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

$$\|h\| := \max_{0\le t\le T}|h(t)| + \max_{0\le t\le T}|h'(t)|,\quad \forall h\in\mathcal H_T.$$

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. $\|\cdot\|$? 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.