Let $T:l^2 \rightarrow l^2$ be defined as $y = Tx, x = (\zeta_j), y = (\eta_j), \eta_j = \alpha_j\zeta_j$ where $(\alpha_j)$ is dense in $[0,1]$. What is the continuous, residual and the point spectrum?
Now I have computed both the point and the residual is empty since $T$ is a bounded self adjoint linear operator. Since $(\alpha_j)$ is dense in $[0,1]$, I computed that $[0,1] = \sigma(T)$. I am having a tough time computing the continuous spectrum $\sigma_c(T)$. I have a guess though. If $\lambda \in [0,1]-(\alpha_j)$, is $\lambda \in \sigma_c(T)$?. Thanks for your help.
Because, as you say, the residual spectrum is empty, all elements of the spectrum are approximate eigenvalues. For any approximate eigenvalue $\lambda$ that is not an eigenvalue, you have that $T-\lambda I$ is injective which implies dense range; and this range cannot be closed because otherwise $T-\lambda I$ would be invertible. It follows that the continuous spectrum is $[0,1]\setminus \{\alpha_j\}$, as you say.