where has the Radon-Nikodym property. For every , we can find a bounded linear functional defined as

for all . For each , there exists a unique measurable function E(g) such that

for each measurable function f for which the left integral exists. E(g) is called the conditional expectation ^[11] of g with respect to . The operator P_T defined as

is called Frobenius Perron and is the Radon-Nikodym derivative of with respect to μ. It satisfies the property

Let T be a non-singular measurable transformation on Ω then the composition operator C_T from into the space of strongly measurable functions is given by

for all . An operator T is called Fredholm if R(T) is closed, dim N(T) < ∞ and dim N(T^*) < ∞ where R(T), N(T) and N(T^*) denote the range, kernel and cokernel of T. B(X) denotes the space of all bounded linear operators on X. Multiplication operators on this space are already studied in ^[5] and on different spaces in ^{[1, 2, 3, 6, 7, 8, 9, 12]}. In this paper, we discuss about the composition operators on the Lorentz-Karamata-Bochner space and study its various properties like boundedness, closedness and compactness.

2. Composition Operators

Theorem 2.1. A non-singular transformation induces the composition operator C_T if and only if for some k > 0,

for .

Proof. Suppose that the composition operator is bounded on . Then there exists K > 0 such that

Let x₀ be the fixed element of X with . Define the characteristic function for each measurable subset A of by,

Then we find that

and

This gives

and

Thus, we get

Conversely, suppose the given condition holds. Then

and

Thus

Also

Therefore

Thus, C_T is a bounded operator on .

Corollary 2.2. A measurable transformation T induces the composition operator C_T on if and only if is absolutely continuous with respect to μ and belongs to .

Theorem 2.3. If C_T is the composition operator on . Then C_T is measure preserving if and only if C_T is an isometry.

Proof. Suppose that T is measure preserving them

for all .

The distribution function of C_T becomes

and

Also

This gives

Converse of the theorem is obvious.

Example 1. Let with Lebesgue measure and X be any Banach space. Define

Then T is a non-singular transformation on Ω which is not measure preserving. Hence C_T is not an isometry on .

Theorem 2.4. If C_T is a composition operator on . Then C_T has closed range if and only if there exists such that for almost all , the support of .

Proof. Suppose is bounded away from zero then there exists a positive real number , such that

for almost all .

where

and

and

This gives

Thus, C_T has closed range.

Conversely, C_T has closed range then there exists such that

for all .

Choose a natural number n such that . Let if possible, where . Then . Then

and

This gives

which is a contradiction. Hence is bounded away from zero.

Theorem 2.5. If is a composition operator on . Then C_T has dense range in .

Proof. We will consider two cases:

Case 1. When . Then so we can obtain B ∈ such that

Thus C_T belong to the range of C_T and hence all simple functions of belong to where denotes the range of C_T. Hence, range of C_T is dense in .

Case 2. When . Let then there is a sequence of functions in converging to g in . Since

Clearly, each g_n is measurable and hence g is also measurable. Now suppose . By adjusting f on a set of measure zero, suppose for some . Since is a σ-finite space

where for each n and is an increasing sequence of measurable sets.

This gives

which converges to zero. Thus is dense in

Theorem 2.6. T inducing the composition operator C_T on is a surjection if and only if is bounded away from zero on its support and

Proof. Suppose C_T is a surjection. Then from the last theorem C_T has closed range if and only if is bounded away from zero on its support. Let be of finite measure. Since C_T is a surjection there exist such that . Let

Then

Hence,. Thus Thus Converse is obvious.

Corollary 2.7. A composition operator C_T on B(Ω, X), has dense range if and only if .

Theorem 2.8. If T induces a composition operator on (Ω, X), then , the adjoint of C_T is P_T.

Proof. Let be such that . Then for

By identifying with the functional , we get

Theorem 2.9. If T induces a composition operator on B(Ω, X), then is either zero dimensional or infinite dimensional.

Proof. Suppose and . Let

then . Let be a sequence of disjoint measurable subsets of A such that

where . For each , let . For each n,

Therefore is a linearly independent subset of . Hence, if is not zero dimensional, it is infinite dimensional.

Theorem 2.10. If T induces a composition operator on B(Ω, X). Then C_T is invertible if and only if C_T is Fredholm.

Proof. If C_T is invertible then C_T is Fredholm. Conversely, let C_T be Fredholm then and are both finite dimensional and are of zero dimension. Therefore C_T is injective and has dense range. Since is closed, therefore C_T is surjective. Thus C_T is invertible.

Definition 2. For a strongly measurable function B(X), the set

is called the essential range of f.

Theorem 2.11. ^[11] If C_T is a composition operator on , then the following are equivalent:

(i) C_T is injective.

(ii) f and have the same essential ranges for every .

(iii) μ is absolutely continuous with respect to .

(iv) is different from zero almost everywhere.

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