Question :
Let $a$,$b$ be real numbers with $0\lt b\lt 1$ . Consider the subset $X\subset C[0,b]$ consisting of the functions $f$ s.t $f(0)=a$ .Then $X$ is closed in $C[0,b]$. Define , $$Tf(x) := a+\int_0^x |f(t)|dt ,(0\le x\le b).$$
Prove that $T$ is contraction of $x$ and there is a unique $f\in X$ satisfying $$f'=|f| $$ on $(0,b)$
My attempt :
$$T \ :\ X\subset C[0,b] \rightarrow \mathbb R$$ To prove contraction , I have to show $$|Tf(x)-Tg(x)|\lt ||f(x)-g(x)||_{\infty}$$ where $||\ .\ ||_{\infty}$ is the supremum norm on $C[0,b]$.
$$|Tf(x)-Tg(x)|=|(\int_0^x |f(t)|dt-\int_0^x |g(t)|dt)|$$ $$= |\int_0^x(|f(t)|-|g(t)|)dt\ \ |$$ $$\le |\ \int_0^x |f(t)-g(t)|dt\ |$$ $$\le \int_0^x |f(t)-g(t)|dt $$ $$\le sup_{t\in [0,x]} |f(t)-g(t)|.x$$ $$\lt ||f(t)-g(t)||_\infty ;\ \ as\ \ x\le b\lt 1\ $$
So this is a contraction . But I don't know what "contraction of x " means .
As for the second part , by Banach Contraction Principle , there is a $f\in X$ s.t $$f(x)=a+\int_0^x |f(t)|dt,(0\le x\le b)$$ $$or \ \ ,\ \ f'(x)=|f(x)|$$
Is my proof allright $?$
I am a little skeptical about the last step ,if it is correct or not . Correct me if I have made other mistakes also.
Thanks for any help.