I am referring to this paper, page 9.
The background of my question is the equation $$ \partial_tF-\partial_v(\partial_v+v)F=0,\quad F_{|t=0}=F^0, $$ where $F=F(t,v)$ is unknown for $t>0$.
An equilibrium is given by $$ \mu(v)=\frac{1}{\sqrt{2\pi}}\,e^{-v^2/2}. $$
Now consider $F(t,\cdot)\in\mu L^2(\mu \,\mathrm dv)$, where $\mu \,\mathrm dv=\mu(v) \,\mathrm dv$.
It is said on page 9 (five lines after $(13)$) that the operator $L=-\partial_v (\partial_v+v)$ is self-adjoint in $\mu L^2(\mu \,\mathrm dv)$.
Why? I do not see that.
I think my problem dates back to the problem that I do not understand what is meant by the function space $\mu L^2(\mu\,\mathrm dv)$ and, in particular, its inner product.
Intuitively, I think the inner product is supposed to be something like $$ \langle f,g\rangle_{\mu L^2(\mu dv)}=\mu(v)\int_{-\infty}^\infty f\,g\,\mu\,\mathrm dv $$ but I am really not sure if this makes any sense; at least, using this definition, my computations did not show $$ \langle Lf,g\rangle_{\mu L^2(\mu dv)}=\langle f,Lg\rangle_{\mu L^2(\mu dv)}. $$
EDIT
Supposing that it's a typo and $L^2(\mu\, dv)$ is meant, i.e., $$ \langle f,g\rangle_{L^2(\mu\, dv)}=\int_{-\infty}^{\infty}fg\, \mu\, dv, $$ I get $$ \begin{align*} \langle Lf,g\rangle_{L^2(\mu\, dv)} &= -\int_{-\infty}^\infty (f_{vv}g+fg+v f_v g)\, \mu\, dv\\ &=-\left(\overbrace{\int f_{vv}g\,\mu\, dv}^{=:A}+\overbrace{\int fg\,\mu\, dv}^{=:B}+\overbrace{\int f_v v g\,\mu\, dv}^{=:C}\right) \end{align*} $$ and then by several integrations by part, $$ \begin{align*} A&=\langle f,g_{vv}-g-v^2g\rangle_{L^2(\mu\, dv)}\\ B&=\langle f,g\rangle_{L^2(\mu\, dv)}\\ C&=\langle f,-g-v g_v+v^2g\rangle_{L^2(\mu\, dv)} \end{align*} $$ so that $$ A+B+C=\langle f,g_{vv}-g,-vg_v\rangle_{L^2(\mu\, dv)} $$ and thus $$ \begin{align*} \langle Lf,g\rangle_{L^2(\mu\, dv)}&=-(A+B+C)\\ &=\langle f,-g_{vv}+g+vg_v\rangle_{L^2(\mu\, dv)}\\ &\neq\langle f,-g_{vv}-g-vg_v\rangle_{L^2(\mu\, dv)}\\ &=\langle f,Lg\rangle_{L^2(\mu\, dv)} \end{align*} $$
It seems indeed to be an unclear and undefined notation of the paper.
I suppose it means the following $$ \mu \,L^2(\mu\,\mathrm d v) = \{g = \mu\,f: f\in L^2(\mu\,\mathrm d v)\} $$ and so I would expect that $$ \|g\|_{\mu \,L^2(\mu\,\mathrm d v)}^2 = \|g/\mu\|_{L^2(\mu\,\mathrm d v)}^2 = \int |g|^2\,\mu^{-1}\,\mathrm d v. \\ \langle f,g\rangle_{\mu \,L^2(\mu\,\mathrm d v)} = \int f\,g\,\mu^{-1}\,\mathrm d v. $$ Then since $\partial_v\mu(v)^{-1} = v\,\mu(v)^{-1}$, $$ \langle Lf,g\rangle_{\mu \,L^2(\mu\,\mathrm d v)} = -\int \partial_v(\partial_v+v)f\,g\,\mu^{-1}\,\mathrm d v \\ = \int (\partial_v+v)f\,\partial_v(g\,\mu^{-1})\,\mathrm d v \\ = \int (\partial_v+v)f\,(\partial_v+v)g\,\mu^{-1}\,\mathrm d v = \langle f,Lg\rangle_{\mu \,L^2(\mu\,\mathrm d v)} $$ so $L^*=L$.