A set $U$ is a convex set if whenever $\mathbf{x},\mathbf{y}$ are points in $U$, the line segment joining $\mathbf{x}$ to $\mathbf{y}$,$$\mathcal l(\mathbf{x},\mathbf{y})=\left\{t\mathbf x+(1-t)\mathbf y):0\leq t\leq 1 \right\},$$ is also in $U$.
Let $A_i\subset\mathbb R_m$ be a convex set for $i=1,\dots,n$. Prove that the set$$\sum_{i=1}^n\alpha_iA_i=\left\{\sum_{i=1}^n\alpha_ia_i|\alpha_i\in\mathbb R,a_i\in A_i\right\}$$ is also convex set.
MY TRY:
Choose any two $a_{i,1},a_{i,2}$ from $A_i$, then $(t(a_{i,1})+(1-t)(a_{i,2}))\in A_i$.Then,$\sum_{i=1}^n\alpha_i(t(a_{i,1})+(1-t)(a_{i,2}))=t\sum_{i=1}^n\alpha_ia_{i,1}+(1-t)\sum_{i=1}^n\alpha_ia_{i,2}$.
Best Answer
And this is $s=\sum\limits_i\alpha_ib_i$ where $b_i=$ $____$. For each $i$, by convexity of $A_i$, $b_i$ belongs to $____$, hence $s$ belongs to $____$, QED.