, of T. is injective but does not have dense range. That is, there is some element x in X and a neighborhood N of x such that is never in N. In this case, the map may be bounded or unbounded, but in any case does not admit a unique extension to a bounded linear map on all of X. Then λ is said to be in the residual spectrum of T, . So σ(T) is the disjoint union of these three sets, In addition, when does not have dense range, whether is injective or not, then λ is said to be in the compression spectrum of T, σcp(T). The compression spectrum consists of the whole residual spectrum and part of point spectrum. For unbounded operators The spectrum of an unbounded operator can be divided into three parts in the same way as in the bounded case, but because the operator is not defined everywhere, the definitions of domain, inverse, etc. are more involved. Examples Multiplication operator Given a σ-finite measure space (S, Σ, μ), consider the Banach space Lp(μ). A function h: S → C is called essentially bounded if h is bounded μ-almost everywhere. An essentially bounded h induces a bounded multiplication operator Th on Lp(μ): The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts. If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies . So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets . Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly This shows is not bounded below, therefore not invertible. If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th as follows. Let f be the characteristic function of