“…( Proof: To show that A is well defined it suffices to show that (Aϕ) (0) = 0 and α η 0 [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds, and [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds dt [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds dt [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds dt, which implies [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds [1] (s), ϕ [2] (s), ..., ϕ [n] (s) ds…”