Prove that $T$ satisfies the hypotheses of the contraction mapping principle.

Question

Let $T: C([0,1]) \to C([0,1])$ be defined by $Tf(x) = x + \int_{0}^{x}tf(t)\,dt$. Prove that $T$ satisfies the hypotheses of the contraction mapping principle. Show that the fixed point is a solution to the differential equation $f'(x) = xf(x) + 1$.

How is my proof? I'm pretty iffy on it, though it seems right. I am specifically concerned about when I drop the sup term and then reintroduce it.

\begin{align*} d(Tf,Tg) &= \sup_{x\in [0,1]} \Biggl\lvert \biggl( x + \int_{0}^{x}tf(t)\,dt \biggr) - \biggl( x + \int_{0}^{x}tg(t)\,dt \biggr) \Biggr\rvert \\ &= \sup_{x\in [0,1]} \Biggl\lvert \int_{0}^{x} t\,\bigl(f(t)-g(t)\bigr)\,dt \Biggr\rvert \\ &\leq \sup_{x\in [0,1]} \int_{0}^{x} t\, \bigl\lvert f(t)-g(t) \bigr\rvert\,dt \\ &= \int_{0}^{1} t\, \bigl\lvert f(t)-g(t) \bigr\rvert\,dt \\ &\leq \int_{0}^{1} t\, \biggl(\, \sup_{t\in [0,1]} \lvert f(t)-g(t) \rvert \biggr) \, dt \\ &= \int_{0}^{1} t\, d(f,g) \,dt \\ &= d(f,g) \int_{0}^{1} t \,dt \\ &= \frac{1}{2} d(f,g) \end{align*} Now, suppose we have a fixed point $f$. We know that $$ T(f) = T\Bigl(\, \lim_{n \to \infty} T^n(f) \Bigr) = \lim_{n \to \infty} T^{n+1}(f) = f, $$ and so $$ f(x) = T(f(x)) = x + \int_{0}^{x}tf(t)\, dt \quad\implies\quad f'(x) = 1 + xf(x) $$ by the Fundamental Theorem of Calculus and the fact that $f$ is continuous.

Answer 1

All of the inequalities you have stated are correct. One method to justify the inequality \begin{equation} \sup_{x\in [0,1]} \int_{0}^{x}t|f(t)-g(t)| \, dt = \int_{0}^{1}t|f(t)-g(t)| \, dt \end{equation} is to use that the continuous map $t\mapsto t|f(t) - g(t)|$ takes positive values on $[0,1]$ to obtain \begin{equation} \int_{0}^{x}t|f(t) - g(t)| \, dt \leq \int_{0}^{1}t|f(t)-g(t)| \, dt \end{equation} for all $x\in [0,1]$ and then apply the definition of the supremum.

However, making use of a suggestion from a comment below by Ryszard Szwarc, there is a shorter way to obtain the inequality \begin{equation} \|Tf - Tg\|_{\infty} \leq \frac{1}{2} \|f - g\|_{\infty} \tag{$\ast$} \end{equation} for all $f,g\in C[0,1]$. We show this now. Fix $f,g\in C[0,1]$ and $x\in [0,1]$. We have \begin{align*} |(Tf)(x) - (Tg)(x)| &= {} \Big| \int_{0}^{x} t (f(t) - g(t) ) \, dt \Big| \\ &\leq \int_{0}^{x} t |f(t) - g(t)| \, dt \\ &\leq \sup_{s\in [0,x]} |f(s) - g(s)| \int_{0}^{x} t \, dt \\ &\leq \|f - g\|_{\infty} \,{x^2\over 2} \\ &\le \frac{1}{2} \|f - g\|_{\infty} . \end{align*} As $x\in [0,1]$ was arbitrary we deduce $(\ast )$.