Is $M_t:=\exp\left(-\int_0^tc(Y_s)\:{\rm d}s\right)$ differentiable at $0$? Let

*

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$E$ be a topological space

*$(Y_t)_{t\ge0}$ be an $E$-valued right-continuous process on $(\Omega,\mathcal A,\operatorname P)$

*$c:E\to[0,\infty)$ be Borel measurable with $$\int_0^tc(Y_s)\:{\rm d}s<\infty\;\;\;\text{for all }t\ge0$$ and $$M_t:=\exp\left(-\int_0^tc(Y_s)\:{\rm d}s\right)\;\;\;\text{for }t\ge0$$
Note that $(M_t)_{t\ge0}$ is a $[0,1]$-valued continuous nondecreasing process on $(\Omega,\mathcal A,\operatorname P)$ with $M_0=0$. We can infer that every path of $(M_t)_{t\ge0}$ is differentiable Lebesgue almost everywhere. But are we able to show that $(M_t)_{t\ge0}$ is differentiable at $0$ and $$\operatorname E\left[\frac{M_t-M_0}t\right]\xrightarrow{t\to0+}\operatorname E\left[\left.\frac{\rm d}{{\rm d}t}M_t\right|_{t=0+}\right]\tag2?$$ If not can we somehow turn the claim to be true?
EDIT: What I actually want to show is $$\operatorname E\left[\frac{M_t-M_0}t\right]\xrightarrow{t\to0+}-\operatorname E\left[c(Y_0)\right]\tag3$$ which might be weaker than what I'm asking for above.
EDIT 2: If we assume that $c$ is continuous, we easily see that $$f_\omega(t):=\int_0^tc(X_s(\omega))\:{\rm d}s\;\;\;\text{for }t\ge0$$ is differentiable at $0$ for all $\omega\in\Omega$. In order to conclude $(2)$, a sufficient condition should be that $c$ is bounded, since then $$\frac{\left|f_\omega(t)\right|}t\le\left\|c\right\|_\infty<\infty\;\;\;\text{for all }t>0\text{ and }\omega\in\Omega\tag4.$$ Using $e^{-x}\ge1+x$ for all $x\in\mathbb R$, we see that $$0\le\frac{M_0-M_t}t\le\left\|c\right\|_\infty$$ and hence Lebesgue`s dominated convergence theorem is applicable. But are there weaker assumptions?
 A: If we only assume $c$ is continuous but don't assume $c$ is bounded, this doesn't hold in general.  To see this, let $c$ be any continuous, integrable, and non-negative function with $\limsup_{t \rightarrow \infty} c(t) = \infty$.
Define the sequences $(a_n)$ and $(b_n)$ by $b_1 = 0$ and
\begin{align*}
b_{n+1} &= e^{\frac{b_n^2}{2}}+b_n, \\
a_n &= \frac{2 N(b_{n+1})-\sqrt{\frac{2}{\pi}}\left(\frac{e^{-(n+1)}}{1-e^{-1}}\right)}{n}, 
\end{align*}
where $N(x)$ is the normal CDF.  Note that $(b_n)$ is increasing and $(b_n) \rightarrow \infty$, and that $(a_n)$ is decreasing (at least, for sufficiently large values of $n$) and $(a_n) \rightarrow 0$.
Since $\limsup_{t \rightarrow \infty}c(t) = \infty$, there exists a sequence $(y_n)$ such that $c(y_n) > \frac{n}{a_{n-1}-a_n}$ for all $n$.  For convenience, we define functions $a, g$ by $a(y) = a_n$ for $y \in [b_n,b_{n+1})$ and $g(y) = y_n$ for $y \in [b_n,b_{n+1})$.
