as a functional on D$_{0}$ ( R d$_{x}$ ) that provides for any ˜ ψ ∈ D$_{0}$ ( R d$_{x}$ ) The following lemma shows that the two representations are compatible and each can be easily obtained from the other. Lemma. Suppose that F$_{x,y}$ ∈ Φ$_{c}$ . Then the value of the functional given by (13) for ψ ∈ D$_{0}$ (0 , 1) d$_{x}$ is the same as the value of the functional given by (15) for ˜ ψ ( x ) = f$_{x}$ ( x ) ψ ( F ( x )) ∈ D$_{0}$ ( R d$_{x}$ ) ; and vice versa: given (15) the value of (13) for ψ ( F$_{x}$$_{1}$, ..., F$_{x}$$_{d}$$_{x}$ ) = ˜ ψ ( x 1 ,...,x$_{d}$$_{x}$ ) f$_{x}$ ( x 1 ,...,x$_{d}$$_{x}$ $_{)}$, where x$_{i}$ is uniquely determined by the value of F$_{x}$$_{i}$ : x$_{i}$ = F − 1 x$_{i}$ ( F$_{x}$$_{i}$ ( x$_{i}$ )) , is the same. Proof. For any ψ ∈ D (0 , 1) d$_{x}$ define ˜ ψ on R d$_{x}$ by ˜ ψ ( x ) = f$_{x}$ ( x ) ψ ( F ( x )) , then ( F$_{y}$$_{|}$$_{x}$, ψ ) defined by (12) by differentiability of F$_{x,y}$ in x is equal to Denote by z$_{i}$ the value F$_{x}$$_{i}$ ( x ) , i = 1 , ..., d$_{x}$ ; then (for clarity we subscript the operator ∂ by the variable(s) with respect to which we differentiate): ( F$_{y}$$_{|}$$_{x}$, ˜ ψ ) = ∫ ... ∫ ∂ $^{d$_{x}$}$F$_{x,y}$ ( x, y ) f$_{x}$ ( x ) ˜ ψ ( x ) dx$_{1}$...dx$_{d}$$_{x}$. ∂ d$_{x}$ z F$_{x,y}$ ( F − 1 x 1 ( z$_{1}$ ) , ..., F − 1 x$_{d}$$_{x}$ ( z$_{d}$$_{x}$ ) , y ) f$_{x}$ ( x ) = ∂ d$_{x}$ x F$_{x,y}$ ( x, y ) . ( F$_{y}$$_{|}$$_{x}$, ˜ ψ ) = ∫ ... ∫ ∂ $^{d$_{x}$}$F$_{x,y}$ ( x, y ) f$_{x}$ ( x ) ˜ ψ ( x ) dx$_{1}$...dx$_{d}$$_{x}$. (15) 21