φ is uniquely expressed as γ − $^{1}$ε$_{1}$ . Substitute this expression for φ into the differential equations in (16), and denote the continuous functions γ ′ $_{k}$γ − 1 by κ$_{k}$ ( ξ ) to obtain equations where the left-hand side is defined and equals zero in S $^{∗}$. κ$_{k}$ ( ξ ) ε$_{1}$ − iε$_{2}$$_{k}$ = 0 , k = 1 , ..., d (19) We can show that the function κ$_{k}$ is uniquely determined in the class of continuous functions on W by the equation (19) . Proof is by contradiction. Suppose that there are two distinct continuous functions on supp( γ ) , κ$_{k}$$_{1}$ = κ$_{k}$$_{2}$ that satisfy ((19)). Then κ$_{k}$$_{1}$ (¯ x ) = κ$_{k}$$_{2}$ (¯ x ) for some ¯ x ∈ supp ( γ ) . Without loss of generality assume that ¯ x is in the interior of W ; by continuity κ$_{k}$$_{1}$ = κ$_{k}$$_{2}$ everywhere for some closed convex U ⊂ W. Consider now D ( U ) $^{∗}$; we can write for any ψ ∈ D ( U ) . A generalized function that is zero for all ψ ∈ D ( U ) coincides with the ordinary zero function on U and is also zero for all ψ ∈ D$_{0}$ ( U ), where D$_{0}$ ( U ) denotes the space of continuous test functions on U. For the space of test functions D$_{0}$ ( U ) multiplication by continuous ( κ$_{k}$$_{1}$ − κ$_{k}$$_{2}$ ) = 0 is an isomorphism. Then we can write ( ε$_{1}$ ( κ$_{k}$$_{1}$ − κ$_{k}$$_{2}$ ) , ψ ) = 0 0 = ([ ε$_{1}$ ( κ$_{k}$$_{1}$ − κ$_{k}$$_{2}$ )] , ψ ) = ( ε$_{1}$, ( κ$_{k}$$_{1}$ − κ$_{k}$$_{2}$ ) ψ ) 31