This is a homework question and I find myself ending up with the "opposite" of what I am supposed to prove. Below is the question and my attempt:
Suppose that $\lambda \in \mathbb{R}$. Show that if $\lambda \in \sigma_p(T)$ and $\lambda \notin \sigma_p(T^*)$, then $\lambda \in \sigma_r(T^*)$. I assume $T$ is a bounded linear operator on $X$ and maps to $X$, where $X$ is Banach. $\sigma_p(T)$ is the point spectrum of $T$, and $\sigma_r(T)$ is the residual spectrum.
I thought $\lambda \in \sigma_p(T)$ implies $\lambda I - T$ is not injective and $\exists v \neq 0$ such that $\lambda v = Tv$, and $\lambda \notin \sigma_p(T^*)$ implies $\lambda I - T^*$ is injective.
Since $\sigma(T) = \sigma(T^*)$ (reference: Spectrum for a bounded linear operator and its adjoint on a Banach space are same., or Conway, "A Course in Functional Analysis", pg.208 Prop 6.1), then $\lambda$ must be in either $\sigma_c(T^*)$ or $\sigma_r(T^*)$, where $\sigma_c(T^*)$ is the continuous spectrum. However, as $Ker((\lambda I - T)^*) = Ker(\lambda I - T^*) = \{0\}$, and $Ran(\lambda I - T)^{\perp} = Ker(\lambda I - T^*)$, so $Ran(\lambda I - T)$ must be dense in $X$, so $\lambda \in \sigma_c(T^*)$ by definition. However, I'm supposed to show $\lambda \in \sigma_r(T^*)$.
Can anyone point the mistake I made?
Well I don't know how you obtained $\lambda\in\sigma_c(T^*)$, as you showed was that $\operatorname{Ran}(\lambda I-T)$ is dense, not that $\operatorname{Ran}(\lambda I-T^*)$ is dense. In any case, the calculation you want is $$^\perp\operatorname{Ran}(\lambda I-T^*)=\ker(\lambda I-T)\neq\{0\}.$$ Thus $\lambda I-T$ does not have dense range. But by hypothesis, $\lambda\notin\sigma_p(T^*)$, and thus $\lambda\in\sigma_r(T)$.