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This question is a continuation of Is this a contraction mapping for small $T$?

Set, for $T, m>0$, $H^m_T:=\{h:[0,T]\to [0,m]:~ h,~h' \mbox{ are both continuous on } [0,T]\}$ endowed with the norm

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

We aim to show the existence of fixed point(s) of $F$ for $T$ small enough, where $F$ is defined on $H_T$ 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 $G$ is given for $0\le t<s$ and $x,y\in\mathbb R$

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

$\rho: \mathbb R_+\to \mathbb R_+$ is a probability density and $A: \mathbb R_+\to\mathbb R_+$ denotes the inverse of the function

$$\mathbb R_+\ni t\mapsto \int_0^t \big(1+h(r)\big)^2dr\in \mathbb R_+$$

Iosif has shown that $F$ is contracting for small $T$ with $m=1$. Similarly, we can extend the result for any $m>0$. It remains to show the existence of some closed subspace $H$ s.t. $F(H)\subset H$. Does such $H$ exist?

From the physical interpretation, the fixed point of $F$ should be non-increasing and with its derivative vanishing at infinity. So I think a suitable choice should be $H=H_T^m$ for some $m>0$, or $H=\{h\in H_T^m: h'\le 0\}$, or $H=\{h\in H_T^m: \sup_{0\le t\le T}|h'(t)|\le n\}$ for some $n>0$. While I don't know how to show it rigorously.

Any answer, comments or references are appreciated.

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  • $\begingroup$ You may want to consider the set of all nonincreasing absolutely continuous functions $h\colon[0,T]\to[0,\infty)$ with norm $\|h\| := |h(0)| + \text{esssup}_{0<t\le T}|h'(t)|$, where $h'(t)$ is the left derivative of $h$ at $t$. $\endgroup$ Commented Dec 29, 2021 at 0:35
  • $\begingroup$ Thanks Iosif for the nice comment. I believe your feeling is correct, while the difficulty for me is to show that, for a non-increasing function $h$ (which may satisfy some other conditions), $F[h]$ is also non-increasing. Due to the error function $erf$, it is not trivial for me. Anyway I will try to go into details. Feel free to let me know if you get the proof. Thanks so much! $\endgroup$
    – GJC20
    Commented Dec 29, 2021 at 6:04
  • $\begingroup$ I see. I will think about this. $\endgroup$ Commented Dec 29, 2021 at 15:43
  • $\begingroup$ I think knowing more about a priori known properties of the supposed fixed point may help. In particular, the problem seems to have a probabilistic meaning, which may be helpful. $\endgroup$ Commented Dec 29, 2021 at 16:00
  • $\begingroup$ Thank you very kindly for your consideration. Yes. you really have an amazing intuition. The fixed point of $F$ is indeed corresponds to the unique solution $m$ appearing in this post mathoverflow.net/questions/410069/… (while the equation is $dX_t=dt + \sqrt{2} dW_t/(1+m(t))$ instead of $dX_t=dt + dW_t/(1+m(t))$). As we know the existence & uniqueness of the solution, we aim to study further its regularity. Assuming its regularity, we can derive the equation satisfied by $m$. So $h$ should be non-increasing functions and take values in $[0,1]$ $\endgroup$
    – GJC20
    Commented Dec 29, 2021 at 18:00

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