Characteristic value of a kernel

Remarks

A characteristic value is never $0$, since for $\lambda = 0$ the equation forces $\varphi = 0$. So $\lambda$ is a characteristic value exactly when $1/\lambda$ is an eigenvalue of the integral operator $T_K$, and the nonzero solutions $\varphi$ are the eigenfunctions for $1/\lambda$.

Example: for $K(x, y) = xy$ on $[0, 1]$, the number $\lambda = 3$ is a characteristic value with solution $\varphi(x) = x$, since $3 \int_0^1 x y \cdot y \, dy = 3 \cdot x/3 = x$.

Non-example: for the same kernel, $\lambda = 1$ is not a characteristic value. Every solution has the form $\varphi(x) = c x$ with $c = \int_0^1 y \varphi(y) \, dy = c/3$, so $c = 0$ and $\varphi = 0$.

Many classical texts call $\lambda$ an eigenvalue of the kernel; the name characteristic value avoids confusion with the eigenvalue $1/\lambda$ of the operator. A kernel may have no characteristic value at all: the Volterra kernel, equal to $1$ for $y \le x$ and $0$ otherwise, is an example.