 Solution

Under $Mh\le k$, 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 iterationproved inductively by integrating the preceding bound between $a$ and $x$. ThusFor some $n$ the coefficient is less than one, so $T^n$ is a contraction. Its unique fixed point $y_{\ast }$ is also fixed by $T$: $T^n(T{y}_{\ast })=T(T^n{y}_{\ast })=T{y}_{\ast }$ and uniqueness forces $T{y}_{\ast }=y_{\ast }$. Conversely any fixed point of $T$ is one of $T^n$. This iterated contraction argument gives 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 $Mh\le k$ keeps the constructed solution inside that rectangle throughout the interval.