Domaine of an adjoint operator Let $A$ be the derivative operator which defined on $L^2(0,1)$ with domain $$D(A) = \left\{ {v \in {H^1}(0,1),v(0) = 0} \right\}.$$
It is obvious that with an integration by part we find that $A^*=A$ with
$$D(A^*) = \left\{ {v \in {H^1}(0,1),v(1) = 0} \right\}.$$
My question is: For $A^*$ to be the adjoint of $A$, they should have the same domaine, but herre I don't see that. Can we say that $A^*=A$?
Thank you.
 A: For an unbounded operator, the domain of the operator and the domain of the adjoint do not necessarily agree. You have the correct domain for $A^*$ in the statement of your problem. $A\ne A^*$. That is, this is not a selfadjoint operator. The only selfadjoint version of $i\frac{d}{dt}$ has a domain with periodic conditions of the form $v(2\pi)=e^{i\theta}v(0)$ for some $\theta$. You can check that this leads to a symmetric operator:
\begin{align}
  \langle if',g\rangle-\langle f,ig'\rangle & =  \int_{0}^{2\pi}if'(t)\overline{g(t)}-f(t)\overline{ig'(t)}dt \\
   & = i\int_{0}^{2\pi}f(t)\overline{g'(t)}+f'(t)\overline{g(t)}dt \\
   & = if\overline{g}|_{0}^{2\pi} \\ & = if(2\pi)\overline{g(2\pi)}-if(0)\overline{g(0)} \\
   & = if(0)\overline{g(0)}-if(0)\overline{g(0)}=0.
\end{align}
And this is selfadjoint on $H^1(0,1)$ with the given endpoint condition. These are the only selfadjoint versions of $i\frac{d}{dt}$ on $H^1(0,1)$.
Your $A$ is not selfadjoint, and that happens because the domain of $A^*$ is not the same. But you have correctly identified $\mathcal{D}(A^*)$ for your $A$.
