I'm trying to follow a proof for solving the ODE
$$\frac{df}{dx} = (f(x)+x)x$$
For $0 \leq x \leq 1$ with the initial condition $ f(0)=0$.
The proof I am following goes like this
Define $F:C[0,1]\rightarrow C[0,1]$ as $F(f)(x)=\int_0^x (f(u)+u)u du$ then the solution to the ODE is the fixed point of F. To apply contraction mapping theorem we need to be working in a complete space so we use the suprememum metric.
$d_\infty(F(f_1),F(f_2))=\sup_{t\in [0,1]}|F(f_1)(t)-F(f_2)(t)| $
$=\sup_{t\in [0,1]}|\int_0^t(f_1(u)+u)udu-\int_0^t(f_2(u)+u)udu |$
$=\sup_{t \in [0,1]}|\int_0^t(f_1(u)-f_2(u))udu |$
$\leq \sup_{t \in [0,1]}\int_0^t|f_1(u)-f_2(u)|udu$
$\leq \int_0^1|f_1(u)-f_2(u)|udu$ ($*$)
$\leq \sup_{t\in [0,1]} |f_1(t)-f_2(t)| \int_0^1 udu $ ($**$)
$= \sup_{t \in [0,1]}|f_1(t)-f_2(t)|/2$
$=d_\infty (f_1,f_2)/2$
Okay so firstly couldn't we use = instead of $\leq$ at ($*$)? I know it doesn't really matter in the proof but we have = everywhere so it would seem inconsistent making me doubt my belief we could use a =. If we can't use an = then why not?
My main issue with this proof is ($**$). It looks kind of like a theorem I know
$$|\int_a^bf(x)dx| \leq |b-a| ||f||_\infty$$
And although I believe the step, and it seems nearly intuitive I can't think of a justification for it.
Although I understand the rest of the proof I'll include it for completeness. Since F satisfies the contraction mapping theorem we can use initial condition with the zero function to get
$f_1(x)=F(f_0) = \frac{x^3}{3}$
$F(f_1)=\frac{x^5}{15} + \frac{x^3}{3}$
And in general $$f(x)=\sum_{i=1}^\infty \frac{x^{2i+1}}{(2i+1)!!}$$
Source for this proof is https://greenlees.staff.shef.ac.uk/mas331/MAS331notes7.pdf example 7.15