I have the following properties about convex subsets of vector spaces that I need to prove. I use the following definitions. Let $X$ be a vector space and all other sets below are understood to be subsets of $X$. Define:
$$
\begin{align}
A+B &:= \{a+b \;|\; a\in A, b\in B \}\\
\alpha A &:= \{\alpha a\; | \; a\in A\}
\end{align}
$$
Then I want to prove that
1) If $A$ and $B$ are convex, then $A+B$ is also convex
2) A is convex if and only if $(s+t)A = sA+tA \quad \forall s,t \geq 0$
I suspect I am struggling mostly with the notation since I am used to a different definition for the sets above, but I have to use these to follow some theorems I am trying to understand. Since I am learning my proof writing on-the-go, any comment on style is also very much appreciated. Here are my attempts
1) Assume $A,B\subset X$ are convex. Let $a_1,a_2\in A$ and $b_1,b_2\in B$. Because $A$ and $B$ are convex, we have that $a_1t + (1-t)a_2 \in A\;$ and $\;b_1s + (1-s)b_2 \in B \;$ for all $t,s \in[0,1]$. If we fix $t = s = t_0\in[0,1]$, then the sum is also convex. Indeed, we have
$$
\begin{align}
a_1t_0 + (1-t_0)a_2 + b_1t_0 + (1-t_0)b_2 &= (a_1+b_1)t_0 + (1-t_0)(a_2+b_2) \\
&= c_1t_0 + (1-t_0)c_2 \in A+B,
\end{align}
$$
where I have defined $c_1 = a_1+b_1$ and $c_2=a_2+b_2$ and it is clear from the definition of $A+B$ that $c_1,c_2\in A+B$. Therefore I have established that $A+B$ is also convext.
Question: Do I lose any generality by assuming the statement in bold letters above? I felt this was necessary because otherwise $s_0+t_0 \in[0,2]$
2) Assume $A$ is convex, then we want to prove inclusion both ways. Let $x\in (t+s)A$, and fix $t,s\in[0,1]$ then by definition of the set, $x=(t+s)a$ for some $a\in A$. Since $x$ is an element of $A\subset X$ (a vector space) we have $(t+s)a = ta + sa \; \cdots$
then I become uneasy. Superficially I say that clearly $ta\in tA$ and $sa\in sA$ therefore $x=ta + sa\in tA+sA$ and done. However, if I stare at the set definition of $tA+sA$ I think it means $\{ta_1 + sa_2|a_1,a_2\in A\}$ since both elements need not to be equal. Is this a problem?
Then for the other way around the set inclusion. Recalling our hypothesis that $A$ is convex, and letting $x\in(t_0A+s_0A)$ where we fixed $t_0,s_0\in[0,1]$, then we have $x=t_0a_1 + s_0a_2$ for some $a_1+a_2\in A$ not necessarily equal. Then I must show somehow that $x\in (t_0+s_0)A$, but that would require $a_1$ to be equal to $a_2$? I am not sure what can I deduce from the equality of the sets.
This is the kind of confusion these definitions bringing me, I am not sure whether I am being general or not; or how to manipulate the elements and the inclusions. Lastly, I have no idea how to get started for the converse of the property. I have more questions regarding other properties, but I think if I see how these come together in the above properties I should be ready to be on my own for now. Thank you for your help.
Best Answer
In part $1)$ you are starting out wrong. You need to show that if $x\in A+B$ and $y\in A+B,$ then $tx+(1-t)y\in A+B,$ for $0\le t\le 1$. Then you say that by definition $x=a_1+b_1, \ y=a_2,b_2,$ for some $a_1,a_2\in A,$ etc.
The first part of $2)$ is correct. You have shown $(s+t)A\subseteq sA+tA.$ This doesn't need convexity at all. Also, you need not have $s,t\in [0,1].$ The remarks after, "Then I become uneasy," are confused. You are right when you say done. What you say about what the definition means is also correct. It's also true that $a_1$ and $a_2$ can be different. But they need not be.
So you need to show that if $A$ is convex, then $sA+tA\subseteq (s+t)A$ Now if $0\le s\le1$ and $t=1-s$ this is certainly true, by the definition of convexity. So the problem is, how to reduce the general case to the case where $0\le s\le 1,\ t=1-s?$ Well, if $s$ and $t$ are nonnegative, consider$$s'=\frac{s}{s+t}\\t'=\frac{t}{s+t}$$
If $a_1,a_2\in A,$ then since $A$ is convex, we have that $sa_1'+t'a_2\in A$. Then multiply both sides by $s+t$ to see that $sa_1+ta_2\in (s+t)A.$
You still have to handle the case where it is not true that $s\ge0,\ t\ge0,$ in particular when they have opposite signs.