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:

  1. $\rho(A) = \overline{\rho(A^*)}$ (Hunter and Nachtergaele 239)
  2. $\sigma_r(A)\subseteq\overline{\sigma_p(A^*)}$ (Hunter and Nachtergaele 223)
  3. $\lambda\in\sigma_p\left(A^*\right) \Longrightarrow \overline{\lambda}\notin\sigma_c(A)$ (can't find a source that directly states this but it comes from a question in the same textbook as the previous two citations)

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?


  1. Hunter, John K., and Bruno Nachtergaele. Applied Analysis. World Scientific, 2007.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.