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

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

A complex number ${\displaystyle \lambda }$ is said to be a regular value if the following three statements are true:

1. ${\displaystyle L_{\lambda ))$ is injective, that is, the corestriction of ${\displaystyle L_{\lambda ))$ to its image has an inverse ${\displaystyle R(\lambda ,L)}$;
2. ${\displaystyle R(\lambda ,L)}$ is a bounded linear operator;
3. ${\displaystyle R(\lambda ,L)}$ is defined on a dense subspace of X, that is, ${\displaystyle L_{\lambda ))$ has dense range.

The resolvent set of L is the set of all regular values of L:

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

The spectrum is the complement of the resolvent set:

${\displaystyle \sigma (L)=\mathbb {C} \setminus \rho (L).}$

The spectrum can be decomposed into the point/discrete spectrum (where condition 1 fails), the continuous spectrum (where conditions 1 and 3 hold but condition 2 fails) and the residual/compression spectrum (where condition 1 holds but condition 3 fails).

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

Properties

• The resolvent set ${\displaystyle \rho (L)\subseteq \mathbb {C} }$ 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.