= Solution
Under $Mh\leq k$, the same <Picard iteration> operator maps all of $C_h$ into itself. It need not be a <contraction mapping> in one step, since $Lh$ can exceed one. Instead repeated integration gives the <factorial estimate for Picard iteration>
$$
|(T^ny)(x)-(T^nz)(x)|\leq\frac{L^n|x-a|^n}{n!}\|y-z\|_\infty,
$$
proved inductively by integrating the preceding bound between $a$ and $x$. Thus
$$
\|T^ny-T^nz\|_\infty\leq\frac{(Lh)^n}{n!}\|y-z\|_\infty.
$$
For some $n$ the coefficient is less than one, so $T^n$ is a contraction. Its unique fixed point $y_*$ is also fixed by $T$: $T^n(Ty_*)=T(T^ny_*)=Ty_*$ and uniqueness forces $Ty_*=y_*$. Conversely any fixed point of $T$ is one of $T^n$. This <iterated contraction> argument gives \b[a unique solution on the full interval $[a-h,a+h]$], without the unjustified extra assumption $Lh<1$.
The uniqueness statement is for solutions in the specified rectangle, where $f$ is given and controlled. The bound $Mh\leq k$ keeps the constructed solution inside that rectangle throughout the interval.
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