Proposition 15

Let $u$ be a closed operator with dense domain on a complex Banach space E and $\lambda$ a complex number belonging to the resolvent set of $u$.

a) The subset Sp(R($u, \lambda$ ))$-\{0\}$ of $\mathbf{C}$ is the image of the spectrum of $u$ under the mapping $\mu\mapsto (\lambda -\mu)^{-1}$ of $\mathbf{C}-\{\lambda \}$ into $\mathbf{C}^*$;

b) For every $\mu\not =\lambda$ in $\mathbf{C}$, one has

Ker($\mu1_E-u$) $=$ Ker(($\lambda -\mu$)$^{-1}1_E-R(u, \lambda )$).

Let us prove assertion a). For every $\mu\not =\lambda$, we compute

Since $\lambda \notin$ Sp($u$) and $\mu\not =\lambda$, the linear mapping $(\lambda -\mu)(\lambda 1_E-u)$ is a bijection of dom($u$) onto E. Consequently, this formula implies that $\mu1_E-u$ is a bijection of dom($u$) onto E if and only if $(\lambda -\mu)^{-1}1_E-R(u, \lambda )$ is a bijection of E onto E, which implies the assertion.

Let us prove b). If $\mu\not =\lambda$ and $x\in$ Ker(($\lambda -\mu$)$^{-1}1_E-R(u, \lambda )$), then $x\in$ dom($u$) and the formula $1_E= (\lambda 1_E-u)\circ R(u, \lambda )$ implies that $x\in$ Ker($\mu1_E-u$). Conversely, if $x\in$ Ker($\mu1_E-u$) and $\mu\not =\lambda$, the formula $1_{dom(u)}= R(u, \lambda )\circ (\lambda 1_E-u)$ implies $R(u, \lambda )(x) = (\lambda -\mu)^{-1}x$.