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?](https://mathoverflow.net/questions/419673/how-does-the-integral-of-pseudo-gaussian-kernel-on-0-infty-depend-on-its-va) 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 $(0,\infty)$.



Questions on the integral of pseudo Gaussian kernel and its derivative on $(0,\infty)$
$\newcommand{\De}{\Delta}\newcommand{\vpi}{\varphi}$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).


