On the Uniqueness of Solutions of the Functional Equation q)(x + f (x)) = q~ (x) + r# (f (x)) MAREK ZDUN (Katowice, Poland) 229 We consider the Cauchy functional equation q~ (x + y) = q~ (x) + <p (y), (1) where the function ~p(x) is defined in the space of real numbers R and assumes values in R. If we suppose, e.g., that the function cp(x) is measurable, then the only solutions are the functions ¢p(x)=ax (cf. for instance, Banach [2], Sierpifiski [5], Acz61 [1]). Let us restrict the domain of the variation of the independent variables x, y so that the couple (x, y) lie on a certain curve. We may put forward the following question: What possibly weakest regularity must then be imposed on the class of solutions of this equation in order that the unique solutions be the functions ~p(x)=ax. If we assume only continuity, then usually we obtain the solution depending on an ar- bitrary function (cf. Kuczma [3]). In the present paper we consider equation (1) when the couple (x, y) varies along the graph of a strictly increasing continuous function going out from the origin of the system of coordinates. Let y =f(x). Equation (1) will have the form ~o(x + f (x)) = tp (x) + tp (f (x)). (2) We shall prove the following THEOREM 1. If the real-valued function f (x), defined in the interval [0, oo), is continuous and strictly increasing in [0, oo) and f (O)=O, then the function ¢(x)=ax is the unique function which is defined in the interval [0, oo), is right-sided differentiable at zero, satisfies equation (2) and fulfils the condition q~' (0)=a. Proof. Write g (x) = x +f (x). The function f (x) is strictly increasing and contin= uous, hence g(x) is also strictly increasing and continuous and maps the interval ['0, oo) onto itself. Let g(x)=y, whence x=g-l(y). Since f(x)+x=y, we have f (x)=y-g-1 (y). Therefore equation (2) will have the form Let us note that ~P (y) = q~ (g-1 (y)) + 9 (Y - g-1 (y)). y_g-l(y)_f(x)= f(x) y g (x) x + f (x)" ReceivedMarch 27, 1971