Convex set. Proof.

Question

Prove that if A is convex set and $\alpha, \beta ≥ 0$ then $(\alpha + \beta)A = \alpha A + \beta A$

What came first on my mind is that I have to show that $(\alpha + \beta)A\subset \alpha A + \beta A$ and $\alpha A + \beta A\subset (\alpha + \beta)A$. But I don't know how to use convexity of A to show this.

Answer 1

One direction doesn't require convexity. Let $(\alpha+\beta)a\in (\alpha+\beta)A$. Then $\alpha a\in \alpha A$ and $\beta a\in \beta A$, and $(\alpha+\beta)a=\alpha a+\beta a$, so $(\alpha+\beta)a\in \alpha A + \beta A$.

For the other direction, let $\alpha a_1 + \beta a_2\in \alpha A + \beta A$. We need to find some $a_3\in A$ such that $\alpha a_1+\beta a_2=(\alpha+\beta)a_3$. To find such an $a_3$, divide both sides by $(\alpha+\beta)$. (footnote*). We set $a_3=\frac{\alpha}{\alpha+\beta}a_1+\frac{\beta}{\alpha+\beta}a_2$. Because $A$ is convex and these fractions are in $(0,1)$ and add up to 1, $a_3\in A$ and we're done.

(footnote*) This proof frays at the edges if $\alpha=\beta=0$, so this special case must be handle separately; it's easy though since $(0+0)A=0A+0A=\{0\}$ for any $A$.

Answer 2

The "$\subset$" part is easy: $(\alpha + \beta)x=\alpha x + \beta x \in \alpha A +\beta A \ \forall x \in A$; on the other hand $\alpha x + \beta y=(\alpha+ \beta)(\frac{\alpha x}{\alpha + \beta} + \frac {\beta y}{\alpha + \beta})\in(\alpha+\beta)A$ where the membership relation is due to the convexity of A.

Answer 3

For the open half-space (I suppose in an n-vector space over the real numbers), consider that up to an isomorphism they're $\{ (x_1,...,x_n): x_n >0\} $ so that each convex combination remains into that set.