This is an exercise in functional analysis:
For $k=1,2,3$, let $A_k: D(A_k)\subset L^2([0,1])\to L^2[(0,1)]$ be the first-order differential operators $A_ku=iu'$ with domains $$ D(A_1) = H^1((0,1)), \\ D(A_2) = \{u\in H^1((0,1))|u(0)=u(1)\}, \\ D(A_3) = \{u\in H^1((0,1))|u(0)=0\}. $$ Show that the spectrum of $\sigma(A_1) =\Bbb{C}$, $\sigma(A_2)=\{2n\pi|n\in{\Bbb Z}\}$, and $\sigma(A_3)=\emptyset$.
In $L^2([0,1])$, the derivative is "weak derivative". If one deals with the eigenvalue problem above in the strong derivative sense, and solve the ODE $$ iu'=\lambda u $$ with (without) different boundary conditions, then one can formally get the spectrum. My question:
How can I justify that this formal technique indeed gives the correct results which is in the sense of weak derivative?
Suppose that $u \in H^1$ with $iu' = \lambda u$. Then we have that $u' = -i\lambda u$ is also in $H^1$, which means that actually $u \in H^2$. But then $u' = -i\lambda u$ is in $H^2$ so actually $u \in H^3$ Continuing on in this fashion yields that $u \in \bigcap_k H^k$. Thus the Sobolev Embedding Theorem implies that $u \in \bigcap_k C^k = C^\infty$. So if $u$ is an eigenfunction, then $u$ is smooth. Now you can solve the ODE and proceed however you wish.