I am currently dealing with an unbounded operator

$T:\{f \in L^2(-2\pi,2\pi); f \in AC^1((-2\pi,2\pi)), \lim_{x \rightarrow \pm 2 \pi} f(x)g(x)=0\} \subset L^2(-2\pi,2\pi)\rightarrow L^2(-2\pi,2\pi)$

$g \in C^{\infty}: g(-2\pi) = g(2\pi)=0$ 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 $\{f \in L^2;f \in AC^1((-2\pi,2\pi)), \lim_{x \rightarrow \pm 2 \pi}g(x) f'(x)=0\}.$

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.

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.