Let $$\Gamma(f,g):=\frac12f'g'\;\;\;\text{for }f,g\in C^1(\mathbb R),$$ $\mu$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$ with a continuously differentiable and positive density $\varrho$ with respect to the Lebesgue measure $\lambda$, $$\mathcal E(f,g):=\int\Gamma(f,g)\:{\rm d}\mu$$ and $$Af:=\frac12f''+\frac12(\ln\varrho)'f'\;\;\;\text{for }f\in C^2(\mathbb R).$$ (For simplicity of notation, $\Gamma(f):=\Gamma(f,g)$ and $\mathcal E(f):=\mathcal E(f,f)$.)

Now, let $v\in C^2(\mathbb R)$ with $v\ge1$. Suppose we know that $$-\mu\left(\frac{Av}v|g|^2\right)\le\mathcal E(g)\tag1\;\;\;\text{for all }g\in C_c^\infty(\mathbb R).$$ Moreover, assume that there is a $(\zeta_k)_{k\in\mathbb N}\subseteq C_c^\infty(\mathbb R)$ with $$0\le\zeta_k\le\zeta_{k+1}\;\;\;\text{for all }k\in\mathbb N\tag2$$ and $$\zeta_k\xrightarrow{k\to\infty}1\;\;\;\mu\text{-almost surely}\tag3$$ as well as $$\Gamma(\zeta_k)\le\frac1k\;\;\;\text{for all }k\in\mathbb N.\tag4$$

Now, let $f\in C_c^\infty(\mathbb R)$ and $m\in\mathbb R$. How can we show $(1)$ for $g:=f-m$?

The idea is that $g_k:=\zeta_kg$ belongs to $C_c^\infty(\mathbb R)$ (in contrast to $g$) and hence $$-\mu\left(\frac{Av}v|g_k|^2\right)\le\mathcal E(g_k)\tag5\;\;\;\text{for all }g\in C_c^\infty(\mathbb R)\;\;\;\text{for all }k\in\mathbb N.$$ Now, $$|\Gamma(g_k)|\le|\Gamma(g)|+2|g|\sqrt{\Gamma(g)}+g^2\tag6$$ and $$|\Gamma(g_k)-\Gamma(g)|\le|\zeta_k^2-1||\Gamma(g)|+\frac2{\sqrt k}|g_k|\sqrt{\Gamma(g)}+\frac1kg^2\xrightarrow{k\to\infty}0\tag7$$ and hence $$\mathcal E(g_k)\xrightarrow{k\to\infty}\mathcal E(g)\tag8$$ by the dominated convergence theorem.

The left-hand side of $(5)$ seems to be more complicated. Clearly, $\frac{Av}v|g_k|^2\xrightarrow{k\to\infty}\frac{Av}v|g|^2$, but $\frac{Av}v|g_k|^2$ doesn't seem to be dominated by an integrable function.