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I am trying to show that for a convex set $A$ and $s,r>0$ positive real numbers we have $rA+sA = (r+s)A$.

Clearly $(r+s)A$ is contained in $rA+sA$ but I am having trouble showing the other inclusion.

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Here's one trick—If you divide each of $r$ and $s$ by $r+s$, you get a number between 0 and 1, which allows you to make a convex sum:

  1. Suppose $z \in rA + sA$. Then $z = rx + sy$ for some $x,y \in A$.

  2. Because $r$ and $s$ are positive, the numbers $t_1 \equiv \frac{r}{r+s}$ and $t_2 \equiv \frac{s}{r+s}$ are between 0 and 1.

  3. Because $A$ is convex and $x,y\in A$, we know that $t_1 x + t_2 y \in A$.

  4. But $z = (r+s)(t_1x + t_2 y)$, hence $z \in (r+s)A$.

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To show the other inclusion $ rA+sA \subset (r+s)A $:

let $\zeta \in rA+sA $, then there exist $ x,y \in A$, such that $ \zeta=rx+sy$.

Now we set $\eta =\frac{r}{r+s}x+\frac{s}{r+s}y \in A $(convexity), obviously, $$ \zeta =(r+s)\eta \in (r+s)A $$ that is, $$ rA+sA \subset (r+s)A $$

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let $a \in rA + sA$

then $\exists b \in A, c \in A, a=rb+sc$

$$a= (r+s) \left(\frac{r}{r+s}b + \frac{s}{r+s}c \right)$$

Note that we have $\left(\frac{r}{r+s}b + \frac{s}{r+s}c \right) \in A$

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