Let $G$ be the heat kernal, i.e. for $0\le t<s$ and $x,y\in\mathbb R$


$$G(t,x;s,y):=\frac{1}{\sqrt{4\pi(s-t)}}\exp\left(-\frac{(y-x)^2}{4(s-t)}\right).$$


For $T>0$, let $\mathcal H\_T:=\{h:[0,T]\to [0,1]:~ h,~h' \mbox{ are both continuous on } [0,T]\}$ be endowed with the norm


$$\|h\|\_T := \max\_{0\le t\le T}|h(t)| + \max\_{0\le t\le T}|h'(t)|,\quad \forall h\in\mathcal H\_T.$$


Define the map $F$ on $\mathcal H\_T$ as follows: $F[h]=\big(F[h](s): 0\le s\le T\big)$ with


$$F[h](s):=\int\_{-s}^{\infty}\left(\int\_0^s G\big(A(u), -u;A(s),y\big)h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A(u)\big)^2} du\right)dy + \int\_{-s}^{\infty}\left(\int\_0^{\infty} G\big(0,x;A(s),y\big)\rho(x)dx\right)dy,$$


where $A: \mathbb R\_+\to\mathbb R\_+$ denotes the inverse of the function


$$\mathbb R\_+\ni t\mapsto \int\_0^t (1+h(r))^2dr\in \mathbb R\_+$$


and $\rho: \mathbb R\_+\to \mathbb R\_+$ is a probability density, i.e.


$$\int\_0^{\infty}\rho(x)dx =1.$$


Can we prove the existence of $T>0$ s.t. $F$ is a contraction w.r.t. $\|\cdot\|\_T$? Any answer, comments or references are appreciated.


PS : I strongly believe the answer is yes, while I am not familiar with the inequalities related to the heat kernal, especially when estimating the derivative $F[h]'$. Bty, $\circ$ stands for the composition of functions.


PS 2 : A straightforward computation yields for all $h\_1, h\_2 \in\mathcal H\_T$,


$$|A\_1(t)-A\_2(t)|\le 4t\max\_{0\le r\le t}|h\_1(r)-h\_2(r)|,\quad \forall 0\le t\le T.$$



Is this a contraction mapping for small $T$?
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\De}{\Delta}\newcommand\R{\mathbb R}$**Edit:** This answer is insufficient, even though (almost) all the reasoning appears relevant to the problem. I will try to come back and fix it.




---


Suppose that $T\in(0,1]$.


Integrating in $y$, simplify the expression for $F$ as follows:
\begin{equation\*}
 F[h](s)=I\_1(h)+I\_2(h), \tag{1}
\end{equation\*}
where
\begin{equation\*}
 I\_1(h):=\int\_0^s \frac{1}{2} \Big(\text{erf}\Big(\frac{s-u}{2 \sqrt{A\_h(s)-A\_h(u)}}\Big)+1\Big) h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A\_h(u)\big)^2}\, du,
\end{equation\*}
\begin{equation\*}
 I\_2(h):=\int\_0^{\infty} \frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A\_h(s)}}\Big)+1\Big)\rho(x)\,dx.
\end{equation\*}
Here the notation $A\_h$ reminds us that $A$ depends on $h$.


Let $H\_T:=\mathcal H\_T$. Let $B\_T$ be the set of all bounded real-valued functions on $[0,T]$, endowed with the sup-norm.


Since $0\le h\le1$, the operator that maps any $h\in H\_T$ to the function $[0,T]\ni u\mapsto\big(1+h(u)\big)^2$ in $L\_T$ is Lipschitz with a universal Lipschitz constant.


Take any $h\_0,h\_1$ in $H\_T$ with
\begin{equation\*}
 \de:=\|h\_1-h\_0\|<1/16. \tag{2}
\end{equation\*}
Letting
\begin{equation\*}
 J\_h(t):=\int\_0^t (1+h(r))^2\,dr,
\end{equation\*}
we see that $J\_h$ is $4$-Lipschitz, $t\le J\_h(t)\le4t$, $A\_h$ is $1$-Lipschitz, $|J\_{h\_1}(t) -J\_{h\_0}(t)|\le8t\de$, and hence
\begin{equation\*}
 \frac u4\le A\_{h\_j}(u)\le\frac{A\_{h\_{1-j}(u)}}{1-8\de}\le\frac u{1-8\de}, \tag{3}
\end{equation\*}
\begin{equation\*}
 |A\_{h\_1}(u)-A\_{h\_0}(u)|\le\frac{8\de u}{1-8\de}\le16\de u; \tag{4}
\end{equation\*}
here in what follows, $0\le t\le T$, $0<u<s\le T$, and $j\in\{0,1\}$, unless specified otherwise.


