Remark

— Let $A$ be a convex subset of a vector space $E$; for any scalars $\alpha > 0$ and $\beta > 0$ we have $\alpha A + \beta A = (\alpha + \beta) A$. In other words, for any $x \in A, y \in A$, there exists $z \in A$ such that $(\alpha + \beta) z = \alpha x + \beta y$; in fact this relation can be written $z = \frac{\alpha}{\alpha + \beta} x + \frac{\beta}{\alpha + \beta} y$ and we have $\frac{\alpha}{\alpha + \beta} > 0, \frac{\beta}{\alpha + \beta} > 0$ and $\frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta} = 1$, from which the assertion follows, on using def. 1.