Is this a contraction mapping for small $T$?

Let $$G$$ be the heat kernal, i.e. for $$0\le t and $$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 $$h_1, h_2 \in\mathcal H_T$$,

$$|A_1(t)-A_2(t)|\le 4t\max_{0\le r\le t}|h_1(r)-h_2(r)|,\quad \forall 0\le t\le T.$$

• Do you want to re-check your post, in particular the expression for $F$? The second double integral in that expression does not seem to depend on $h$. Also, what is $m(r)$? Is it $\ge0$? Is it bounded, in any sense? Dec 27, 2021 at 16:53
• Absolutely right! It is $h$ instead of $m$. Now corrected. Many thanks Dec 27, 2021 at 17:13


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, $$0\le t\le T$$, $$0, and $$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.

• Amazing solution! Thanks so much Iosif. As I look for the fixed points of the operator $F$ for small $T$, it suffices to show the existence of some suitable subspace $H$ s.t. $F(H)\subset H$. Do you think it is possible? I post this question independently here mathoverflow.net/questions/412652/… Dec 28, 2021 at 9:12
• Just a typo above equation (2): I think it should be $B_T$ or even $H_T$ instead of $L_T$ (which is not defined) Dec 28, 2021 at 16:13
• @GJC20 : Indeed, I will have to revisit this answer -- will try to find time for that. Dec 28, 2021 at 20:06
• Sure. There is no rush. Anyway I get the idea and succeed in adopting your arguments to calculate the derivative $F[h]'$. However, I am thinking whether there exists a suitable subset s.t. the fixed point theorem. Could you please take a look at my question at mathoverflow.net/questions/412652/… ? Many thanks Dec 28, 2021 at 20:15
• @GJC20 : A reason why the above argument is insufficient is that we also need to bound the derivative of $F[h]'(s)$ in $h$ -- that is, the derivative in $h$ of the derivative of $F[h](s)$ in $s$. Also, having looked at your newer question at mathoverflow.net/questions/412652/…, I think you can use here a slightly simpler norm, as suggested in my comment to that question. Dec 29, 2021 at 0:46