Suppose that $A$ is a bounded linear operator on a Hilbert space $\mathcal{H}$ over the complex numbers $\mathbb{C}$, and that $A^*$ denotes its adjoint. The residual spectrum of $A^*$, $\sigma_r(A^*)$, has a nice relationship with the point spectrum of $A$ in that $\sigma_r(A^*) \subseteq \overline{\sigma_p(A)}$, where
$$ \begin{align*} \overline{\sigma_p(A)} = \left\{\overline{\lambda}:\lambda\in\sigma_p(A)\right\} \end{align*} $$
Is there a similar relationship between the continuous spectrum of $A^*$ and the spectrum of $A$? I know that $\sigma(A^*) = \overline{\sigma(A)}$ so the conjugates of the elements of the continous spectrum of $A^*$ are contained somewhere in $\sigma(A)$, but can we say more than that?
Here is an answer that I believe works: it can be shown that
$$ \sigma_c(A) = \overline{\sigma_c\left(A^*\right)} $$
For a bounded linear operator $A$ on a Hilbert space, we have the following facts:
Let $\lambda\in\sigma_c(A)$. As a corollary to (1), $\sigma(A) = \overline{\sigma(A^*)}$, so $\overline{\lambda}\in\sigma(A^*)$. By (3), $\overline{\lambda}\notin\sigma_p(A^*)$, and so $\overline{\lambda}\in\sigma_c(A^*)\cup\sigma_r(A^*)$. But $\overline{\lambda}\in\sigma_r(A^*)\Longrightarrow\overline{\overline{\lambda}}=\lambda\in\sigma_p(A^{**}) = \sigma_p(A)$ by (2), contradicting $\lambda\in\sigma_c(A)$.
It follows that $\lambda\in\sigma_c(A)\Longrightarrow\overline{\lambda}\in\sigma_c(A^*)$, and since $\overline{\lambda}\in\sigma_c(A)\Longrightarrow\overline{\overline{\lambda}}=\lambda\in\sigma_c(A^{**}) = \sigma_c(A)$, the continuous spectrum of $A$ is the same as the conjugate of the continuous spectrum of its adjoint, i.e. $\lambda\in\sigma_c(A)\Longleftrightarrow\overline{\lambda}\in\sigma_c(A^*)$, QED.
Can anybody verify this?
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