$\overline{\lambda}$ in adjoint point spectrum implies $\lambda$ in residual spectrum or point spectrum

Question

I'm solving some exercises about spectrum and resolvent and there is one that I don't achieve to prove:

Let $T:\mathcal{D}(T)\rightarrow\mathcal{H}$ be a linear operator whose Hilbert-adjoint operator $T^{*}$ exists. If $\overline{\lambda}\in\sigma_{p}(T^{*})$ then $\lambda\in\sigma_{r}(T)\cup\sigma_{p}(T).$

I was using that $\overline{\lambda}\in\sigma_{p}(T^{*}),$ also I supposed $\lambda\notin\sigma_{p}(T).$ The goal was to prove $\lambda\in\sigma_{r}(T),$ but I don't get any useful.

Any kind of help is thanked in advanced.

Answer 1

If $\overline{\lambda}\in\sigma_{p}(T^{*})$, then $T^*v=\overline{\lambda}v$ for some $v\in\mathcal{D}(T^*)$. Hence for all $u\in\mathcal{D}(T)$ we have $$\langle Tu,v\rangle=\langle u,T^*v\rangle=\langle u,\overline{\lambda}v\rangle=\lambda\langle u,v\rangle,$$ and therefore $$\langle (T-\lambda)u,v\rangle=0,\mbox{ for all}\ u\in\mathcal{D}(T)$$ i.e. the range of $T-\lambda$ is not dense and you have $\lambda\in\sigma_{r}(T)\cup\sigma_{p}(T).$