In the context of this paper, where $$ T:=v\cdot\nabla_x-\nabla_xV\cdot\nabla_v $$ is a transport operator, it is said on page 3, that $T$ is skew-symmetric with respect to $\langle\cdot,\cdot\rangle$, where $$ \langle f,g\rangle=\int\int_{\mathbb{R}^d\times\mathbb{R}^d}fg\, d\mu,\quad d\mu=d\mu(x,v):=\frac{dvdx}{F(x,v)}. $$
What does skew-symmetric mean here? Does this mean $$ \langle Tf,f\rangle=0\quad\forall f\in\textrm{dom}(T)\tag{*}? $$ If yes, I do not see $(*)$:
Let's assume $d=1$ for simplicity, then $$ \begin{align*} \langle Tf,f\rangle &= \iint_{\mathbb{R}\times\mathbb{R}} \left(v\frac{\partial f}{\partial x}-\frac{dV}{dx}\frac{\partial f}{\partial v}\right)f\, d\mu\\ &=\iint_{\mathbb{R}\times\mathbb{R}}\frac{v\frac{\partial f}{\partial x}f}{F(x,v)}\, dv\, dx-\iint_{\mathbb{R}\times\mathbb{R}}\frac{\frac{dV}{dx}\frac{\partial f}{\partial v}}{F(x,v)}\, dv\, dx \end{align*} $$
It means that:
$$ \langle T f, g \rangle = -\langle f, T g\rangle $$
And this can be shown straightforwardly by noting that $\frac{\partial}{\partial x} [v f] = v \frac{\partial f}{\partial x}$ and $\frac{\partial}{\partial v}\left[ \frac{\partial V}{\partial x} f\right] = \frac{\partial V}{\partial x} \frac{\partial f}{\partial v}$, then integrating each of the terms in the operator by parts. The boundary terms from integrating by parts are zero as a result of the normalization condition that they state:
$$ \iint_{\mathbb{R}^d \times \mathbb{R}^d} F \text{d}v \text{d}x = 1 $$
It is called skew-symmetry because the adjoint of $T$ is $-T$, like how a skew symmetric matrix equal to its negation upon transposing it (i.e. adjoint if the matrix has real values).