Let $X$ be a complex Banach space, $T\in\mathfrak L(X)$ and $\lambda\in\mathbb C$.
Are we able to show $\lambda\in\sigma_p(T)$ (the point spectrum of $T$) if and only if $\overline\lambda\in\sigma_r(T')$ (the residual spectrum of $T':X'\to X'$)?
I'm only able to (partially) prove this claim under the assumption that $X$ is a Hilbert space and $T'$ is replaced by the Hilbert space adjoint $T^\ast:X\to X$:
In that situation, if $S\in\mathfrak L(X)$, then $$\mathcal N(S)={\mathcal R(S^\ast)}^\perp\tag1$$ from which we infer that $S$ is injective if and only if $\mathcal R(S^\ast)$ is dense. Substituting $S=\lambda-T$ and noting that $(\lambda-T)^\ast=\overline\lambda-T^\ast$, it is only left to show that if $\lambda-T$ is not injective (hence $\mathcal R(\overline\lambda-T^\ast)$ is not dense), then $\overline\lambda-T^\ast$ is injective.
How can we show this? And are we able to prove the desired equivalence in the general Banach space case well?
No. If $X_1=X_2$, and $T$ is given by $T(x)=\lambda x$, with $\lambda\in \mathbb C \setminus \mathbb R$, then $$ T'(\phi)=\lambda \phi, \quad\forall \phi\in X_2', $$ so the spectra of both operators coincide with their point spectra and $$ \sigma (T) = \sigma (T')=\{\lambda \} \neq \{\bar \lambda \}. $$