# If $A$ is convex then $rA+sA = (r+s)A$

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.

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 $z \in (r+s)A$.

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, $$rA+sA \subset (r+s)A$$

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