adjoint of operator? let $H=L^2(0,1)$ (Hilbert space with usual scalar product )and  the operator $A$ defined by :
$D(A)=\{u\in C^1[0,1]:u(0)=\lambda u(1)\}$ where $\lambda\in\mathbb C$ and 
$Au=iu'$
my questions is : 
1) how to find the adjoint A* of A
2)how to find the values of $\lambda$ for wich A is essentially self-adjoint ?
thank you very much
 A: $g\in \mathcal{D}(A^{\star})$ iff the following holds for all $f\in\mathcal{D}(A)$:
$$
 \int_{0}^{1}g(\overline{Af})dt=\int_{0}^{1}(A^{\star}g)\overline{f}dt \\
 \int_{0}^{1}g(-i\overline{f}')dt=\int_{0}^{1}(A^{\star}g)\overline{f}dt.
$$
Therefore, if $f\in C^1[0,1]$ and $f(0)-\overline{\lambda}f(1)=0$, then
$$
              \int_{0}^{1}gf'dt=\int_{0}^{1}(iA^{\star}g)fdt.
$$
In particular, the above holds for all $f\in \mathcal{C}_{c}^{\infty}(0,1)$. So $g$ has a weak derivative at $g'=-iA^{\star}g$, or $A^{\star}g=ig'$, which means that $g$ may be modified on a set of measure $0$ to become absolutely continuous on $[0,1]$, and $A^{\star}g=ig'$. So $\mathcal{D}(A^{\star})\subseteq\mathcal{AC}[0,1]$ with $g'\in L^2[0,1]$. However $\mathcal{D}(A^{\star})\ne \mathcal{AC}[0,1]$ because
\begin{align}
    0 = (Af,g)-(f,A^{\star}g) & =i\int_{0}^{1}\{f'\overline{g}+f\overline{g}'\}dt \\
       & =i\{f(1)\overline{g(1)}-f(0)\overline{g(0)}\}.
\end{align}
If $(g(0),g(1))=\alpha(1,\overline{\lambda})$ for some scalar $\alpha$, then the right side reduces to $i\{ \lambda f(1)-f(0)\}=0$. Equivalently,
$$
             \overline{\lambda}g(0)-g(1)=0.
$$
Therefore,
$$
  \mathcal{D}(A^{\star})=\{ g \in \mathcal{AC}[0,1] : g(1)=\overline{\lambda}g(0) \} \\
  \mathcal{D}(A^{\star\star})=\{ f \in \mathcal{AC}[0,1] :
f(0)=\lambda f(1) \}
$$
These domains are the same iff $|\lambda|=1$, which is equivalent to $A$ being essentially selfadjoint.
