Resolvent set

In linear algebra and operator theory, the resolvent set of a linear operator is a set of complex numbers for which the operator is in some sense "well-behaved". The resolvent set plays an important role in the resolvent formalism.

Definitions
Let X be a Banach space and let $$L\colon D(L)\rightarrow X$$ be a linear operator with domain $$D(L) \subseteq X$$. Let id denote the identity operator on X. For any $$\lambda \in \mathbb{C}$$, let


 * $$L_{\lambda} = L - \lambda\,\mathrm{id}.$$

A complex number $$\lambda$$ is said to be a regular value if the following three statements are true: The resolvent set of L is the set of all regular values of L:
 * 1) $$L_\lambda$$ is injective, that is, the corestriction of $$L_\lambda$$ to its image has an inverse $$R(\lambda, L)=(L-\lambda \,\mathrm{id})^{-1}$$ called the resolvent;
 * 2) $$R(\lambda,L)$$ is a bounded linear operator;
 * 3) $$R(\lambda,L)$$ is defined on a dense subspace of X, that is, $$L_\lambda$$ has dense range.


 * $$\rho(L) = \{ \lambda \in \mathbb{C} \mid \lambda \mbox{ is a regular value of } L \}.$$

The spectrum is the complement of the resolvent set


 * $$\sigma (L) = \mathbb{C} \setminus \rho (L),$$

and subject to a mutually singular spectral decomposition into the point spectrum (when condition 1 fails), the continuous spectrum (when condition 2 fails) and the residual spectrum (when condition 3 fails).

If $$L$$ is a closed operator, then so is each $$L_\lambda$$, and condition 3 may be replaced by requiring that $$L_\lambda$$ be surjective.

Properties

 * The resolvent set $$\rho(L) \subseteq \mathbb{C}$$ of a bounded linear operator L is an open set.
 * More generally, the resolvent set of a densely defined closed unbounded operator is an open set.