Suppose $V:\mathbb{R}^n\to \mathbb{R}$ is a $C^1$ function and let $DV$ the gradient of $V$. Let $A\subset \mathbb{R}^m$ compact and $f:\mathbb{R}^n\times A\to \mathbb{R}^n$ be a function such that $f$ is continuous, bounded (ie. exist $M_f>0$ such that $\left\lVert f(z,a)\right\rVert \leq M_f$) and there exist $L_f>0$ such that $$\left\lVert f(x_1,a)-f(x_2,a)\right\rVert \leq L_f\left\lVert x_1-x_2\right\rVert$$ for all $x_1,x_2 \in \mathbb{R}^n$ and $a\in A$.
Finally let $\alpha:[0,+\infty)\to A$ a measurable function and $y:[0,+\infty)\to \mathbb{R}^n$ such that $$y(t)=x+\int_{0}^{t}f(y(s),\alpha(s))ds \qquad t\geq 0,$$ for a fixed $x \in \mathbb{R}^n$.
I have to prove that for all $t\geq 0$
$$\int_{0}^{t}DV(y(s))\cdot f(y(s),\alpha(s))ds=\int_{0}^{t}DV(x)\cdot f(x,\alpha(s))ds + o(t),\qquad [1]$$
where $\cdot$ is the usual euclidean scalar product and $o(t)$ indicates a function $h(t)$ such that $\lim_{t\to 0^+}\frac{|h(t)|}{t}=0$.
Can anyone help me prove it?
From the hypothesis on $y$ I can estimate that $$\left\lVert y(s)-x\right\rVert=\left\lVert \int_{0}^{s}f(y(\tau),\alpha(\tau))d\tau\right\rVert\leq \int_{0}^{s}\left\lVert f(y(\tau),\alpha(\tau))\right\rVert d\tau\leq \int_{0}^{s}M_f d\tau=sM_f$$ for all $s\geq 0$. I think I have to use some Taylor formula for $V$ in order to prove $[1]$. But I'm not sure how to proceed.
For instance, this statement is used here in Bardi, Capuzzo-Dolcetta with more general hypothesis on $f$.
We begin by adding $0$
$$\int_{0}^{t}\nabla V(y(s))\cdot f(y(s),\alpha(s))ds - \int_{0}^{t}\nabla V(x)\cdot f(x,\alpha(s))ds \\ = \int_{0}^{t} \nabla V(y(s))\cdot f(y(s),\alpha(s)) - \nabla V(y(s))\cdot f(x,\alpha(s)) ds \\ + \int_{0}^{t} \nabla V(y(s))\cdot f(x,\alpha(s)) - \nabla V(x)\cdot f(x,\alpha(s))ds.$$ There is a constant $C > 0$ such that $\|V(y(s))\|\leq C$ for $s \in [0,t]$, so the first integral can be estimated as $$ \int_{0}^{t} \| \nabla V(y(s))\cdot f(y(s),\alpha(s)) - \nabla V(y(s))\cdot f(x,\alpha(s)) \| ds \\ \leq C\int_{0}^{t} \|f(y(s),\alpha(s)) - f(x,\alpha(s))\|ds \\ \leq L_f C\int_{0}^{t} \|y(s) - x\| \leq L_f C M_f \int_{0}^{t} s ds = \mathcal{o}(t). $$ For the other integral, $$ \frac{1}{t}\int_{0}^{t} \|(\nabla V(y(s)) - \nabla V(x))\cdot f(x,\alpha(s)) \| ds \\ \leq \frac{M_f}{t}\int_{0}^{t} \|(\nabla V(y(s)) - \nabla V(x))\| ds \\ \leq M_f \sup_{s\in[0,t]} \|(\nabla V(y(s)) - \nabla V(x))\|. $$ Note that this goes to $0$ as $t \to 0$ because $s \mapsto \nabla V(y(s))$ is continuous.