I got stuck with my exercise.
Let $H = L^2(0, 2 \pi; \mathbb{C})$, define the operators $A_1 , A_2$ as follows. $$A_j : H \supset D(A_j) \to H , u \mapsto Au := iu' \quad (j=1,2)$$ Where $D(A_1) := {u\in H^1(0, 2 \pi) : u(0) = 0} \quad \text{and} \quad D(A_2) := H^1_{per}(0, 2 \pi)$
a) Show $A_1$ and $A_2$ are closed operators.
b) Find the hilbert space adjoint operator, which of the operators is self adjoint?
c) Show $[A_2u , u ] \in \mathbb{R}$ for every $u \in D(A_2)$. Use this to show $$ \Vert(\lambda -A_2)u \Vert \cdot \Vert u \Vert \geq \vert \langle (\lambda-A_2)u,u \rangle \vert \geq \vert Im \langle(\lambda-A_2)u,u \rangle \vert = \vert Im\lambda \vert \cdot \Vert u \Vert^2$$ for $\lambda \in \mathbb{C} $\ $ \mathbb{R} $ and conclude, that $A_2$ has a real spectrum.
d) Show $R(i,A_2)= (i - A_2) ^{-1}$ is a normal, compact operator.
Hint: use $R(\lambda,A_2) - R(\mu, A_2) = (\mu - \lambda) R(\lambda, A_2)R(\mu,A_2) \quad (\lambda, \mu \in \rho(A_2))$.
You're allowed to use that the embedding $H^1(0,2 \pi) \subset L^2(0, 2 \pi)$ is compact
e) Prove that $\sigma(A_2) = \sigma_p(A_2)$
f) Show $\sigma(A_2) = \mathbb{Z}$
I worked out a) and b) but now im helplessly stuck at the other exercises. The following are my thoughts on the exercises.
c) We may use, $\langle A_2u, u \rangle = \int_0^{2\pi} i u'(t)\overline{u(t)} dt$ and split this up into real and imaginary part. But then I just get $Im(\langle A_2u, u \rangle) = \int_0^{2\pi} Re(u'(t))\cdot Re(u(t))-Im(u'(t))\cdot Im(u(t))$. But i see no way to conclude that it equals 0.
d) This one is the hardest for me. To show that it's normal, I need to show $R(i,A_2)^*R(i,A_2) = R(i,A_2)R(i,A_2)^*$. But I got no idea what $R(i,A_2)^*$ looks like.
e) Here I need to show, that $(\lambda - A_2)$ is surjecive for every $\lambda \in \sigma(A_2) \subset \mathbb{R}$. Which we can break down to $x \mapsto x-ix' $ is surjective. And here i get no further.
f) This seems to be easy as we just need to prove that $(\lambda - A_2)$ is not injective for $\lambda \in \mathbb{Z}$. But at the same time it looks pretty weird, that it should not be injective on $\mathbb{Z}$ but on $\mathbb{Q}$.
It's easy to see that its not injective for $\lambda = 0 $ because the constant functions 1 and 2 have the same derivative but are not equal.