Resolvent of the differentiation operator on the torus

Question

Suppose we have the space $L^2(\mathbb T)$, that is, the space of periodic functions that are in $L^2$. Let $L^0$ be the operator of differentiation, i.e $L^0 f = f'$ where the domain of $L^0$ is the Sobolev space $H^1(\mathbb T)$. How does one prove that $L^0$ has compact resolvent?

Answer 1

Consider $L^{2}(\mathbb{T})$ to be the space of square-integrable periodic functions on $[0,2\pi]$ with inner-product $(f,g)=\frac{1}{2\pi}\int_{0}^{2\pi}f(\theta)\overline{g(\theta)}\,d\theta$. Let $(Lf)(\theta) = \frac{d}{d\theta}f(\theta)$. $L^{2}(\mathbb{T})$ has an orthonormal basis of eigenfunctions $\{ e^{in\theta}\}_{n=1}^{\infty}$ of $L$ because $Le^{in\theta}=ine^{in\theta}$. The resolvent $(L-\lambda I)^{-1}$ of $L$ exists for all $\lambda \notin \{0,\pm i,\pm 2i,\pm 3i,\cdots\}$, and is given by $$ (L-\lambda I)^{-1}f = \sum_{n=-\infty}^{\infty}\frac{(f,e^{in\theta})}{in-\lambda}e^{in\theta},\;\;\;\;\; f \in L^{2}(\mathbb{T}). $$ For fixed $\lambda \notin \{0,\pm i,\pm 2i,\pm 3i,\cdots\}$, $$ \left\|(L-\lambda I)^{-1}f -\sum_{n=-N}^{N}\frac{(f,e^{in\theta})}{in-\lambda}\right\|^{2}= \sum_{|n| > N}\frac{|(f,e^{in\theta})|^{2}}{|\lambda-in|^{2}} \le \frac{1}{\inf_{n\ge N}|\lambda -in|^{2}}\|f\|^{2}. $$ So, for fixed $\lambda\notin\{0,\pm i,\pm 2i,\pm 3i,\cdots\}$, the resolvent $(L-\lambda)^{-1}$ is a uniform limit of finite-rank operators.