Liouville–Neumann series

Remarks

Proof idea: $\abs{K_n} \le M^n (b - a)^{n-1}$, so the $n$-th term is bounded by $\max \abs{f} \cdot (\abs{\lambda} M (b - a))^n$, and the Weierstrass M-test gives uniform convergence. Substituting the series into the equation and integrating term by term shows that $\varphi$ is a solution. If $\psi$ is a second continuous solution, then $\max \abs{\varphi - \psi} \le \abs{\lambda} M (b - a) \max \abs{\varphi - \psi}$, so $\varphi = \psi$. This is the Neumann series $(I - \lambda K)^{-1} = \sum_{n \ge 0} \lambda^n K^n$ for the integral operator on continuous functions with the maximum norm.

Example: for $K = 1$ on $[0, 1]$ and $\abs{\lambda} < 1$, all $K_n = 1$, and the series gives $\varphi = f + \frac{\lambda}{1 - \lambda} \int_0^1 f$. The condition on $\lambda$ cannot be dropped in general: for $\lambda = 1$ the equation $\varphi = f + \int_0^1 \varphi$ has no solution when $\int_0^1 f \ne 0$. For Volterra kernels the bound improves to $\abs{K_n(x, s)} \le M^n (x - s)^{n-1} / (n-1)!$, and the series converges for every $\lambda$.

Going deeper

For a student

The partial sums of the series are the iterates $\varphi_0 = f$, $\varphi_{k} = f + \lambda K \varphi_{k-1}$ of the method of successive approximations. Collecting the terms gives the resolvent kernel $R = \sum_{n \ge 1} \lambda^{n-1} K_n$ and the formula $\varphi = f + \lambda \int R f$. The series in $\lambda$ converges for $\abs{\lambda}$ below the smallest absolute value of a characteristic value of $K$, and for all $\lambda$ if there is none. Beyond that disc one needs Fredholm determinants or the theory of compact operators.