Let $a, b: \mathbb R\_+ \to [0,1]$ be continuous functions. Let $k: \mathbb R\_+\times\mathbb R \to [1,2]$ be $1-$Lipschitz. Set, for $0<s<t$ and $y>0$,


$$A(s,t,y):=\int\_s^t\frac{k(u,y)}{1+a(u)}du \quad\mbox{and} \quad B(s,t,y):=\int\_s^t\frac{k(u,y)}{1+b(u)}du.$$


Define further


$$f(s,x,t,y):=\frac{1}{\sqrt{2\pi A(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2A(s,t,y)}\right)\quad\mbox{and} \quad g(s,x,t,y):=\frac{1}{\sqrt{2\pi B(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2B(s,t,y)}\right).$$


Given a probability density $p$ on $(0,\infty)$ which can be assumed to be as good as possible, can we show the existence of some $C>0$ (depending only on $p$) s.t.


$$\left|\int\_0^\infty p(x)dx \int\_0^\infty f(s,x,t,y)dy - \int\_0^\infty p(x)dx\int\_0^\infty g(s,x,t,y)dy\right |\le C(t-s)^{1/2}\|a-b \|\_t,$$


where $\|a-b \|\_t:=\max\_{0\le u\le t}|a(u)-b(u)|$.


PS : It is straightforward that $|A(s,t,y)-B(s,t,y)|\le C(t-s)\|a-b \|\_t$ for some $C>0$. If $k$ is independent of $y$, i.e. $k(u,y)\equiv k(u)$, then $f,g$ are both Gaussian densities and the above inequality holds by computation.



How does the integral of pseudo Gaussian kernel on $(0,\infty)$ depend on its variance?
$\newcommand{\si}{\sigma}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}\newcommand{\De}{\Delta}$Let us show that the desired bound holds if
\begin{equation\*}
 \int\_0^\infty dx\,|p'(x)|<\infty, \tag{1}\label{1}
\end{equation\*}
which in particular implies that there exists the limit
\begin{equation\*}
 p\_0:=p(0+)\in[0,\infty). \tag{2}\label{2}
\end{equation\*}
One may note that, for condition \eqref{1} to hold, it is enough that e.g. the density $p$ be continuously differentiable and bounded on $(0,\infty)$ with only finitely many modes.


Let
\begin{equation\*}
 A:=A(y):=A(s,t,y),\quad B:=B(y):=B(s,t,y),
\end{equation\*}
\begin{equation\*}
 h(x,y):=h(s,x,t,y):=f(s,x,t,y)-g(s,x,t,y),
\end{equation\*}
\begin{equation\*}
 I:=\int\_0^\infty dx\,p(x) \int\_0^\infty dy\,h(x,y). \tag{3}\label{3}
\end{equation\*}


We want to show that
\begin{equation\*}
 |I|\ll(t-s)^{1/2}\De a, \tag{$\clubsuit$}\label{\*}
\end{equation\*}
where $E\ll F$ means that $|E|\le cF$ for some real constant $c$ depending only on $p$ and
\begin{equation\*}
 \De a:=\|a-b\|\_t. 
\end{equation\*}


Let $\Phi$ and $\vpi$ denote the standard normal cdf and pdf, respectively.
By \eqref{3} and \eqref{1},  

\begin{equation\*}
\begin{aligned}
 I&=\int\_0^\infty dx\,\Big(p\_0+\int\_0^x d\xi\,p'(\xi)\Big) \int\_0^\infty dy\,h(x,y) \\ 
 &=p\_0 J+K, 
\end{aligned}
\tag{4}\label{4}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
 J&:=\int\_0^\infty dy\,\int\_0^\infty dx\, h(x,y), \\ 
 K&:=\int\_0^\infty dx\,\int\_0^x d\xi\,p'(\xi) \int\_0^\infty dy\,h(x,y) \\ 
 &{\color{red}{\,\,=}}\int\_0^\infty dy\,\int\_0^\infty d\xi\,p'(\xi)\int\_\xi^\infty dx\, h(x,y) \\ 
 &=\int\_0^\infty dy\,\int\_0^\infty d\xi\,p'(\xi)
 \Big[\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
 -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big] \\ 
 &=\int\_0^\infty d\xi\,p'(\xi)
 \int\_0^\infty dy\,
 \Big[\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
 -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big]. 
\end{aligned}
 \tag{5}\label{5}
\end{equation\*}
The red equality in \eqref{5} holds by the Fubini theorem -- which is the crucial point of the entire proof, as it allows one to deal, instead of $\big|\vpi\big(\frac z\si\big)'\_\si\big|$, with $\big|\Phi\big(\frac z\si\big)'\_\si\big|$ as in \eqref{!} below and thus get the crucial additional factor $|z|$ in the numerators there, which alleviates the possible smallness of the denominator $t-s$ of the ratio $\frac{2|z|}{t-s}$ in \eqref{!}.


Note that $\{A,B\}\subset[\frac{t-s}2,2(t-s)]$, and hence for any real $z$ and any $\si$ between $\sqrt A$ and $\sqrt B$ we have
\begin{equation\*}
 \Big|\Phi\Big(\frac z\si\Big)'\_\si\Big|
 =\frac{|z|}{\si^2}\,\vpi\Big(\frac z\si\Big)
 \le\frac{2|z|}{t-s}\,\vpi\Big(\frac z{\sqrt{2(t-s)}}\Big) \tag{$\heartsuit$}\label{!}
\end{equation\*}
and
\begin{equation\*}
 |\sqrt A-\sqrt B|=\frac{|A-B|}{\sqrt A+\sqrt B}\ll (t-s)^{1/2}\De a, 
\end{equation\*}
so that (by, say, the mean value theorem)
\begin{equation\*}
 \Big|\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
 -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big|
 \ll\frac{|y-\xi|}{t-s}\,\vpi\Big(\frac{y-\xi}{\sqrt{2(t-s)}}\Big) (t-s)^{1/2}\De a 
\end{equation\*}
and  

\begin{equation\*}
\begin{aligned}
&\int\_0^\infty dy\,
 \Big|\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
 -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big| \\ 
&\ll\int\_{-\infty}^\infty dy\,
 \frac{|y-\xi|}{t-s}\,\vpi\Big(\frac{y-\xi}{\sqrt{2(t-s)}}\Big) (t-s)^{1/2}\De a \\ 
&\ll(t-s)^{1/2}\De a. 
\end{aligned}
\end{equation\*}


So, by \eqref{5} and \eqref{1},
\begin{equation\*}
 |K|\ll (t-s)^{1/2}\De a. \tag{6}\label{6}
\end{equation\*}
Similarly and a bit easier, we get
\begin{equation\*}
 |J|\ll (t-s)^{1/2}\De a. \tag{7}\label{7}
\end{equation\*}
Now \eqref{\*} follows from \eqref{4}, \eqref{2}, \eqref{6}, and \eqref{7}.


