I'd like to know if one can define a pertinent Hilbert space where the operator $$A_p v := -\frac{1}{2} v" + (vF + v\int_\mathbb{R} Sp + p\int_\mathbb{R} Sv )'$$ is symmetric. Here, $p$ satisfies the following differential equation with $p \in L^2_1(\mathbb{S})\cap C^\infty(\mathbb{S}) $ $$\frac{p'}{2p} = F + \int Sp $$ and $u,v$ and $F,S$ can be taken to satisfie at least the following weak regularity conditions:

- $u,v \in L^2_0(\mathbb{R})\cap C^2(\mathbb{R})$
- $F,S \in C^\infty(\mathbb{R})$ and odd

but the regularity will probably be restricted much more by choosing the nice Hilbert space that we wish. We note that $\int_\mathbb{R} Sp= \int_\mathbb{R} S(x) p(x) dx$.

My motivation for a possible positive answer for this question comes from the following similar problem. Given a fixed $p \in L^2_1(\mathbb{S})\cap C^\infty(\mathbb{S}) $ ($\mathbb{S} = \mathbb{R}/(2\pi\mathbb{Z})$ and $p \neq 0$), solution of $\frac{p'}{2p}= S*p$ ($S(\theta) = \sin(\theta)$) for, the operator

$$B_p v := -\frac{1}{2} v" + ( v\int_\mathbb{S} S*p + p\int_\mathbb{S} S*v )'$$

a such Hilbert space is $H_ {-1,1/p}$ provideded with $\left\langle f,g\right\rangle_{H_ {-1,1/p}} := \int \frac{\mathcal{F}\mathcal{G}}{p} $ where $\mathcal{F}$ and $\mathcal{G}$ are respectively the primitives in $ L^2(\mathbb{R})$ of $f$ and $g$ such that $\int \frac{\mathcal{F}} {p} = \int \frac{\mathcal{G}} {p} =0$ . Here $H_ {-1,1/p}:= H_{1,p}'$ (notation:$V'$ is the dual space of $V$), $H_{1,p}:= ${$ \overline{ f \in C^1(\mathbb{S}): \int f =0 } ^H$} (wich is also a Hilbert space with scalar product $\left\langle f,g\right\rangle_{H_ {1,p}} := \int (f'g'p)$ ) and $H =L^2 _0 (\mathbb{S})$.

We can prove for this case that $$\left\langle f,B_pg\right\rangle_{H_ {-1,1/p}} = \left\langle B_pf,g\right\rangle_{H_ {-1,1/p}}=-\frac{1}{2}\int\frac{fg}{p}+\int fS'*g$$ then $B_p$ is clearly symetric in $H_ {-1,1/p}$

I expecte had well motivated my question. I've tried several integral and derivative restrictions under spaces similar to $H_ {-1,1/p}$ but I've not been sucsecefull yet. I'd be glad for some advice. If you visit the topic please leave a message with your opinion.