= Solution
For $f,g\in X$, the Lipschitz condition gives
$$
\begin{aligned}
t^{-c}\|\phi(f)(t)-\phi(g)(t)\|
&\leq Mt^{-c}\int_1^t\|f(l)-g(l)\|\,dl\\
&\leq M\|f-g\|_c\,t^{-c}\int_1^t l^c\,dl\\
&=\frac{M}{c+1}(t-t^{-c})\|f-g\|_c\\
&\leq\frac{Mb}{c+1}\|f-g\|_c.
\end{aligned}
$$
Choose $c$ so that $c+1>Mb$. Then $\phi$ is a <contraction mapping>. The space $X$ is complete in the sup norm, and therefore also in the equivalent weighted norm.
For initial value $y_0$, define
$$
\Phi_{y_0}(f)(t)=y_0+\int_1^tF(l,f(l))\,dl.
$$
It has the same contraction constant. The <Banach fixed-point theorem> gives a unique fixed point $f$, which satisfies the integral equation. The <fundamental theorem of calculus> then gives
$$
f'(t)=F(t,f(t)),
\qquad f(1)=y_0.
$$
Conversely every solution satisfies that integral equation, proving uniqueness on $[1,b]$. This weighted-norm argument is the <Bielecki norm method>.
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