To define our process, let $Y_0 = |X|$ where $X \sim N(0,1)$, and
\begin{align*}
Y_t &= \begin{cases} Y_0 & t \in [0,a(Y_0)) \\
g(Y_0) & t \in [a(Y_0),\infty).
\end{cases}
\end{align*}
We will now show that, for this $Y$, we have $\liminf_{t \rightarrow 0}\mathbb{E}[(M_t-M_0)/t] = -\infty$.  To do this, we will show $\lim_{N \rightarrow \infty} \mathbb{E}[(M_{a_N}-M_0)/a_N] = -\infty$.
Note that, for $Y_0 \in [b_n,b_{n+1})$ and $t \le a_n$ we have $\int_0^t c(Y_s)ds = t c(Y_0)$ and for $t \ge a_n$ we have $\int_0^t c(Y_s)ds = a_nc(Y_0) + (t-a_n)c_n$.  Therefore, we can write $\int_0^t c(Y_s)ds = h(t,Y_0)$, and evaluate $\mathbb{E}[e^{-h(t,Y_0)}]$ by integrating with respect to the normal density.  We let $t = a_N$ and compute
\begin{align*}
\mathbb{E}\left[e^{-h(a_N,Y_0)}\right] &= \sqrt{\frac{2}{\pi}} \int_0^\infty e^{-y^2/2}e^{-h(a_N,y)}dy \\
&= \sqrt{\frac{2}{\pi}} \sum_{n=1}^\infty \int_{b_n}^{b_{n+1}} e^{-y^2/2}e^{-h(a_N,y)}dy \\
&= \sqrt{\frac{2}{\pi}} \left(\sum_{n=1}^{N} \int_{b_n}^{b_{n+1}} e^{-y^2/2}e^{-a_Nc(y)}dy + \sum_{n=N+1}^{\infty} \int_{b_n}^{b_{n+1}} e^{-y^2/2}e^{-(a_n c(y) + (a_N-a_n)c_n)}dy \right) \\
&\le \sqrt{\frac{2}{\pi}} \left(\sum_{n=1}^{N} \int_{b_n}^{b_{n+1}} e^{-y^2/2} dy + \sum_{n=N+1}^{\infty} \int_{b_n}^{b_{n+1}} e^{-b_n^2/2}e^{-(a_N-a_n)c_n}dy \right)  \\
&= \sqrt{\frac{2}{\pi}} \left(\int_{0}^{b_{N+1}} e^{-y^2/2} dy + \sum_{n=N+1}^{\infty} (b_{n+1}-b_n) e^{-b_n^2/2}e^{-(a_N-a_n)c_n} \right) \\
&= 1 - 2 N(b_{N+1}) +  \sqrt{\frac{2}{\pi}} \left(\sum_{n=N+1}^{\infty} e^{-(a_N-a_n)c_n} \right) \\
&\le 1 - 2 N(b_{N+1}) +  \sqrt{\frac{2}{\pi}} \left(\sum_{n=N+1}^{\infty} e^{-(a_{n-1}-a_n)c_n} \right) \\
&\le 1 - 2 N(b_{N+1}) +  \sqrt{\frac{2}{\pi}} \left(\sum_{n=N+1}^{\infty} e^{-n} \right) \\
&= 1 - 2 N(b_{N+1}) +  \sqrt{\frac{2}{\pi}} \left( \frac{e^{-(N+1)}}{1-e^{-1}} \right) \\
&= 1-N \cdot a_N.
\end{align*}
Hence, for sufficiently large $N$, $$\mathbb{E}\left[\frac{M_{a_N}-M_0}{a_N}\right] \le \mathbb{E}\left[\frac{1-Na_N-1}{a_N}\right] = -N.$$  Sending $N \rightarrow \infty$ thus gives that $\liminf_{t \rightarrow 0}\mathbb{E}[(M_t-M_0)/t] = -\infty$ (it's probably easy to show that we can replace the liminf with just lim, but we don't need that).  However, $$\mathbb{E}[c(Y_0)] = \sqrt{\frac{2}{\pi}} \int_0^\infty c(y)e^{-y^2/2}dy \le \sqrt{\frac{2}{\pi}} \int_0^\infty c(y)dy < \infty,$$ so it is not the case that $\liminf_{t \rightarrow 0}\mathbb{E}[(M_t-M_0)/t] = -\mathbb{E}[c(Y_0)]$.
