So this is a question from Teschl G. - Mathematical methods in quantum mechanics- Problem 2.19 \
Compute the resolvent of $$Af=f' \enspace D(A)=\{f\in H^1[0,1] \enspace | \enspace f(0)=0\}$$
and show that unbounded operators can have empty spectrum. Where $$D(A)\subseteq L^2[a,b] \enspace \text{and} \enspace Ran(A)\subseteq L^2[a,b]$$
I tried to calculate directly or to imitate the example in the book but they all just stayed as a restatement of definitions.Like if this then this then this and nothing.
So thank you for your answers
Okay, so you are considering $\frac{d}{dx} : \mathcal{D}(\frac{d}{dx})\subset L^2[0,1]\rightarrow L^2[0,1]$, where $\mathcal{D}(\frac{d}{dx})$ consists of all functions $f\in H^1[0,1]$ that vanish at $0$. Given $g\in L^2[0,1]$, the resolvent equation is $$ \left(\frac{d}{dx}-\lambda I\right)f=g,\;\;\; f\in H^1[0,1], f(0)=0, \lambda\in\mathbb{C}. \\ \frac{d}{dx}(e^{-\lambda x}f)=e^{-\lambda x}g \\ \int_0^x \frac{d}{dx'}(e^{-\lambda x'}f(x'))dx'=\int_0^x e^{-\lambda x'}g(x')dx' \\ e^{-\lambda x}f(x) = \int_0^x e^{-\lambda x'}g(x')dx' \\ f(x) = e^{\lambda x}\int_0^x e^{-\lambda x'}g(x')dx'. $$ So, the resolvent $R(\lambda)=(\frac{d}{dx}-\lambda I)^{-1}$ is defined for all $\lambda\in\mathbb{C}$ and is given by $$ R(\lambda)g = \int_0^x e^{\lambda(x-x')}g(x')dx'. $$ It's not hard to verify that $R(\lambda) : L^2[0,1]\rightarrow L^2[0,1]$ is a bounded operator for all $\lambda \in \mathbb{C}$. That means $\frac{d}{dx}$ has no spectrum.