Let $E$ be a $\mathbb R$-Banach space and $(\mathcal D(A),A)$ be a linear operator on $E$ (i.e. $\mathcal D(A)$ is a subspace of $E$ and $A$ is a homomoprihsm between $\mathcal D(A)$ and $E$).
Now, let $$A_\lambda:=A-\lambda\operatorname{id}_{\mathcal D(A)}$$ for $\lambda\in\mathbb R$. In the Wikipedia article, they say that $\lambda\in\mathbb R$ is called regular for $A$ if
- $A_\lambda$ is injective
- $R_\lambda(A):=A_\lambda^{-1}$ is bounded
- the range $A_\lambda\mathcal D(A)$ of $A_\lambda$ is dense
Clearly, since $A_\lambda$ is injective, we can talk about its inverse $A_\lambda^{-1}$ and $R_\lambda(A)$ is an isomorphism between $\mathcal R(A_\lambda)=A_\lambda\mathcal D(A)$ and $\mathcal D(A)$. If $R_\lambda(A)$ is bounded, then it has an extension to $E$ and if $\mathcal R(A_\lambda)$ is dense, the extension is unique.
So, $R_\lambda(A)$ can be considered as a bounded linear operator on $E$ with $R_\lambda(A)x=Ax-\lambda x$ for all $x\in\mathcal D(A)$. Does this somehow imply that $A_\lambda$ has a (unique) extension to $E$?