the measurable set h−1(λ), then by considering two cases, we find so λ is an eigenvalue of Th. Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of Th. To show this, we must show that has dense range. Given , again we consider the sequence of sets . Let gn be the characteristic function of . Define Direct calculation shows that fn ∈ Lp(μ), with . Then by the dominated convergence theorem, in the Lp(μ) norm. Therefore, multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum. Shifts In the special case when S is the set of natural numbers and μ is the counting measure, the corresponding Lp(μ) is denoted by lp. This space consists of complex valued sequences {xn} such that For 1 < p < ∞, l p is reflexive. Define the left shift T : l p → l p by T is a partial isometry with operator norm 1. So σ(T) lies in the closed unit disk of the complex plane. T* is the right shift (or unilateral shift), which is an isometry on l q, where 1/p + 1/q = 1: For λ ∈ C with |λ| < 1, and T x = λ x. Consequently, the point spectrum of T contains the open unit disk. Now, T* has no eigenvalues, i.e. σp(T*) is empty. Thus, invoking reflexivity and the theorem given above (that σp(T) ⊂ σr(T*) ∪ σp(T*)), we can deduce that the open unit disk lies in the residual spectrum of T*. The spectrum of a bounded operator is closed, which implies the unit circle, { |λ| = 1 } ⊂ C, is in σ(T). Again by reflexivity of l p and the theorem given above (this time, that ), we have that σr(T) is also empty. Therefore, for a complex number λ with unit norm, one must have λ ∈ σp(T) or λ ∈ σc(T). Now if |λ| = 1 and then which cannot be in l p, a contradiction. This means the unit circle must lie in the continuous spectrum of T. So for the left shift T, σp(T) is the open unit disk and σc(T) is the unit circle, whereas for the right shift T*, σr(T*) is the open unit disk and σc(T*) is