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 TL(X).

A complex number λ is in the spectrum of T, denoted σ(T), if Tλ does not have bounded inverse. If Tλ 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 Tλ is either not one-to-one or not onto. One can easily check that if Tλ 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 Tλ 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. Tλ is not injective.
  2. Tλ is injective, and has dense range. But Tλ is not bounded below, and we fail to get that the range is all of X. Equivalently, the densely-defined linear map (Tλ) xx is not bounded, therefore can not be extended to all of X.
  3. Tλ is injective and does not have dense range. In this case, the map (Tλ) xx 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 TL(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 σp(T). λ ∈ σ(T) is in the point spectrum if and only if Tλ is not injective.
  2. If λ ∈ σ(T) is not an eigenvalue and the range of Tλ, Ran(Tλ), is dense in X, λ is said to be in the continuous spectrum, σc(T), of T.
  3. If Tλ is injective and Ran(Tλ) is not dense, λ is in the residual spectrum of T, σr(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, σr(T) ⊂ σp(T*) ⊂ σr(T) ∪ σp(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 ∈ σr(T). So Ran(T - λ) is not dense in X. By the Hahn–Banach theorem, there exists a non-zero φX* that vanishes on Ran(T - λ). For all xX,

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(Tλ) is dense, then φ must be the zero functional, a contradiction. The claim is proved.

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

Read more about this topic:  Continuous Spectrum

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