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

Use the convention that the inner product is linear in its first argument. By the defining identity for the adjoint operator,For the last implication choose $y=Tx$. This proves the first image-kernel orthogonality for an adjoint identity. Taking orthogonal complements and using the double orthogonal complement theorem for the vector subspace $\mathrm{Im}{T}^{\ast }$ gives

Put $A=\lambda I-T$. Since $T$ is a self-adjoint operator, $A^{\ast }=\overline{\lambda }I-T$. Applying the preceding image-kernel orthogonality for an adjoint identities to $A^{\ast }$ givesIf $\lambda$ is real, $A^{\ast }=A$ and $\ker A=\{0\}$ because $\lambda$ is not an eigenvalue. If $\lambda$ is nonreal and $A^{\ast }x=0$, then $Tx=\overline{\lambda }x$. The scalar $⟨Tx,x⟩$ is real by self-adjointness, while it equals $\overline{\lambda }\Vert x{\Vert }^2$. Its imaginary part forces $x=0$. Thus $\ker A^{\ast }=\{0\}$ in both cases, andThe conjugation on $\lambda$ in the adjoint operator is essential when $\lambda$ is nonreal.

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.