Solution of a functional equation arising from utility that is both separable and additive

The problem of determining all utility measures over binary gambles that are both separable and additive leads to the functional equation

The following conditions are more or less natural to the problem: $f$ strictly increasing, $Q$ strictly decreasing; both map their domains onto intervals ($f$ onto a $[0,K)$, $Q$ onto $[0,1]$); thus both are continuous, $k>1$, $f(0)=0$, $f(1)=1$, $Q(1)=0$, $Q(0)=1$. We determine, however, the general solution without any of these conditions (except $f: [0,k) \to \mathbb{R}_{+}:= [0, \infty )$, $Q:[0,1]\to \mathbb{R}_{+}$, both into). If we exclude two trivial solutions, then we get as general solution $f(v) = \alpha v^{\beta }$ ($\beta > 0$, $\alpha > 0$; $\alpha = 1$ for $f(1) = 1$), which satisfies all the above conditions. The paper concludes with a remark on the case where the equation is satisfied only almost everywhere.