In this note we determine the lower and upper estimates for the essential norm of a composition operator on the Orlicz spaces under certain conditions. | Turk J Math 34 (2010) , 537 – 542. ¨ ITAK ˙ c TUB doi: The essential norm of a composition operator on Orlicz spaces M. R. Jabbarzadeh Abstract In this note we determine the lower and upper estimates for the essential norm of a composition operator on the Orlicz spaces under certain conditions. Key Words: Orlicz spaces, composition operator, compact operators, essential norm. 1. Introduction and preliminaries Let ϕ : [0, ∞) → [0, ∞) be a non-decreasing and convex function such that limx→∞ ϕ(x) = ∞ and ϕ(x) = 0 if and only if x = 0 . Such a function is known as an Orlicz function. Let (X, Σ, μ) be a complete sigma finite measure space. We identify any two functions that are equal μ-almost everywhere on X . Let Lϕ (μ) be the set of all measurable functions such that X ϕ(α|f|)dμ 0 . The space Lϕ (μ) is called an Orlicz space and is a Banach space with the Luxemburg norm defined by f ϕ = inf{δ > 0 : ϕ( X |f| )dμ ≤ 1}. δ If ϕ(x) = xp , 1 ≤ p 0 and M ≥ 0 such that ϕ(2t) ≤ kϕ(t) for all t ≥ M . It is known fact that if ϕ satisfies the Δ2 -condition, then simple functions are dense in Lϕ (μ). Let τ : X → X be a non-singular measurable transformation, that is, μ ◦ τ −1 (A) := μ(τ −1 (A)) = 0 for each A ∈ Σ whenever μ(A) = 0 . This condition means that the measure μ ◦ τ −1 is absolutely continuous with respect to μ. Let h := dμ ◦ τ −1 /dμ be the Radon-Nikodym derivative. In addition, we assume that h is almost everywhere finite valued, or equivalently that (X, τ −1 (Σ), μ) is σ -finite. An atom of the measure μ is an element A ∈ Σ with μ(A) > 0 such that for each F ∈ Σ, if F ⊂ A then either μ(F ) = 0 or μ(F ) = μ(A). Let A be an atom. Since μ is σ -finite, it follows that μ(A) 0 : ϕ(s) > t} is the rightcontinuous inverse of ϕ. Let ϕ1 , ϕ2 be Orlicz functions. Then ϕ1 is said to be essentially stronger than ϕ2 −1 (it is usually denoted ϕ1 ϕ2 ) if ϕ−1 1 (t)/ϕ2 (t) → 0 as t → ∞ . It is well-known that if f ϕ ≤ 1 then Iϕ (f) := ϕ(|f|)dμ ≤ f ϕ .
