# Convex combination

#### Fernando Revilla

##### Well-known member
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if S is a convex set and x1, x2, x3, . . . xn are n elements in S then Their convex combination is also in S .
Firstly, we prove that if $x=\alpha_1x_1+\alpha_2x_2+\alpha_3x_3$ for $\alpha_i\ge 0$ and $\sum_i\alpha_i=1$, then $x\in S$. Consider $$\alpha'_i=\frac{\alpha_i}{\alpha_1+\alpha_2} \quad (i=1,2)$$ Note that $\sum_i\alpha'_i=1$ and $\alpha'_i\ge 0.$ We have $$x=\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=(\alpha_1+\alpha_2)(\alpha'_1x_1+\alpha'_2x_2)+\alpha_3x_3$$ By hypothesis $\alpha'_1x_1+\alpha'_2x_2\in S.$ Denote $x_4=\alpha'_1x_1+\alpha'_2x_2$, then $$x=(\alpha_1+\alpha_2)x_4+\alpha_3$$ But $x$ is a convex combination of two elements of $S$, so $x\in S.$ This argument can be generalized for any $i$.