---
tag: 4W0EG94RZ5
kind: statement
role: theorem
name: Fredholm series for the resolvent
msc: [45B05]
uses: [ZM5G144PZG, PPX0AK81QV, HP5VJ9EBST, DEYYNVCV9J, ARTJV1XCRX, DC7HMHVREP]
about: [ARTJV1XCRX, PPX0AK81QV, HP5VJ9EBST]
sources:
  - url: https://en.wikipedia.org/wiki/Fredholm_theory
---

## Statement

Let $a < b$ and let $K \colon [a, b] \times [a, b] \to \C$ be a [[continuous kernel|ZM5G144PZG]].
Let $D(\lambda)$ be its [[Fredholm determinant|PPX0AK81QV]] and $D(x, y; \lambda)$ its first [[Fredholm minor|HP5VJ9EBST]].
Let $\lambda \in \C$ with $D(\lambda) \ne 0$, and put
$$R(x, y; \lambda) = \frac{D(x, y; \lambda)}{D(\lambda)}, \qquad x, y \in [a, b].$$
Then:

1. $R(\cdot, \cdot; \lambda)$ is continuous on $[a, b]^2$ and satisfies
$$R(x, y; \lambda) - K(x, y) = \lambda \int_a^b K(x, t) R(t, y; \lambda) \, dt = \lambda \int_a^b R(x, t; \lambda) K(t, y) \, dt;$$
2. for every continuous $f \colon [a, b] \to \C$, the [[Fredholm equation of the second kind|DEYYNVCV9J]] $\varphi(x) - \lambda \int_a^b K(x, y) \varphi(y) \, dy = f(x)$ has exactly one continuous solution, namely
$$\varphi(x) = f(x) + \lambda \int_a^b R(x, y; \lambda) f(y) \, dy;$$
3. $\lambda$ is not a [[characteristic value|DC7HMHVREP]] of $K$.

So $R$ is the [[resolvent kernel|ARTJV1XCRX]] of $K$, and it is a meromorphic function of $\lambda$ whose poles lie among the zeros of $D$.

## Remarks

The proof rests on two identities between entire functions of $\lambda$.
They come from expanding the determinants in the series along the first row or the first column:
$$D(x, y; \lambda) = K(x, y) D(\lambda) + \lambda \int_a^b K(x, t) D(t, y; \lambda) \, dt,$$
and the same with $K$ and $D$ exchanged in the integral.
Dividing by $D(\lambda)$ gives part 1, and part 1 shows by direct substitution that the formula in part 2 solves the equation.
Uniqueness comes from the same identities applied to a solution of the homogeneous equation, which must then be $0$.

Example: for $K(x, y) = xy$ on $[0, 1]$, $D(\lambda) = 1 - \lambda/3$ and $D(x, y; \lambda) = xy$.
So $R(x, y; \lambda) = \frac{3xy}{3 - \lambda}$ for $\lambda \ne 3$.
For $|\lambda| < 3$ this agrees with the sum $\sum_{n \ge 0} \lambda^n xy / 3^n$ of the [[Liouville–Neumann series|H1HWSWME6B]].

The Neumann series converges only for small $\abs{\lambda}$, while Fredholm's formula is valid for every $\lambda$ with $D(\lambda) \ne 0$.

## Explanations

### L1 human

This is Cramer's rule for integral equations.
The solution of a linear system is a ratio of determinants, and here the solution kernel is a ratio of two everywhere convergent power series in $\lambda$.
The formula works as long as the denominator does not vanish.

### L2 human

The result turns the [[Fredholm equation of the second kind|DEYYNVCV9J]] into an explicit computation whenever $D(\lambda) \ne 0$.
It extends the [[resolvent kernel|ARTJV1XCRX]] beyond the disc where the Neumann series converges, as a meromorphic function of $\lambda$.
At a zero $\lambda_0$ of $D$, the homogeneous equation has nonzero solutions, and these can be read off from the first Fredholm minor that does not vanish identically at $\lambda_0$.
In operator language, $R$ is the kernel of $\lambda^{-1}((I - \lambda T_K)^{-1} - I)$, where $T_K$ is the integral operator with kernel $K$.