Contraction mapping principle for integral equations

Remarks

Proof idea: let $T$ be the right side, viewed as a map of $C[a, b]$ with the maximum norm. This space is a complete metric space, by the completeness of the space of bounded continuous functions. The map $T$ sends continuous functions to continuous functions, since $H$ is uniformly continuous on compact sets. The Lipschitz condition gives $\max \abs{T \varphi - T \psi} \le \abs{\lambda} L (b - a) \max \abs{\varphi - \psi}$, so $T$ is a contraction. The Banach fixed point theorem gives the unique fixed point and the convergence of the fixed-point iteration.

Example: the linear case $H(x, s, u) = K(x, s) u$ with a continuous kernel $K$ has $L = \max \abs{K}$, and the theorem gives unique solvability of $\varphi = f + \lambda \int_a^b K \varphi$ for $\abs{\lambda} \max \abs{K} (b - a) < 1$.

The smallness condition matters: for $H(x, s, u) = u$ on $[0, 1]$ and $\lambda = 1$, so $\abs{\lambda} L (b - a) = 1$, the equation $\varphi = f + \int_0^1 \varphi$ has no solution when $\int_0^1 f \ne 0$. For the Volterra form, with $\int_a^x$ in place of $\int_a^b$, the conclusion holds for every $\lambda$, because some power of $T$ is a contraction. The same proof works for complex-valued functions.

Going deeper

For a student

This is the standard first existence theorem for nonlinear Fredholm-type equations of Urysohn form. The error after $k$ steps is at most $q^k (1 - q)^{-1} \max \abs{\varphi_1 - \varphi_0}$ with $q = \abs{\lambda} L (b - a)$, so the method also gives numerical bounds. The same argument, applied on a short time interval, proves the Picard–Lindelöf theorem for ordinary differential equations through their integral form. For larger $\lambda$ one needs other tools, such as Schauder's fixed point theorem or Fredholm theory in the linear case.