How does the integral of pseudo Gaussian kernel on $(0,\infty)$ depend on its variance? 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.
 A: $\newcommand{\si}{\sigma}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}\newcommand{\De}{\Delta}$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}.
