Multiplication and Composition operators on wp(f)

In this paper we characterize the boundedness, closed range, invertibility of the multiplication operators acting on sequence spaces wp(f) defined by a modulus function. We also make an efforts to study some properties of composition operators on these spaces. Subjclass [2000] : Primary 47B20, Secondary 47B38.

For any sequence x, write G. G. Lorentz [6] A straight forward calculation then show that If f is a modulus function and homogeneous of degree 1, then we define a sequence space as The space w p (f ) with the norm Let v : N → N and u : N → C be two mappings.Then the bounded linear transformations ) and (M u h)(x) = u(x)h(x) are called composition and multiplication operators respectively.By B(w p (f )), we denote the set of all bounded linear operators from w p (f ) into itself and [z(u)] denote, the set {n ∈ N : u(n) = 0}.For more details about the study of multiplication and composition operators see ( [1], [4], [5], [8], [10], [11]).
In this paper we study multiplication and composition operators acting on sequence spaces w p (f ) defined by a modulus function.

Multiplication operators acting on sequence spaces defined by a modulus function
In this section we characterize multiplication operators acting on w p (f ).
Theorem 2.1.Let M u : w p (f ) → w p (f ) be a linear transformation.Then M u is a bounded operator if and only if there exists M > 0 such that Proof.Suppose that the condition of the theorem is true.For x ∈ w p (f ), we have This proves the continuity of M u at the origin and hence everywhere in view of linearity of M u .
Conversely, if the condition of the theorem were false, then for every integer k > 0 there exists n k ∈ N and This proves that M u is not bounded.Hence the condition must be true.
We now prove that V induces a multiplication operator.If V does not induce a bounded operator, then for every k ∈ N , there exists ´. Let Then which contradicts the continuity of A. Hence A must be a bounded operator and A = M V .
Theorem 2.3.Let M u ∈ B(w p (f )).Then M u is invertible if and only if there exists > 0 such that Proof.We first assume that there exists > 0 such that This proves that M V is a bounded operator, where Conversely, suppose that M u is invertible with M V as its inverse.Clearly V = 1 u .Hence by continuity of M V , there exists M > 0 such that Or equivalently ´. Hence the condition must be true.

Taking
Proof.If [Z(u)] is a finite set, then kerM u is finite dimensional.From the condition (ii), M u has closed range.
Moreover dim(w p (f )/ranM u ) is finite.This proves that M u is Fredholm.
The converse of the theorem is obvious.
Corollary 2.5.Let M u ∈ B(w p (f )).Then M u has closed range if and only if there exists δ > 0 such that 0 and y ∈ R + .
Proof.Assume that the condition of the theorem is true.Let h ∈ ranM u .
Then there exists a sequence is a Cauchy sequence.Therefore for every > 0 there exists n 0 ∈ N such that Then from (1) it follows that { hn } is a Cauchy sequence in w p (f ).But w p (f ) is complete.
Therefore there exists h ∈ w p (f ) such that hn → h.Hence by continuity of M u , we get M u h n = M u hn → M u h.Hence h = M u h so that h ∈ ranM u .Thus M u has closed range.
Conversely, if the condition of the theorem were false, then for every positive integer k there exists n k ∈ N and Then This proves that M u is not bounded away from zero so that M u does not have closed range.

Composition operators acting on sequence spaces defined by a modulus function
In this section we study some properties of composition operators on w p (f ).
Theorem 3.1.Let T v : w p (f ) → w p (f ) be a linear transformation.Then T v is a bounded operator if there exists M > 0 such that ´.
Proof.Suppose that the condition of the theorem is true.
The continuity of T v at origin follows from the inequality (2).Since T v is linear, so it is continuous everywhere.
Proof.We assume that the condition (3) is true.We have to show that T v has closed range.Let x ∈ ranT v and let {x i } be a sequence in w p (f ) such that T v x n → x.Then for every > 0 there exists positive integer n 0 such that from (4) it follows that {x i } is a Cauchy sequence in w p (f ).In view of completeness of w p (f ), there exists y ∈ w p (f ) such that x i → y.From the continuity of T v , T v x i → T v y.Hence x = T v y so that x ∈ ranT v .Hence ran T v is closed.
Theorem 3.3.Let T v ∈ B(w p (f )).Then T v is an isometry if ´.
Proof.If the condition of the theorem is satisfied, then for every x ∈ w p (f ), we have ´.
Proof.The proof is trivial.
Hence T v is an isometry.Theorem 3.4.Let T v ∈ B(w p (f )).If T v is an isometry, then sup n X k∈v −1 (m) k p−1 f ³ |t nk (x)| p ´= sup n m p−1 f ³ |t mn x| p