(b) all κ$_{kn}$ = ( φ$_{n}$ ) ′ k φ − 1 n , and κ$_{k}$ = ( φ ) ′ k φ − 1 are such that ( κ$_{kn}$ , ε$_{1}$$_{n}$ ) , ( κ$_{k}$ , ε$_{1}$ ) belong to some product pair, the products ε$_{n}$φ − 1 n exist in S ∗ and F t − $^{1}$( ε$_{1}$$_{n}$φ − 1 n ) → g in S $^{∗}$. Proof. (a) We have that φ, φ$_{n}$ ∈ O$_{M}$ . It follows that ( φ ) ′ k , ( φ$_{n}$ ) ′ k ∈ O$_{M}$ . Also ( φ ) ′ k , ( φ$_{n}$ ) ′ k ∈ Φ( m $^{′}$, V ) , where m ′ = m + ι, with ι a vector of ones. From Theorem 3 it follows that for every n the functions γ$_{n}$ and φ$_{n}$ are uniquely identified on W. From now on we consider all functions and function spaces restricted to W, even when W does not coincide with R $^{d}$, but keep the same notation. The functions belong also to Φ( ˜ m, V ) where ˜ m = m + m $^{′}$. Without loss of generality assume that each κ$_{k}$ is also in the same Φ( ˜ m, V ) , and so all κ$_{kn}$ , κ$_{k}$ are in a bounded set in S $^{∗}$. Since from condition ( a ) it follows that κ$_{kn}$ = ( φ$_{n}$ ) ′ k φ − 1 n ∈ O$_{M}$ , products are defined and from equations ε$_{1}$$_{n}$ κ$_{kn}$ − (( ε$_{1}$$_{n}$ ) ′ k − iε$_{2}$$_{kn}$ ) = 0 and convergence of ε$_{in}$ to ε$_{i}$ we get that ε$_{1}$$_{n}$ κ$_{kn}$ − ε$_{1}$ κ$_{k}$ converges to zero in S $^{∗}$. For functions in O$_{M}$ products with any elements from S ∗ exist, thus ε$_{1}$$_{n}$ κ$_{kn}$ − ε$_{1}$ κ$_{kn}$ exists; moreover ( ε$_{1}$$_{n}$ − ε$_{1}$ ) κ$_{kn}$ converges to zero in S ∗ by the hypocontinuity property (Schwartz, p.246). It follows that ε$_{1}$ ( κ$_{kn}$ − κ$_{k}$ ) converges to zero in S $^{∗}$. Since ε$_{1}$ is supported on W and ( κ$_{kn}$ − κ$_{k}$ ) ∈ O$_{M}$ by continuity of the functional ε$_{1}$ it follows that κ$_{kn}$ − κ$_{k}$ converges to zero on W. It then follows that φ$_{n}$ − φ → 0 in S ∗ as well as pointwise and uniformly on bounded sets in W , the product φ − $^{1}$φ − 1 n is in a bounded set in S $^{∗}$, thus φ − 1 n − φ − 1 = φ − $^{1}$φ − 1 n ( φ − φ$_{n}$ ) converges to zero in S $^{∗}$. Consider ε$_{1}$$_{n}$φ − 1 n − ε$_{1}$φ − 1 = ε$_{1}$$_{n}$ ( φ − 1 n − φ − $^{1}$) + ( ε$_{1}$$_{n}$ − ε$_{1}$ ) φ − $^{1}$; this difference 42