I am currently dealing with an unbounded operator 


$T:\{f \in L^2(-2\pi,2\pi); f \in AC((-2\pi,2\pi)), T(f) \in L^2, \lim\_{x \rightarrow \pm 2 \pi} f(x)g(x)=0\} \subset L^2(-2\pi,2\pi)\rightarrow L^2(-2\pi,2\pi)$


$g \in C^{\infty}: g(-2\pi) = g(2\pi)=0$ and $g|\_{(-2\pi,2\pi)} >0.$


Then I want to show that $T(f) = -i(gf)'$ is closed and has an adjoint operator $T^\*(f) = -igf'$ defined on the domain $\{f \in L^2;f \in AC((-2\pi,2\pi)), T^\*(f) \in L^2, \lim\_{x \rightarrow \pm 2 \pi}g(x) f'(x)=0\}.$


Unfortunately, this is not my normal field of interest and so I thought that some people here might immediately know how to approach this problem.



adjoint of this closed (?) operator
Let's first of all recall that the minimal operator of differentiation
$$
D(S) = \{ f\in L^2\cap AC : f'\in L^2, f(\pm 2\pi)=0 \} , \quad\quad Sf = -if'
$$
is closed and has the maximal operator as its adjoint:
$$
D(S^\*) =\{ f\in L^2\cap AC: f'\in L^2\} , \quad\quad S^\*f=-if'
$$
Now since $g$ is smooth and $g>0$, your condition that $f\in AC$ is the same as asking that $gf\in AC$. This means that your $T=Sg$ is the composition of $S$ with the (bounded) operator of multiplication by $g$.


Thus $T$ is indeed closed, and $T^\*=gS^\*$. In particular, $D(T^\*)=D(S^\*)$.


