 Solution

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one constant $0\le q<1$. Choose $x_0\in X$ and define $x_{n+1}=T{x}_n$. Successive distances satisfy $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for every $m\ge 1$,Thus $(x_n)$ is a Cauchy sequence. Completeness gives $x_n\to x_{\ast }$. Since a contraction mapping is a continuous function, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ were another fixed point, $d(x_{\ast },y_{\ast })=d(T{x}_{\ast },T{y}_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves both existence and uniqueness.

Now suppose $T=f^k$ is a contraction mapping, where $k$ is a positive integer. Let $y$ be its unique fixed point. Since $f$ commutes with its own iterate, $T(fy)=f(Ty)=fy$. Thus $fy$ is also a fixed point of $T$, so $fy=y$. Any fixed point of $f$ is a fixed point of $T$, and must equal $y$. A contractive iterate therefore gives a unique fixed point of $f$, without assuming continuity of $f$ itself.

The space $X=C([0,1],\mathbb{R})$ is complete for the uniform norm. Indeed, a uniformly Cauchy sequence converges pointwise by completeness of $\mathbb{R}$; passing to the limit in its uniform Cauchy bound shows uniform convergence, and the uniform limit of continuous functions is a continuous function. The operator $F$ maps this space to itself: $s\mapsto \varphi (h(s),s)$ is a continuous function, and its integral plus the continuous $g$ is a continuous function.

For $n=0$ the desired bound is the definition of $\Vert h-k{\Vert }_{\infty }$. Suppose it holds for $n$. The Lipschitz condition and the fact that the same $g$ cancels giveThis proves the estimate by induction. Taking the supremum over $0\le t\le 1$, the Lipschitz constant of $F^n$ is at most $M^n/n!$. These constants tend to zero, since the ratio of successive terms is $M/(n+1)$. Some iterate is consequently a contraction. The preceding result gives a unique fixed point of $F$, which is exactly $\boxed{\text{the unique continuous solution of the integral equation on }[0,1]}$. No assumption $M<1$ is needed.