For brevity, let us refer to the Lipschitz maps with Lipschitz constants depending only on $\de$ as good-Lipschitz. In particular, the map $H\_T\ni h\mapsto A\_h$ is good-Lipschitz. Therefore, the maps $H\_T\ni h\mapsto\big(1+h\circ A\_h(u)\big)^2$ and
\begin{equation\*}
 H\_T\ni h\mapsto h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A\_h(u)\big)^2} \tag{5}
\end{equation\*}
are good-Lipschitz, for each $u\in[0,T]$. The latter map is also bounded by a constant depending only on $\de$; let us refer to such maps as well-bounded.


For $h\_t:=h\_0+t(h\_1-h\_0)$, $\tau\_t:=A\_{h\_t}(s)-A\_{h\_t}(u)$, and $\De\tau:=\tau\_1-\tau\_0$, we have
$0<\frac{s-u}4\le\tau\_t\le s-u$ (since $J\_h$ is $4$-Lipschitz and $A\_h$ is $1$-Lipschitz), $|\De\tau|\le32\de s$ (by (4)), and hence  

the absolute value of the derivative of the map
\begin{equation\*}
 [0,1]\ni t\mapsto 
\dfrac{1}{2} \Big(\text{erf}\Big(\dfrac{s-u}{2 \sqrt{A\_{h\_t}(s)-A\_{h\_t}(u)}}\Big)+1\Big)
\end{equation\*}
is
\begin{equation\*}
\begin{aligned}
& \frac{ s-u }{4 \sqrt{\pi } \tau\_t^{3/2}}\ \exp\Big\{-\frac{(s-u)^2}{4\tau\_t}\Big\}\,|\De\tau| \\ 
& \le\frac1{4^{-1/2} \sqrt{\pi } (s-u)^{1/2}}\ \exp\Big\{-\frac{s-u}4\Big\}\,32\de s. 
\end{aligned} 
\end{equation\*}
The integral in $u\in(0,s)$ of the latter expression is $\le CT$; here and in what follows, $C$ will denote various well-bounded expressions.


Therefore and because $|\text{erf}|\le1$ and the map (5) is good-Lipschitz and well-bounded, we conclude that $I\_1$ is $CT$-Lipschitz.


It is similar but easier to show that $I\_2$ is $CT$-Lipschitz as well.
Indeed, still with $h\_t:=h\_0+t(h\_1-h\_0)$, let $\rho\_t:=A\_{h\_t}(s)$, and $\De\rho:=\rho\_1-\rho\_0$.
Then
$0<\frac s4\le\rho\_t\le s$ (since $J\_h$ is $4$-Lipschitz and $A\_h$ is $1$-Lipschitz), $|\De\rho|\le16\de s$ (by (4)), and hence  

the absolute value of the derivative of the map
\begin{equation\*}
 [0,1]\ni t\mapsto 
\frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A\_{h\_t}(s)}}\Big)+1\Big)
\end{equation\*}
is
\begin{equation\*}
\begin{aligned}
&\frac{|s+x|}{4 \sqrt{\pi } \rho\_t^{3/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho\_t}\Big\}\,|\De\rho| \\ 
&=\frac{|s+x|}{4 \sqrt{\pi } \rho\_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho\_t}\Big\}\,
\frac{|\De\rho|}{\rho\_t} \\ 
&\le\frac{|s+x|}{4 \sqrt{\pi } \rho\_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho\_t}\Big\}\,
64\de\le C; 
\end{aligned} 
\end{equation\*}
so, the integral in $u\in(0,s)$ of the latter expression is $\le CT$. So, $I\_2$ is $CT$-Lipschitz as well.


We conclude that $F$ is $CT$-Lipschitz and hence a contraction if $T$ is small enough.


