Let $l^2(\mathbb{Z})$ be the space of square summable bi-infinite sequences. Let $\textbf{L denote the operator of multiplication by $n$}$ and let $\mathcal{H}=\{\{a_n\}\in l^2(\mathbb{Z}):\{na_n\}\in l^2(\mathbb{Z})\} \subset l^2(\mathbb{Z})$ and $||\{a_n\}||_{\mathcal{H}}^2=\sum_{n\in \mathbb{Z}} (1+n^2)|a_n|^2$. Also let $R \in \mathcal{L}(l^2(\mathbb{Z}),l^2(\mathbb{Z}))$ i.e, bounded linear function from $l^2(\mathbb{Z}) \to l^2(\mathbb{Z})$.
a) We have to show that the spectrum of $L+R$ is discrete. Precisely, there is a discrete set $D \subset \mathbb{C}$ such that $L+R-\lambda I:\mathcal{H}\to l^2(\mathbb{Z})$ is invertible for $\lambda \notin D$.
b) As an application of the above we have:
Let $\mathbb{T}=\mathbb{R}\backslash (2\pi\mathbb{Z})$ and $V \in C(\mathbb{T})$. Show that the set $D=\{\lambda \in \mathbb{C}: \text{there is f}\in C^1(\mathbb{T}) \text{ not identically $0$ and solving } f'+Vf=\lambda f\}$ is discrete.
$\textbf{Thoughts}:$
I think b) should follow directly from a) by taking the fourier transform for the given equation.
For a) I think the point spectrum will be $0$ since for large enough n, $R\{\{a_n\}\}_n$ will be small. I am not so sure about the residual spectrum.
In general, I am only aware of Reisz Schauder which gives a discrete set spectrum but it works only for compact operators. In this case I don't see why $L+R$ would be compact, at least directly. One thing that I thought could help was that $H^1(\mathbb{T})$ is compactly embedded in $L^2(\mathbb{T})$ but it doesn't seem to be of any help.
Any help will be appreciated. Thanks.
Best Answer
In the following for a linear operator $T$ in $\ell^2(\Bbb Z)$, $\rho(T)$ denotes the resolvent set of $T$, i.e., the set of all $\lambda\in\Bbb C$ for which the operator $T-\lambda = T-\lambda I$ is invertible. Also, $\sigma(T)$ denotes the spectrum of $T$, i.e., $\sigma(T) = \Bbb C\setminus\rho(T)$.
For $\lambda\in\rho(L)$ you have $L+R-\lambda = (I + R(L-\lambda)^{-1})(L-\lambda)$. We know that $\|(L-\lambda)^{-1}\| = \tfrac 1{dist(\lambda,\sigma(L))}$. So, for $dist(\lambda,\sigma(L))$ large enough (note that $\sigma(L) = \{n : n\in\Bbb Z\}$) we have $\|R(L-\lambda)^{-1}\| < 1$ and thus $L+R-\lambda$ is invertible for these $\lambda$. So, $\rho(L+R)\neq\emptyset$. Now, for $\lambda\in\rho(L+R)\cap\rho(L)$, $$ (L+R-\lambda)^{-1} = (L-\lambda)^{-1}[I + R(L-\lambda)^{-1}]^{-1}. $$ Now, we also (hopefully) know that $(L-\lambda)^{-1}$ is a compact operator. So $(L+R-\lambda)^{-1}$ is compact. Hence, the resolvent of $L+R$ is compact, which means that $\sigma(L+R)$ is discrete.
For part (b) invoke the Fourier transform $\mathcal F : C(\Bbb T)\to\ell^2(\Bbb Z)$.