I'm trying to solve the below problem to prepare for the mid-term exam.
Let $E=\mathcal C^1([0,1],\mathbb R)$. We define a mapping $T:E \to {\mathbb R}^{[0,1]}$ by $$\forall f \in E, \forall x \in [0,1]:T(f) (x) =1+ \int_{0}^{x} f\left(t-t^{2}\right) \mathrm{d}t$$
Prove that $T[E] \subseteq E$.
Show that $$T \circ T(f)(x)=1+x+\int_{0}^{x} \int_{0}^{t-t^{2}} f\left(u-u^{2}\right) \mathrm{d}u \, \mathrm{d}t$$
Show that $$\|(T \circ T(f))' - (T \circ T(g))' \|_{\infty} \le \frac{1}{4}\|f-g\|_{\infty}$$
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
By Fundamental Theorem of Calculus, we have $T(f)'(x) = f(x-x^2)$. Because both $f$ and $x-x^2$ are continuous, so is $T(f)'$. Hence $T(f) \in E$.
We have $$\begin{aligned} T \circ T(f)(x) &= 1+\int_{0}^{x} \left( 1+ \int_{0}^{t-t^{2}} f\left(u-u^{2}\right) \mathrm{d}u \right ) \mathrm{d}t \\&= 1 + \int_{0}^{x} \mathrm{d}t+ \int_{0}^{x} \int_{0}^{t-t^{2}} f\left(u-u^{2}\right) \mathrm{d}u \, \mathrm{d}t \\ &=1+x + \int_{0}^{x} \int_{0}^{t-t^{2}} f\left(u-u^{2}\right) \mathrm{d}u \, \mathrm{d}t \end{aligned}$$
First, we have $\max_{x\in [0,1]} (x-x^2) = 1/4$. By Fundamental Theorem of Calculus, we have $(T \circ T(f))' (x) = 1 +\int_0^{x-x^2} f(u-u^2) \, \mathrm{d}u$ and $(T \circ T(g))' (x) = 1 +\int_0^{x-x^2} g(u-u^2) \, \mathrm{d}u$. As such, $$\begin{aligned} \|(T \circ T(f))' - (T \circ T(g))' \|_{\infty} &= \max_{x \in [0,1]} \left | \int_0^{x-x^2} \left (f(u-u^2) - g(u-u^2) \right) \, \mathrm{d}u \right| \\ &= \max_{x \in [0,1]} \left | \int_0^{x-x^2} (f-g) \left (u-u^2 \right) \, \mathrm{d}u \right | \\ &\le \max_{x \in [0,1]} \int_0^{x-x^2} \left |(f-g) \left (u-u^2 \right) \right | \mathrm{d}u \\ &= \int_0^{1/4} \left |(f-g) \left (u-u^2 \right) \right | \mathrm{d}u \\ &\le \int_0^{1/4} \max_{x \in [0,1]} |(f-g)(x)| \, \mathrm{d}u \\ &= \int_0^{1/4} \|f-g\|_\infty \, \mathrm{d}u \\&= \frac{1}{4} \|f-g\|_\infty\end{aligned}$$