 Solution

Use Picard iteration on $[a-h,a+h]$:By induction, every iterate stays in the prescribed rectangle: if $u_n$ does, then $|u_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Continuity makes each next integral well defined. The Lipschitz condition gives, for $n\ge 0$,For $n=0$ this is the previous bound; integration of the bound multiplied by $A$ proves the next step, on either side of $a$.

The sum of these bounds converges uniformly on the entire interval. Thus $u_n$ converges uniformly to a continuous $u$, whose graph remains in the closed rectangle. The Lipschitz condition also makes $f(t,u_n(t))$ converge uniformly to $f(t,u(t))$, so passage to the limit givesThe fundamental theorem of calculus makes $u$ a continuously differentiable solution with $u(a)=b$.

For uniqueness, let $u,v$ be two solutions and put $D_0=\sup |u-v|$ on the interval. Repeated use of the Lipschitz condition in their integral equations givesfor every $n$. Letting $n\to \infty$ gives $u=v$. The global bound $M$ prevents any competing solution whose graph lies in $D$ from exiting the rectangle through a horizontal boundary before the end of the interval. Therefore a unique solution exists on the full stated interval. No extra restriction $Ah<1$ is needed; the factorial estimates replace that restriction.