Continuous Spectrum - Operators On Banach Space

Operators On Banach Space

Let X be a Banach space, L(X) the family of bounded operators on X, and T ∈ L(X).

A complex number λ is in the spectrum of T, denoted σ(T), if T − λ does not have bounded inverse. If T − λ is invertible (i.e., if it is one-to-one and onto), then its inverse is bounded; this follows directly from the open mapping theorem. So, λ is in the spectrum of T if and only if T − λ is either not one-to-one or not onto. One can easily check that if T − λ is one-to-one, bounded below (i.e. does not send far apart elements of X too close together), and has dense range, then in fact T − λ must be onto, so λ will not be in σ(T). Therefore, if λ is in σ(T), one of the following must be true:

1. T − λ is not injective.
2. T − λ is injective, and has dense range. But T − λ is not bounded below, and we fail to get that the range is all of X. Equivalently, the densely-defined linear map (T − λ) x → x is not bounded, therefore can not be extended to all of X.
3. T − λ is injective and does not have dense range. In this case, the map (T − λ) x → x may be bounded or unbounded, but in any case does not admit a unique extension to a bounded linear map on all of X.

Correspondingly, for T ∈ L(X), its spectrum σ(T) can be classified as follows:

1. The point spectrum of T consists of eigenvalues of T and will be denoted by σ_p(T). λ ∈ σ(T) is in the point spectrum if and only if T − λ is not injective.
2. If λ ∈ σ(T) is not an eigenvalue and the range of T − λ, Ran(T − λ), is dense in X, λ is said to be in the continuous spectrum, σ_c(T), of T.
3. If T − λ is injective and Ran(T − λ) is not dense, λ is in the residual spectrum of T, σ_r(T).

So σ(T) is the disjoint union

If X* is the dual space of X, and T* : X* → X* is the adjoint operator of T, then σ(T) = σ(T*).

Theorem For a bounded operator T, σ_r(T) ⊂ σ_p(T*) ⊂ σ_r(T) ∪ σ_p(T).

Proof The notation <·, φ> will denote an element of X*, i.e. x → <x, φ> is the action of a bounded linear functional φ. Let λ ∈ σ_r(T). So Ran(T - λ) is not dense in X. By the Hahn–Banach theorem, there exists a non-zero φ ∈ X* that vanishes on Ran(T - λ). For all x ∈ X,

Therefore (T* - λ)φ = 0 ∈ X* and λ is an eigenvalue of T*. The shows the former inclusion. Next suppose that (T* - λ)φ = 0 where φ ≠ 0, i.e.

If Ran(T − λ) is dense, then φ must be the zero functional, a contradiction. The claim is proved.

In particular, when X is a reflexive Banach space, σ_r(T*) ⊂ σ_p(T**) = σ_p(T).