Spectrum of a closed operator - Imported Knowlpedia Content

Definition

Let $T:\mathcal D(T)\subseteq X\to X$ be a densely defined closed operator on a complex Banach space. Its spectrum is

\sigma(T)=\mathbb C\setminus\rho(T),

where $\rho(T)$ is the resolvent set: the set of $\lambda\in\mathbb C$ for which

T-\lambda I:\mathcal D(T)\longrightarrow X

is bijective and its inverse is bounded on $X$ . Because $T$ is closed, boundedness of this everywhere-defined inverse follows from bijectivity and the closed graph theorem. The domain is part of the definition; replacing $\mathcal D(T)$ can change the spectrum even when the differential formula for $T$ is unchanged.

Basic topology

The resolvent set is open, the resolvent depends analytically on $\lambda$ , and $\sigma(T)$ is closed. Unlike the spectrum of a bounded operator, the spectrum of a closed unbounded operator need not be bounded, compact, or even nonempty. For a self-adjoint operator on a Hilbert space, however, the spectrum is a nonempty closed subset of $\mathbb R$ , and every nonreal number lies in the resolvent set.

Spectral parts

If $T-\lambda I$ is not injective, then $\lambda$ is an eigenvalue and belongs to the point spectrum. When it is injective with dense but non-surjective range, $\lambda$ lies in the continuous spectrum under the standard Hilbert space convention. When the range is not dense, $\lambda$ lies in the residual spectrum. These parts can behave very differently for non-self-adjoint operators.

Dependence on closed realizations

An unbounded expression such as a differential operator does not determine a spectrum until its domain, including any boundary conditions, has been fixed. Distinct closed realizations of the same expression can have different eigenvalues and resolvents. Spectral assertions should therefore name the closed operator rather than only its formal formula.

References