= Solution
The space $C([a,b])$ with the <supremum norm> is a <Banach space>. Continuity of the kernel makes $K$ a <bounded linear operator>. Induction on $n$ gives
$$
A^ny=f+Kf+\cdots+K^{n-1}f+K^ny,
$$
so $A^ny-A^nz=K^n(y-z)$. The ordered integration region gives the <factorial bound for a Volterra iterate>:
$$
|(K^ny)(t)|\leq\frac{M^n(t-a)^n}{n!}\|y\|_\infty.
$$
For completeness, if this holds at $n$, then integrating $M\,M^n(s-a)^n/n!$ from $a$ to $t$ gives the bound at $n+1$; the initial case is the defining integral. Thus
$$
\|A^ny-A^nz\|_\infty\leq\alpha_n\|y-z\|_\infty,\qquad \alpha_n=\frac{[M(b-a)]^n}{n!},
$$
and $\sum_{n\geq1}\alpha_n=e^{M(b-a)}-1<\infty$. Part (b) now gives uniform convergence of every iteration sequence to the unique <fixed point>. Since $y=Ay$ is precisely the <Volterra integral equation>, this is its unique continuous solution. No restriction $M(b-a)<1$ is needed: the factorial controls all iterates on the whole finite interval.
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