Paley–Wiener theorem for Volterra resolvents

Remarks

The resolvent gives the solution of the Volterra equation $x + k * x = f$ as $x = f - r * f$. So the theorem decides when the equation is stable 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 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 $(k * r)(t) = a e^{-t} - a e^{-(1 + a) t}$.

Going deeper

For a student

The theorem is due to Paley and Wiener and is the basic tool for the stability of linear Volterra equations of convolution type. 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. Through linearization, the integrability of the resolvent kernel is also the starting point for the stability of nonlinear Volterra equations.