Resolvent kernel

Remarks

Example: $K(x, s) = x s$ on $[0, 1]$, so $M = 1$ and $K_n = x s / 3^{n-1}$. For $\abs{\lambda} < 1$ the series is $x s \sum_{n \ge 1} (\lambda/3)^{n-1} = 3 x s / (3 - \lambda)$.

Non-example: for $K = 1$ on $[0, 1]$ and $\lambda = 1$, all $K_n = 1$ and the series $\sum 1$ diverges. This $\lambda$ violates the condition $\abs{\lambda} M (b - a) < 1$, and indeed the homogeneous equation $\varphi = \int_0^1 \varphi$ has the nonzero solution $\varphi = 1$.

The solution of the integral equation of the second kind $\varphi(x) = f(x) + \lambda \int_a^b K(x, s) \varphi(s) \, ds$ is then $\varphi(x) = f(x) + \lambda \int_a^b R(x, s; \lambda) f(s) \, ds$. In operator terms, $R = K (I - \lambda K)^{-1}$. By Fredholm theory, $R$ extends to a meromorphic function of $\lambda$ on $\C$, the quotient of two entire functions. The resolvent kernel is then defined at every $\lambda$ that is not a pole. The example shows that the extension can go beyond the disc of the definition: $3 x s / (3 - \lambda)$ has its only pole at $\lambda = 3$.