How to prove that $(Tz)(t)=\sin(t)+\int_{0}^{t} e^{-s^2}z(se^t)ds$ is contraction mapping. I'm having troubles proving that the map $(Tz)(t)=\sin(t)+\int_{0}^{t} e^{-s^2}z(se^t)ds$ is a contraction mapping for $t\in[0,\infty)$. Here $z$ is a continuous bounded function on $[0,\infty)$.
The metric space here is $M:=BC([0,\infty),\mathbb{R})$ with the metric $d_\infty(x,y)=\sup\limits_{t\in[0,\infty)}|x(t)-y(t)|$. I know this is a complete metric space with this metric.
$T:M\rightarrow M$ and I have already proven that $T(M)\subset M$
I know that $e^{-s^2}$ is bounded on $[0,\infty)$ and so $e^{-s^2}z(se^t)$ is bounded. But I think it doesn't help me to prove that
$\lvert(Tx)(t)-(Ty)(t)\rvert\leq C\lvert x(t)-y(t)\rvert$, for all $x,y \in BC([0,\infty),\mathbb{R})$ and $t\in[0,\infty)$
That is all the information I have about the problem.
 A: What you actually need to prove is that "There exists a constant $0 \le C < 1$ such that $d(Tx,Ty) \le Cd(x,y)$ for all $x,y \in BC([0,\infty),\mathbb{R})$."
This is equivalent to showing that "There exists a constant $0 \le C < 1$ such that $|(Tx)(t)-(Ty)(t)| \le Cd(x,y)$ for all $t \in [0,\infty)$ and all $x,y \in BC([0,\infty),\mathbb{R})$."
Where you might be having difficulty is that the statement "There exists a constant $0 \le C < 1$ such that $|(Tx)(t)-(Ty)(t)| \le C|x(t)-y(t)|$ for all $t \in [0,\infty)$ and all $x,y \in BC([0,\infty),\mathbb{R})$." is actually a stronger statement than what you need to prove, since $|x(t)-y(t)|$ could be much smaller than $d(x,y)$ for some values of $t$.
To prove the statement that you need to prove, start by writing out what $|(Tx)(t)-(Ty)(t)|$ is and attempt to bound it in terms of $d(x,y)$.
\begin{align*}
|(Tx)(t)-(Ty)(t)| &= \left|\left(\sin t + \int_0^te^{-s^2}x(se^t)\,ds \right) - \left(\sin t + \int_0^te^{-s^2}y(se^t)\,ds \right)\right|
\\
&= \left|\int_0^te^{-s^2}x(se^t)\,ds - \int_0^te^{-s^2}y(se^t)\,ds \right|
\\
&= \left|\int_0^te^{-s^2}\left(x(se^t)-y(se^t)\right)\,ds\right|
\\
&\le \int_0^t\left|e^{-s^2}\left(x(se^t)-y(se^t)\right)\right|\,ds
\\
&= \int_0^te^{-s^2}\left|x(se^t)-y(se^t)\right|\,ds
\end{align*}
Can you continue from here?
