1

If $A$ is convex set and $\alpha,\beta>0$, show that $(\alpha+\beta)Α=\alpha Α+\beta Α$.

I tried to show that, but I am not sure if it was so simple. This is how I did it:

$$(\alpha+\beta)Α := \left\{z:z=(\alpha+\beta)x,\,x\in A\right\}$$

$\alpha Α := \left\{z:z=\alpha x, x\in A\right\}$

$\beta Α :=\left\{z:z=\beta x,\, x\in A\right\}$

Let it be $z\in (\alpha+\beta)Α$, $x\in A$.

$$z=(\alpha+\beta)x=\alpha x+\beta x,$$

$\alpha x \in \alphaΑ$, $\beta x \in\betaΑ$.

Johnny Westerling
  • 1,761
gorica
  • 15
  • 2

1 Answers1

1

The inclusion $(\alpha+\beta)A\subseteq \alpha A+\beta A$ is proved by your arguments, but the other part is yet missing: for that, let $x$ and $y$ be elements of $A$, and we need to prove that $\alpha x+\beta y\in (\alpha+\beta)A$.

For that, observe that, by convexity, we have $$\frac{\alpha}{\alpha+\beta}x+\frac{\beta}{\alpha+\beta}y\in A\,.$$

Berci
  • 90,745
  • Thank you. Can you help me to prove that upper semispace is convex set? I've got this:

    If x and y are in upper semispace, then upper semispace is convex if αx+(1-α)y is it's element, that is if 〈αx+(1-α)y,c〉 >γ Since x and y are in upper semispace, I know that 〈x,c〉>γ and 〈y,c〉>γ. Now I have 〈αx+(1-α)y,c〉=〈αx,c〉+〈(1-α)y,c〉=α〈x,c〉+(1-α)〈y,c〉=α[〈x,c〉-〈y,c〉]+〈y,c〉. I don't know how to show that this is >γ. It is ok if 〈x,c〉>〈y,c〉 or 〈x,c〉=〈y,c〉, but what if 〈x,c〉<〈y,c〉?

    And can you tell me from where can I learn how to write on this site, I think you got a headache reading my questions? :D

     – gorica Mar 10 '13 at 13:14
  • It's probably easier to prove convexity first of a semispace of the form ${x:\langle x,c\rangle>0}$ for a fixed vector $c$, then prove that the shifted version of a convex set is convex again, i.e. if $A$ is convex, then $A+v$ is convex for all vectors $v$. Anyway, if the case $\langle x,c\rangle\ge\langle y,c\rangle$ is indeed verified, then for $\beta:=1-\alpha$ (still in $[0,1]$) and exchanging $x$ and $y$ we arrive to the case $\langle x,c\rangle\ge\langle y,c\rangle$. $$ ,$$ 2. The easiest is to right-click on math formulas, "Show Math as TeX commands", then include formulas in $.
    • – Berci Mar 10 '13 at 20:00