Past exam of the functional analysis II course of the University of Oxford / 2023 / 3 / a / ii

Again put $A=\lambda I-T$. Surjectivity gives, by image-kernel orthogonality for an adjoint,If $\lambda$ is real, $A=A^{\ast }$, so $\ker A=\{0\}$. If $\lambda$ is nonreal and $Ax=0$, then $Tx=\lambda x$; reality of $⟨Tx,x⟩=\lambda \Vert x{\Vert }^2$ again implies $x=0$. Thus $A$ is injective as well as surjective, without needing to assume separately that $\lambda$ is not an eigenvalue.

Since $A$ is a bounded linear operator on the Banach space $X$, the bounded inverse theorem makes $A^{-1}$ bounded. ThereforeThis is exactly membership in the resolvent set of an operator.