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.

score 3 · Answer 1 · answered May 27 '13 at 13:45

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 $A$.

Great, thanks :)
I have one question more, from another problem. I have to prove that open half-space is convex set. — user23709, May 27 '13 at 13:50
@user23709, glad to be of help. You should consider posting your other question separately. — vadim123, May 27 '13 at 14:08

score 0 · Answer 2 · edited Apr 01 '16 at 10:40

Could it be possible to try the definition of convex sets?

$A$ is a convex set and $\alpha,\beta\geq 0$, then $(\alpha+\beta)A=\alpha A+\beta A$. As $\alpha A$ and $\beta A$ in $A$, then according to the definition of convexity $\lambda\alpha A + (1-\lambda)\beta A$ equals $(\lambda\alpha + (1-\lambda)\beta)A$. This implies $A$ inverse $(\lambda\alpha + (1-\lambda)\beta)$ equals $\lambda\alpha + (1-\lambda)\beta$, and that is element $A$ so it exist.

score 0 · Answer 3 · answered May 27 '13 at 13:49

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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.

answered May 27 '13 at 13:49

Edoardo Lanari

3,564

Thanks :) I think I solved problem with half-space, but you can write it too... just to be sure :) – user23709 May 27 '13 at 14:12

score 0 · Answer 4 · answered May 27 '13 at 14:18

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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.

answered May 27 '13 at 14:18

Edoardo Lanari

3,564

Convex set. Proof.

4 Answers4