Questions on the integral of pseudo Gaussian kernel and its derivative on $(0,\infty)$ Consider pseudo Gaussian densities for $0<s<t$ and $x,y\in\mathbb R$
$$f(s,x,t,y):=\frac{1}{\sqrt{2\pi A(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2A(s,t,y)}\right)\quad\mbox{and} \quad g(s,x,t,y):=\frac{1}{\sqrt{2\pi B(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2B(s,t,y)}\right),$$
where $A(s,t,y):=\int_s^t k(u,y)/(1+a(u))du$ and $B(s,t,y):=\int_s^tk(u,y)/(1+b(u))du$. I'm interested in the dependence of the difference $f-g$ on the parameters $a,b$. Assume  $a, b: \mathbb R_+ \to [0,1]$ are continuous, $k: \mathbb R_+\times\mathbb R \to [1,2]$ is $1-$Lipschitz. Does there exist $C>0$ depending only on $T>0$ s.t.
\begin{eqnarray}
\left|\int_0^\infty f(0,0,t,y)dy - \int_0^\infty g(0,0,t,y)dy\right | &\le& Ct^{1/2}\|a-b \|_t,\quad \forall 0<t\le T\quad (\ast)  \\
\int_0^t \left|\int_0^\infty \partial_sf(s,0,t,y)dy - \int_0^\infty \partial_s g(s,0,t,y)dy\right |ds &\le& C(t-s)^{1/2}\|a-b \|_t,\quad \forall 0<s<t\le T\quad (\star),
\end{eqnarray}
where $\|a-b \|_t:=\max_{0\le u\le t}|a(u)-b(u)|$.
PS : $(\star)$ can be shown if
\begin{eqnarray}
\left|\int_0^\infty \partial_sf(s,0,t,y)dy - \int_0^\infty \partial_s g(s,0,t,y)dy\right | \le C(t-s)^{-1/2}\|a-b \|_t.
\end{eqnarray}
Iosif Pinelis has shown in How does the integral of pseudo Gaussian kernel on $(0,\infty)$ depend on its variance? that
$$\left|\int_0^\infty p(x)dx \int_0^\infty f(s,x,t,y)dy - \int_0^\infty p(x)dx\int_0^\infty g(s,x,t,y)dy\right |\le C(t-s)^{1/2}\|a-b \|_t,$$
where $p$ is some probability density on $(0,\infty)$.
 A: $\newcommand{\De}{\Delta}\newcommand{\vpi}{\varphi}$Let
\begin{equation*}
    A:=A(y):=A(0,t,y),\quad B:=B(y):=B(0,t,y),
\end{equation*}
so that
\begin{equation*}
    f(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{A(y)}(y), \quad g(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{B(y)}(y),
\end{equation*}
where
\begin{equation*}
    \vpi_a(u):=\frac1{\sqrt a}\,\exp\Big(-\frac{u^2}{2a}\Big). 
\end{equation*}
Note that
\begin{equation*}
    \frac{\partial}{\partial a}\vpi_a(u)
    =\frac12\Big(\frac{u^2}{a^{5/2}}-\frac1{a^{3/2}}\Big)\exp\Big(-\frac{u^2}{2a}\Big). 
\end{equation*}
So,
\begin{equation*}
    2\sqrt{2\pi}\,\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)]  \tag{1}\label{1}
    =\int_0^1 dv\,I(v),
\end{equation*}
where
\begin{equation*}
    I(v):=I_1(v)-I_2(v), \tag{2}\label{2}
\end{equation*}
\begin{equation*}
    I_1(v):=\int_0^\infty dy\,H(y)
\frac{y^2}{c_v(y)^{5/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big), 
\end{equation*}
\begin{equation*}
    I_2(v):=\int_0^\infty dy\,H(y)
\frac1{c_v(y)^{3/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big), 
\end{equation*}
\begin{equation*}
    H:=B-A, 
\end{equation*}
\begin{equation*}
    c_v:=A+v(B-A). 
\end{equation*}
Next is the crucial step:
\begin{equation*}
    I_1(v)=I_{11}(v)+I_{12}(v), \tag{3}\label{3}
\end{equation*}
where
\begin{equation*}
    I_{11}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\Big(\frac{y}{c(y)}-\frac{y^2c'(y)}{c(y)^2}\Big)
\frac{y}{c(y)^{3/2}}H(y), 
\end{equation*}
\begin{equation*}
    I_{12}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big) 
\frac{y^3c'(y)}{c(y)^{7/2}} H(y), 
\end{equation*}
\begin{equation*}
    c(y):=c_v(y), 
\end{equation*}
and $c'$ is the derivative of the Lipschitz function $c=c_v$; this derivative exists almost everywhere (a.e.), since $k$ is $1$-Lipschitz. Moreover,
\begin{equation*}
    |c'|\le t 
\end{equation*}
a.e.
Note also that for $y\ge0$ and $t>0$
\begin{equation*}
t/2\le c(y)\le2t,\quad  |H(y)|\ll t\,\De a,\quad    |H'(y)|\ll t\,\De a,
\end{equation*}
where
\begin{equation*}
    \De a:=\|a-b\|_t 
\end{equation*}
and $E\ll F$ means that $|E|\le CF$ for some universal real constant $C$.
So,
\begin{equation*}
    |I_{12}(v)|\ll\int_0^\infty dy\,\exp\Big(-\frac{y^2}{4t}\Big) 
\frac{y^3\,t}{t^{7/2}}\, t\,\De a \asymp\sqrt t\,\De a. \tag{4}\label{4}
\end{equation*}
Integrating by parts, we have
\begin{equation*}
\begin{aligned}
    I_{11}(v)&=-\int_0^\infty dy\,\Big[\frac d{dy}\exp\Big(-\frac{y^2}{2c_v(y)}\Big)\Big]
\frac{y}{c(y)^{3/2}}H(y) \\ 
&=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac d{dy}\Big[\frac{y}{c(y)^{3/2}}H(y)\Big] \\ 
&=I_2(v)+I_{111}(v)-\frac32\,I_{112}(v), 
\end{aligned}
\tag{5}\label{5}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
    I_{111}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{y}{c(y)^{3/2}}H'(y), 
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
    I_{112}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{yc'(y)}{c(y)^{5/2}}H(y). 
\end{aligned}
\end{equation*}
Similarly to \eqref{4}, we get
\begin{equation}
    |I_{111}(v)|+|I_{112}(v)|\ll \sqrt t\,\De a. \tag{6}\label{6}
\end{equation}
Collecting \eqref{1}, \eqref{2}, \eqref{3}, \eqref{4}, \eqref{5}, and \eqref{6}, we conclude that
\begin{equation*}
    \Big|\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)]\Big|\ll \sqrt t\,\De a,
\end{equation*}
which proves the first inequality of the two ones in question.
The other inequality can apparently be proved similarly. Since this answer is already rather long and complicated, I will leave the other inequality as an exercise (or to be posted elsewhere).
