I've been puzzling over this for some time now, and can't quite make my intuitions precise. I need to find the resolvent set and spectrum of the operator
$$ Lu=i\frac{du}{dx} $$ taken to be (densely) defined over $L^2([0,\infty))$, and eventually its adjoint, which is the same operator but on the set (assuming I've computed it correctly):
$$ \text{Dom}(L^*)=\{v\in L^2[0,\infty):v(0)=0\} $$
The pointwise spectrum of $L$ is easy to compute: $(\lambda I-L)u=0$ is a simple ODE, whose solution along with the $L^2$ condition implies
$$ \sigma_p(L)=\{\lambda=a+bi\in\Bbb{C}:b<0\}, $$the lower half-plane. Now intuitively, since $\sigma(L)$ is a closed set, I expect $\rho(L)$ to be the upper half-plane, the residual spectrum to to be the real axis, and the continuous spectrum to be empty. However I'm struggling to show this rigorously since $L$ is an unbounded operator, and all the theory I find pertains to bounded linear operators.
So here are my questions:
- Is there a (somewhat) direct way to characterize the residual spectrum of an operator without explicitly computing $(\lambda I-L)^{-1}$? Or am I just being a whimp and should find $(\lambda I-L)^{-1}$?
- The adjoint clearly has no pointwise spectrum because of the domain restriction. Does this mean its resolvent set will be all of $\Bbb{C}$ or am I missing something?
Feel free to direct me to your favorite book on spectral/operator theory if necessary - I can't seem to find answers to my questions after looking in 4 or 5. Thanks!