To begin with it is necessary to discuss the meaning of ϕf where f ∈G^{∗} and ϕ ∈G. What should it mean?
First suppose f ∈ L^{p}

(ℝn)

or measurable with polynomial growth. Then ϕf also has these properties.
Hence, it should be the case that ϕf

(ψ)

= ∫_{ℝn}ϕfψdx = ∫_{ℝn}f

(ϕψ)

dx. This motivates the following
definition.

Definition 10.5.24Let T ∈G^{∗}and let ϕ ∈G. Then ϕT ≡ Tϕ ∈G^{∗}will be defined by

ϕT (ψ ) ≡ T (ϕ ψ).

The next topic is that of convolution. It was just shown that

n∕2 n∕2
F (f ∗ϕ ) = (2π) F ϕF f,F −1(f ∗ϕ) = (2π) F−1ϕF −1f

whenever f ∈ L^{2}

(ℝn )

and ϕ ∈G so the same definition is retained in the general case because it makes
perfect sense and agrees with the earlier definition.

Definition 10.5.25Let f ∈G^{∗}and let ϕ ∈G. Then define the convolution of f with an element of G asfollows.

n∕2 −1 ∗
f ∗ϕ ≡ (2π ) F (FϕF f) ∈ G

There is an obvious question. With this definition, is it true that F^{−1}

(f ∗ϕ)

=

(2π)

^{n∕2}F^{−1}ϕF^{−1}f as it
was earlier?

Theorem 10.5.26Let f ∈G^{∗}and let ϕ ∈G.

F (f ∗ϕ ) = (2π)n∕2F ϕFf, (10.20)

(10.20)

−1 n∕2 −1 −1
F (f ∗ ϕ) = (2π) F ϕF f. (10.21)

(10.21)

Proof: Note that 10.20 follows from Definition 10.5.25 and both assertions hold for f ∈G. Consider
10.21. Here is a simple formula involving a pair of functions in G.