Example of an operator with continuous spectrum

Question

Given $f \in L^2(\mathbb{R})$, let $u \in H^1(\mathbb{R})$ be the unique (weak) solution of the problem$$-u'' + u = f \text{ on }\mathbb{R},$$in the sense that$$\int_{\mathbb{R}} u'v' + \int_\mathbb{R} uv = \int_\mathbb{R} fv \text{ for all }v \in H^1(\mathbb{R}).$$Set $u = Tf$ and consider $T$ as a bounded linear operator from $H =L^2(\mathbb{R})$ into itself. What is $\sigma(T) = [0, 1]$? I hypothesize that it $\sigma(T) = [0, 1]$, but I am at a loss at how to see it.

Answer 1

For $f \in L^2$, you know that $\hat{f} \in L^2$ and $$ u(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\hat{f}(s)}{s^{2}+1}e^{isx}ds $$ does satisfy your weak equation. The opertor $T$ is a multiplication operator in the Fourier domain. That is $Tf = \mathcal{F}^{-1}\frac{1}{s^2+1}\mathcal{F}f$ where $\mathcal{F}$ is the Fourier transform. The operator $T$ has only continuous spectrum equal to the essential range of $f(s)=\frac{1}{s^2+1}$, which is $[0,1]$ as you conjectured.