Show the mapping below is a contraction $E=C([0,1],\mathbb{R})$ is a Banach space with respect to norm
$$
||f||=\sup_{0\le x\le 1}|f(x)|e^{-Mx}\\
$$
for $f\in E$, and $M>0$ is a fixed real number.
The question is to show one can choose the value of $M$ such that:
$$
T:E\to E,\ T(f)(x)=\alpha+\int_0^x af(t^b) \text{d}t\\
\alpha\in\mathbb{R},a>0,b>1
$$
is a contraction mapping.
 A: For any $f\in E$ also $Tf\in E$ by the fundamental theorem of calculus.
By definition of the norm for any $f,g\in E$ it holds:
$$
\|T(f)-T(g)\|_E= \sup _x |\int \limits_0^x a (f(t^b)-g(t^b))|dt e^{-Mx}\leq \sup_x a \int \limits_0^{x^b} |f(u)-g(u)|b u^{b-1}du e^{-Mx} 
$$ with the substitution $u=t^b$ and using that $b>1.$ Now using the non-negativity of the inntegrand and the fact that $u^{b-1}\leq 1$ for $u\in [0,1]$, we get
$$
\|T(f)-T(g)\|_E\leq a \cdot b\sup\limits_{x\in [0,1]} \sup \limits_{u\in [0,1]} |f(u)-g(u)| e^{-Mu}\cdot e^{Mu-Mx}\leq a\cdot b\cdot e^{M}\cdot \|f-g\|_E  
$$ where we used that $e^{Mu}, e^{-Mx}$ takes its maximum at $1, 0$ respectively.
Hence $T$ is a contraction if $a\cdot b \cdot e^M< 1.$
A: We have
$$\displaylines{|T(f)(x)-T(g)(x)|\le a\int\limits_0^x |f(t^b)-g(t^b)|\,dt \\ =
a\int\limits_0^x |f(t^b)-g(t^b)|e^{-Mt^b}\,e^{Mt^b}\,dt \le a\|f-g\|\,\int\limits_0^x e^{Mt^b}\,dt}
$$
Thus $$\displaylines{|T(f)(x)-T(g)(x)|\,e^{-Mx}\le a\|f-g\|\,\int\limits_0^xe^{M(t^b-t)}\,dt \\ \le  a\|f-g\|\,\int\limits_0^1e^{M(t^b-t)}\,dt}
$$
and
$$\|T(f)-T(g)\|\le a I_{b,M}\ \|f-g\|\quad I_{b,M}:=\int\limits_0^1e^{-M(t-t^b)}\,dt$$
The function $t-t^b$ is concave (the second derivative is negative), hence
$$t-t^b\ge \begin{cases} (1-2^{1-b})t & 0\le t\le {1\over 2}\\  (1-2^{1-b})(1-t) & {1\over 2}\le t\le 1
\end{cases}
$$
Thus
$$\displaylines{I_{b,M} \le \int\limits_0^{1/2}e^{-M(1-2^{1-b})t}\,dt + \int\limits_{1/2}^1e^{-M(1-2^{1-b})(1-t)}\,dt \\ =2\int\limits_0^{1/2}e^{-M(1-2^{1-b})t}\,dt 
\le 2\int\limits_0^{\infty}e^{-M(1-2^{1-b})t}\,dt =
{2\over M(1-2^{1-b})}}$$
Therefore  the operator $T$ is a contraction for
$$ M>{2a\over 1-2^{1-b}} $$
