# Linear combination of convex set is convex

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 $A_i\in\mathbb R_m$ 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,$\sum_{i=1}^n\alpha_i(t(a_{i,1})+(1-t)(a_{i,2}))=t\sum_{i=1}^n\alpha_ia_{i,1}+(1-t)\sum_{i=1}^n\alpha_ia_{i,2}$.

• You have the right calculations, but you should start with "Let $x$ and $y$ be any two elements of the given set. Then $x = \ldots, y = \ldots$." – user119191 Jan 5 '14 at 14:49
• The definition is odd. Is $(\alpha_i)$ fixed once and for all (LHS) or not (RHS)? – Did Jan 5 '14 at 14:50
• Is the claim really true, i mean you can take 2 non intersecting discs in $R^2$, and a point lying in each disc but the line segment is not entirely in the union of 2 discs. – derivative Jan 5 '14 at 14:51
• @derivative The claim is not that the union is convex. – Did Jan 5 '14 at 15:01

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.
• Sir, I think, $b_i=(t(a_{i,1})+(1-t)(a_{i,2}))$, and $b_i$ belongs to $A_i$ and, $s$ belongs to $\sum_{i=1}^n\alpha_iA_i=\left\{\sum_{i=1}^n\alpha_ia_i|\alpha_i\in\mathbb R,a_i\in A_i\right\}$, right? – Silent Jan 5 '14 at 14:58