converges to zero and well-posedness obtains without any restrictions on γ . This result could also extend to infinitely differentiable φ with some zeros and applies in the univariate deconvolution to error distributions such as the uniform and triangular (see Schwartz, pp. 123-125). 

Below we provide cases of well-posed deconvolution where less restrictive differentiability conditions are imposed on φ ; this would require additional conditions on the γ ’s. The nature of the conditions is to ensure that the product pair is defined for the Fourier transforms; for a continuous φ this requires a trade-off between the degree of singularity of γ (and correspond- ingly, ε ) and the differentiability of φ, these trade-offs can occur locally as in the example where γ is the derivative of a δ − function and φ is continuously differentiable at 0, but to streamline the proofs we consider global restrictions in the product pairs. 

First, suppose that both φ and γ are continuos functions, then ε is con- tinuous as well. Let Γ be a subspace of all continuous functions on R d such that all γ ∈ Γ belong in a bounded set in S $^{∗}$; then for some ¯ m we have Γ = S ∗ 0 , ¯ $_{m}$( V ) (implying ( (1 + t $^{2}$) − 1 ) ¯ $^{m}$| γ | < V < ∞ ) . Then for any bounded continuous φ the product ε = γφ ∈ S ∗ 0 , ¯ $_{m}$( V ) . 

## Lemma 2. Suppose that γ, γ$_{n}$ ∈ S ∗ 0 , ¯ $_{m}$( V ) , φ is a bounded continuous function; ε$_{n}$ = γ$_{n}$φ, ε = ε$_{0}$ = γφ, 0 ≤ n < ∞ and φ − 1 is a continuous function that satisfies (13). Then if ε$_{n}$ → ε$_{0}$ as n → ∞ in S ∗ we get that γ$_{n}$ = φ − $^{1}$ε$_{n}$ converges to γ = φ − $^{1}$ε$_{0}$ . 

Proof. First note that for any number ξ > 0 for any ψ ∈ S there exists 

