Convolution of Hyperfunctions

Abstract

Let us begin with two ordinary functions φ ₁(x) and φ ₂(x) and suppose that their Fourier transforms

$${\psi _1}(\xi ) = F{\phi _1}(x),{\psi _2}(\xi ) = F{\phi _2}(x)$$

(1.1)

exist. The Fourier transform of the product φ ₁(x) · φ ₂(x) is

$$F\{ {\phi _1}(x){\phi _2}(x)\} = {\text{ }}\int_{ - \infty }^\infty {{\phi _1}(x){\phi _2}(x){e^{ - j\xi x}}d\xi .}$$

(1.2)

Since φ ₁(x) is the inverse Fourier transform of ψ(ξ), we have

$${\phi _1}(x) = {F^{ - 1}}{\psi _1}(\xi ) = {\text{ }}\int_{ - \infty }^\infty {\psi 1(t){e^{ - jxt}}dt}$$

(1.3)

Substituting this into the r.h.s. of (1.2) gives

$$\begin{array}{*{20}{c}} {\int_{ - \infty }^\infty {{\phi _2}(x)dx} \int_{ - \infty }^\infty {\psi (t){e^{ - j(\xi - t)x}}dt = {\mkern 1mu} } } \\ { = \int_{ - \infty }^\infty {{\psi _1}(t)dt} \int_{ - \infty }^\infty {{\phi _2}(x){e^{ - j(\xi - t)x}}dx = {\mkern 1mu} } } \\ { = \int_{ - \infty }^\infty {{\psi _1}(t)} {\psi _2}(\xi - t)dt.} \end{array}$$

.