implying that ε$_{1}$ is defined and is a zero generalized function in D$_{0}$ ( U ) $^{′}$. If that were so ε$_{1}$ would be a zero generalized function in D ( U ) ∗ since D ( U ) ⊂ D$_{0}$ ( U ) but this is not possible since ε$_{1}$ = γφ, which is non-zero by assumption . Next we show that γ is then uniquely determined on W. Indeed, for any t, s ∈ supp( γ ) with the kth coordinate denoted t$_{k}$, write the continuous function where integration is along any arc joining 0 and s in W. This is the unique solution to γ (0) = c, γ − $^{1}$γ ′ k = κ$_{k}$ (see, e.g., Schwartz, 1966, p.61). Then in S ∗ g = F t − $^{1}$( γ ) is uniquely defined. γ ( s ) = c exp ∫ s 0 ∑ d k $_{=1}$κ$_{k}$ ( t ) dt$_{k}$, (b) In view of the result in Theorem 2 it is sufficient to show that φ is uniquely determined on W. Consider the space of generalized functions D $^{∗}$( W ). Since φ is non-zero on W and continuously differentiable, then by differentiating the first equation in (16), substituting from the second equation and multiplying by φ − 1 in D $^{∗}$( W ) (where the product exists as shown in Lemma 1) we get that the generalized function equals zero in the sense of generalized functions, in D $^{∗}$( W ) . Note that by assumption ε$_{1}$ cannot be zero on W and both ε$_{1}$ and ε$_{2}$$_{k}$ are zero outside of W. Define ˜ κ$_{k}$ = φ ′ $_{k}$φ − $^{1}$; then ˜ κ$_{k}$ is continuous on supp( γ ) and is a regular ε$_{1}$φ − $^{1}$φ ′ k − (( ε$_{1}$ ) ′ k − iε$_{2}$$_{k}$ ) 32