For k epsilon N-o, denote by C-k (R) the space of k-times continuously differentiable functions on R, which for k= 0 is the space C (R) of continuous functions. In this paper we consider operator equations for maps T :C-k (R) -> C (R) like e.g. the chain rule equation T(f o g)=Tf o g center dot Tg, f,g epsilon C-k(R), find the general form of the solution operators T and study their dependence on the domain of T, i.e. on the order of smoothness k. Typically, the solution operators can be extended from C-k(R) to C-1 (R) with k >= 1 for just a few values of 1 like 1 epsilon {0, 1) or 1 epsilon {0, 1, 2, 3}, so that rather few spaces C-1 (R) are "natural" domains of definition for T. In the case of the chain rule equation, the natural domains are obtained for 1 epsilon {0, 1} since the general form of the solutions of this equation for k E N is given by Tf = Hof/vertical bar f'vertical bar {sgn f'}, f epsilon C-k (R), for a suitable p >= 0, a strictly positive function H epsilon C (R), and where the term {sgn f'} may be present or not present. Therefore the solution does not depend on k epsilon N; if T satisfies the chain rule equation on C-k(R) for some k > 1, it may be extended by the same formula to C-1 (R). For k = 0, i.e. T : C (R) -> C (R), the only solution is given by T f = Hof/H, cf. [1]. As another example of this phenomenon, consider the Leibniz rule equation for operators T : C-k (R) -> C(R) T(f center dot g)=Tf center dot g+ f center dot Tg, f,g epsilon C-k(R), with a,b epsilon C(R) being continuous functions, cf. [3]. Again, the solution does not depend on k epsilon N, and for k > 1 may be extended by the same formula to C-1 (R). Fork =0, the only solution of (1) is the entropy solution Tf =af In if I with a E C(R). The last fact had been shown by Goldmann and 'emr1 [2]. Thus again, the "natural" domains of T are the spaces C1 (R) for 1 epsilon {0, 1). restricted to C-k(R)-functions with non-vanishing derivative, we may allow p >= 0. For p = 0, the main term in T is the Schwarzian derivative S, S f = ( f'"/f'- 3/2(f"/f')(2).) (C) 2013 Elsevier Inc. All rights reserved.
