Is there a cyclic vector for $-\frac{d^{2}}{dx^{2}}$ on $L^{2}[0,2\pi]$ with periodic conditions?

Question

Let $\mathcal{H}=L^{2}[0,2\pi]$, and let $L=-\frac{d^{2}}{dx^{2}}$ on the domain $\mathcal{D}(L)$ consisting of twice absolutely continuous functions $f$ on $[0,2\pi]$ with $f''\in\mathcal{H}$ and $f(0)=f(2\pi)$, $f'(0)=f'(2\pi)$. Then $L$ is selfadjoint on this domain. Is there a cyclic vector $\phi \in \mathcal{H}$ for the $C^{\star}$ algebra generated by the resolvents $(L-\lambda I)^{-1}$?

Answer 1

No. The reason is: the spectrum of $L$ has multiplicities.

Let $e_n=\frac{1}{\sqrt{2\pi}}e^{inx},\ n\in\mathbb Z.$ It is an ONB in $\mathcal H$ consisting of eigenvalues of $L.$ For a fixed $n\in\mathbb N,$ the subspace $V_n$ spanned by $e_n,e_{-n}$ is an eigenspace for $L$ corresponding to the eigenvalue $n^2.$ Hence $V_n$ is an eigenspace for every $(L-\lambda I)^{-1}$ (corresponding to eigenvalue $(n^2-\lambda)^{-1}$). That is, $V_n$ is invariant under every $(L-\lambda I)^{-1}$ and the restriction of $(L-\lambda I)^{-1}$ onto $V_n$ is a multiple of identity. Hence $V_n$ is an invariant subspace for the $C^*$-algebra and the $C^*$-algebra acts by scalar operators on $V_n.$ If there was a cyclic vector $\psi$ for the $C^*$-algebra, the projection $P_n\psi$ of $\psi$ onto $V_n$ would be cyclic for the subspace $V_n,$ which is not true.