In the following, I am referring to this paper.
Let $M(v)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{\vert v\vert^2}{2}\right)$ and define the measures $$ dM:=M(v)dv,\qquad d\gamma=\gamma(v)dv:=\frac{dv}{M(v)}. $$ Moreover, let $L^2(dM)$ and $L^2(d\gamma)$ denote the spaces of square integrable functions against the measures $dM$ and $d\gamma$, respectively.
At the end of p. 6 in the linked paper, it is said that it is a "straightforward consequence" of the Gaussian Poincaré inequality that for $f\in L^2(d\gamma)$, we have $$ \Vert f-pM\Vert_{L^2(d\gamma)}\leq\left\Vert \partial_v\left(\frac{f}{M}\right)\right\Vert_{L^2(dM)}, $$ where $p=\int f\, dv$.
Could you please give me a hint how to deduce this "straightforward" inequality?
Unfortunately, it is not straightforward for me.
Just factor out the $M$ in the left-hand side : $$\Vert f-pM\Vert_{L^2(d\gamma)}^2 = \left\Vert \left(\frac f M-p\right)M\right\Vert_{L^2(d\gamma)}^2 \tag1$$
Now observe that the RHS in $(1)$ is equal to the $L^2(dM)$ norm of $f/M - p$, indeed : $$\begin{align*}\left\Vert \left(\frac f M-p\right)M\right\Vert_{L^2(d\gamma)}^2:&=\int \left(\frac f M(v)-p\right)^2M(v)^2 d\gamma (v)\\ &=\int \left(\frac f M(v)-p\right)^2M(v)^2 \frac{dv}{M(v)}\\ &=\int \left(\frac f M(v)-p\right)^2M(v) dv\\ &=\left\Vert \frac f M-p\right\Vert_{L^2(dM)}^2 \end{align*} $$ Now observe that $p=\int \frac{f}{M}dM$, and it follows from Poincaré inequality for the Gaussian measure $dM$ (so with Poincaré constant equal to $1$) that $$\left\Vert \frac f M-p\right\Vert_{L^2(dM)}^2 =: \text{Var}_{dM}\left(\frac{f}{M}\right) \le \mathbb E_{dM}\left[\left|\partial_v\left(\frac{f}{M}\right)\right|^2\right]:=\left\Vert \partial_v\left(\frac{f}{M}\right)\right\Vert_{L^2(dM)}^2 $$