of all such sets: Ω = ∪ W. Then supp( b ) = R $^{d}$\ Ω . Note that a generalized function has support in a closed set, for example, support of the δ − function is just one point 0. We start with deconvolution for (2). In the deconvolution problem f, or equivalently its Fourier transform, φ, is assumed to be given. Typically, φ is a characteristic function and thus is continuous and bounded. The convolution equation (2) uniquely identifies g for a known f if it can be shown that the corresponding equation (15) has meaning and can be uniquely solved for γ. Next a useful Lemma is proved. This Lemma shows that if a product between a generalized function γ and a continuous function φ is defined: ε = γφ, then division of ε by φ is uniquely defined on supp( φ ) in D $^{∗}$. Lemma 1. Suppose that ε = γφ in D $^{∗}$; φ is a continuous function. Then the generalized function φ − $^{1}$ε is uniquely defined in D ∗ on supp ( φ ) . Proof. Denote W =supp( φ ) and consider D ( W ) the space of all the functions in D with supports restricted to belong to W ; the space D $^{∗}$( W ) is the dual space for D ( W ) . Next, consider a covering of the open set W by bounded sets: W = ∪ W$_{ν}$ where each W$_{ν}$ is an open bounded set. Similarly, consider D ( W$_{ν}$ ) . Then any generalized function in D ∗ can be restricted to the dual space D $^{∗}$( W$_{ν}$ ) . In D $^{∗}$( W$_{ν}$ ) the generalized function γ solves ε = γφ. Suppose that the solution is not unique and there is ˜ γ = γ such that ˜ γφ = ε. Then ( γ − ˜ γ ) φ is defined and represents a zero element in D $^{∗}$( W$_{ν}$ ) . A zero functional can 27