Fredholm determinant

Remarks

Example: for $K(x, y) = xy$ on $[0, 1]$, $c_1 = \int_0^1 x^2 \, dx = 1/3$. For $n \ge 2$ the matrix $[x_i x_j]$ has rank at most $1$, so its determinant is $0$ and $c_n = 0$. Hence $D(\lambda) = 1 - \lambda/3$.

Non-example: the definition does not apply when $K$ is only square-integrable, because the diagonal values $K(x, x)$ enter $c_1 = \int_a^b K(x, x) \, dx$ and are not defined for an element of $L^2$. For such kernels one uses a modified determinant in which the diagonal terms are removed.

The convergence for all $\lambda$ follows from Hadamard's inequality: if $\abs{K} \le M$, then $\abs{c_n} \le n^{n/2} M^n (b - a)^n / n!$. The coefficient $c_1$ is the integral $\int_a^b K(x, x) \, dx$, the trace of the kernel, and $D(\lambda) = 1 - c_1 \lambda + \cdots$. For a degenerate kernel $K(x, y) = \sum_{i=1}^{m} a_i(x) b_i(y)$, one has $D(\lambda) = \det(I_m - \lambda A)$, where $A_{ij} = \int_a^b b_i(y) a_j(y) \, dy$.