Resolvent

Remarks

Some authors use $(\lambda I - T)^{-1}$ instead, which changes the sign. The resolvent set is open, and $R(\cdot, T)$ is a holomorphic function with values in $\mathcal{B}(X)$. It satisfies the resolvent identity $$R(\lambda, T) - R(\mu, T) = (\lambda - \mu) R(\lambda, T) R(\mu, T) \qquad (\lambda, \mu \in \rho(T)),$$ and $\norm{R(\lambda, T)} \ge 1 / \operatorname{dist}(\lambda, \sigma(T))$.

Example: for $D(x_n) = (x_n / n)$ on $\ell^2$ and $\lambda \in \rho(D)$, $R(\lambda, D)$ is the diagonal operator with entries $1/(1/n - \lambda)$.

Non-example: $\lambda = 0$ is not in $\rho(D)$, since $D$ is not surjective. Indeed $(1/n)$ is in $\ell^2$ but $(1)$ is not, so $(1/n)$ is not in the image of $D$. Hence $R(0, D)$ is not defined.

The same definitions apply to closed operators, with $R(\lambda, T) \colon X \to D(T)$.