Let's consider a real-valued function $V\in L^\infty(\mathbb{R})$ and a family of operators $\{S(t)\}_{t>0}$ defined on $L^2(\mathbb{R})$ as follows:
$$\qquad \qquad\left(S(t)f \right)(x) = \frac{1}{\sqrt{2\pi t}}\int\limits_\mathbb{R}\exp\left[- \frac{t(V(x)+V(x+y))}{2} - \frac{y^2}{2t}\right]f(x+y)dy \qquad \qquad (1)$$ for each $t\in(0,+\infty)$, $f\in L^2(\mathbb{R})$ and $x\in\mathbb{R}$. For every $t$ this is a linear bounded self-adjoint integral operator that maps $L^2(\mathbb{R})$ to $L^2(\mathbb{R})$. It is also continuous with respect to $t$, i.e. for any $f\in L^2(\mathbb{R})$ the function $(0,\infty)\ni t\longmapsto S(t)f\in L^2(\mathbb{R})$ is continuous. This operator is constructed as a composition of Fourier multiplier $$ (B(t)f)(x)=\frac{1}{\sqrt{2\pi t}} \int\limits_{\mathbb{R}} \exp\left[\frac{-(x-y)^2}{2t}\right]f(y)dy $$ and multiplier $$ (F(t)f)(x) = \exp\left[-\frac{t}{2}V(x)\right]f(x) $$ in such way that $S(t)=F(t)\circ B(t)\circ F(t)$.
The problem is to find a sort of derivative of this family of operators in $0$ on $C_0^\infty(\mathbb{R})$ (smooth functions with compact support), i.e. such operator $L\in \mathcal{L}(L^2(\mathbb{R}))$ that $L\varphi = \lim\limits_{t \to 0}\frac{S(t)\varphi-\varphi}{t}$ for every $\varphi\in C_0^\infty(\mathbb{R})$. The limit is considered in terms of $L^2$-convergence.
If we make an assumption that function $V$ is differentiable almost everywhere and its derivative is bounded on $\mathbb{R}$, than we can show that $$(S(t)\varphi)(x) = \varphi(x)+t\Big(\underbrace{\frac{1}{2}\varphi''(x) - V(x)\varphi(x)}_{=(L\varphi)(x)}\Big) + A(t,x),$$ where $A(t,x) = o(t)$ in $L^2(\mathbb{R})$ (i.e. $\lim\limits_{t\to 0}\frac{1}{t}\|A(t,x)\|_{L^2} = 0$), by representing all of participating functions in form of Taylor polynomials with remainder in Lagrange form and placing it into (1): $$ \varphi(x+\sqrt{t}z) = \varphi(x) + \sqrt{t}z\varphi'(x)+\frac{tz^2}{2}\varphi''(x)+\frac{t\sqrt{t}z^3}{6}\varphi'''(\theta), \qquad \theta \in (x, x+\sqrt{t}z) $$ $$V\left(x+\sqrt{t}z\right)=V(x)+\sqrt{t}zV'(\eta), \qquad \eta \in \left(x, x+\sqrt{t}z\right).$$
$$ \exp\left[-\frac{t}{2}V(x)\right] = 1 - \frac{t}{2}V(x)+\frac{t^2}{8}V^2(x)e^{\xi_1}, \qquad \xi_1 \in \left(0, -\frac{t}{2}V(x)\right). $$ $$ \exp\left[-\frac{t}{2}V\left(x+\sqrt{t}z\right)\right] = 1 - \frac{t}{2}V\left(x+\sqrt{t}z\right) + \frac{t^2}{8} V^2\left(x+\sqrt{t}z\right) e^{\xi_2} = 1 - \frac{t}{2}V(x)-\frac{t\sqrt{t}}{2}zV'(\eta) + \frac{t^2}{8} V^2(x)e^{\xi_2} + \frac{t^2\sqrt{t}}{4}zV(x)V'(\eta)e^{\xi_2} + \frac{t^3}{8}z^2V'(\eta)^2e^{\xi_2}, \\ \xi_2 \in \left(0, -\frac{t}{2}V(x+\sqrt{t}z)\right). $$ Then the result is obtained through tedious process of parentheses removing, calculating some of resulted integrals and estimating the others. Estimation is based on the fact that functions $V$ and $\varphi$ are bounded with their derivatives.
I wonder if there's any way to show this fact without having added constraints on function $V$ such as differentiability and boundedness of its derivative?