Fourier transform method for convolution equations

Remarks

If $1 - \lambda \hat k(\xi) \ne 0$ for every $\xi \in \R$, the method gives the unique solution in $L^2(\R)$. Indeed $\hat k$ is continuous and tends to $0$ at infinity by the Riemann–Lebesgue lemma, so $\abs{1 - \lambda \hat k}$ is bounded below by a positive constant. Then $\hat f / (1 - \lambda \hat k)$ lies in $L^2(\R)$, and the Plancherel theorem makes every step reversible.

Example: $k(s) = e^{-\abs{s}}$, $\lambda = \frac14$, $f(x) = \frac12 e^{-\abs{x}}$. Here $\hat k(\xi) = \frac{2}{1 + \xi^2}$, the symbol $1 - \frac{1}{2(1 + \xi^2)}$ is at least $\frac12$, and step 2 gives $\hat\varphi(\xi) = \frac{1}{1/2 + \xi^2}$, so $\varphi(x) = \frac{1}{\sqrt 2} e^{-\abs{x}/\sqrt 2}$.

Non-example: for $k(s) = \frac12 e^{-\abs{s}}$ and $\lambda = 1$ the symbol $\frac{\xi^2}{1 + \xi^2}$ vanishes at $\xi = 0$. Step 2 then gives a function $\hat f(\xi) (1 + \xi^2)/\xi^2$ that need not be in $L^2(\R)$, so the method does not produce a solution for every $f \in L^2(\R)$.

The method needs the integral to run over the whole group $\R$. On a half-line one gets a Wiener–Hopf equation, which needs the Wiener–Hopf method instead. For equations on $[0, \infty)$ of Volterra type, the Laplace transform method plays the same role.