Spectral decompostion of a linear operator in Hilbert space

Question

I am having trouble calculating the countinuous spectrum of the following operator, as defined by their action on an orthonormal base of the Hilbert space: $$ A(\phi_k)=e^{-k^2}\phi_k $$ Being $\{\phi_k\}_{k\in\mathbb{Z}}$ the orthonormal base. I have figured out that $\lambda_k=e^{-k^2}$ are all eigenvalues,and that the inverse is such that the resolvent $R_\lambda=(A-\lambda I)^{-1}$ is such that: $R_\lambda (\phi_k)=(e^{-k^2}-\lambda)^{-1}\phi_k$$, and that it is not bounded in the eigenvalues, and only there. Can I say then that the countinuous spectrum is empty?

Answer 1

Since $A\phi_k = e^{-k^2}\phi_k$ for every basis vector $\phi_k$ you can construct matrix representations of $A$ on the ascending chain of subspaces $span\{\phi_1\} \subset span\{\phi_1,\phi_2\} \subset \cdots$. Yyou can argue by induction that this process will form a complete representation of $A$ in the limit as the number of basis vectors becomes infinite. In particular, since $A$ consists only of its eigenvalues (plus zeros) in any finite projection, induction on $n$ shows the spectrum of this representation will be exactly the set $\{e^{-k^2}\}_k.$