---
tag: 7T9BZV552E
kind: statement
role: theorem
name: Paley–Wiener theorem for Volterra resolvents
msc: [45M05, "45E10", 45D05]
uses: [DND8CBEVXT, 0V49JEZW06, SASP14MA1Z, 8M012XRPZM]
about: [8M012XRPZM, DND8CBEVXT]
sources:
  - url: https://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem
---

## Statement

Let $n \ge 1$ and let $k \in L^1(\R_+; \C^{n \times n})$ be an integrable matrix-valued kernel on $\R_+ = [0, \infty)$.
Write $(k * r)(t) = \int_0^t k(t - s) \, r(s) \, ds$.
Let $r \in L^1_{\mathrm{loc}}(\R_+; \C^{n \times n})$ be the resolvent of $k$, that is, the unique locally integrable solution of
$$r + k * r = k \quad \text{and} \quad r + r * k = k.$$
Let $\hat k(z) = \int_0^\infty e^{-z t} k(t) \, dt$ be the [[Laplace transform|0V49JEZW06]] of $k$, defined for $\operatorname{Re} z \ge 0$.
Then $r \in L^1(\R_+; \C^{n \times n})$ if and only if
$$\det\big(I + \hat k(z)\big) \ne 0 \quad \text{for all } z \in \C \text{ with } \operatorname{Re} z \ge 0.$$

## Remarks

The resolvent gives the solution of the [[Volterra equation|DND8CBEVXT]] $x + k * x = f$ as $x = f - r * f$.
So the theorem decides when the equation is [[stable|8M012XRPZM]] in the sense that the resolvent is integrable.

Proof idea.
If $r$ is integrable, taking Laplace transforms gives $(I + \hat k(z))(I - \hat r(z)) = I$ for $\operatorname{Re} z \ge 0$, so $I + \hat k(z)$ is invertible.
The converse is the deep part.
It is a Wiener-type theorem for the convolution algebra $L^1(\R_+)$ with a unit adjoined.
In that algebra, an element whose transform never vanishes on the closed right half-plane, including at infinity, is invertible.
The point at infinity causes no trouble, because $\hat k(z) \to 0$ as $\abs{z} \to \infty$ in the closed half-plane.
Compare [[Wiener's lemma|NC768TWZ3D]] for absolutely convergent Fourier series.

Example: let $n = 1$ and $k(t) = a e^{-t}$ with $a \in \C$.
Then $\hat k(z) = a / (z + 1)$, and $1 + \hat k(z) = 0$ only for $z = -1 - a$.
The theorem says that $r$ is integrable if and only if $\operatorname{Re} a > -1$.
Indeed $r(t) = a e^{-(1 + a) t}$, as one checks from $(k * r)(t) = a e^{-t} - a e^{-(1 + a) t}$.

## Explanations

### L1 human

A Volterra equation of convolution type feeds the past of the unknown back into the present through a kernel $k$.
Its solution is obtained from the data by a second kernel, the resolvent.
The theorem tells when the resolvent is integrable, which means that bounded or integrable data give bounded or integrable solutions.
The test is a single condition on the Laplace transform of $k$: the matrix $I + \hat k(z)$ must stay invertible on the closed right half-plane.

### L2 human

The theorem is due to Paley and Wiener and is the basic tool for the stability of linear [[Volterra equations of convolution type|SNFYTC1TTA]].
It is the integral equation analogue of the test for linear ordinary differential equations $x' = A x$, where integrability of $e^{tA}$ holds exactly when all eigenvalues of $A$ have negative real part.
With $r$ integrable, Young's inequality shows that the solution map $f \mapsto f - r * f$ is bounded on $L^p(\R_+)$ for $1 \le p \le \infty$, see [[Lp space|SASP14MA1Z]].
Through linearization, the integrability of the [[resolvent kernel|ARTJV1XCRX]] is also the starting point for the stability of nonlinear Volterra equations.