2
$\begingroup$

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

||cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ You have the right calculations, but you should start with "Let $x$ and $y$ be any two elements of the given set. Then $x = \ldots, y = \ldots$." $\endgroup$ – user119191 Jan 5 '14 at 14:49
  • $\begingroup$ The definition is odd. Is $(\alpha_i)$ fixed once and for all (LHS) or not (RHS)? $\endgroup$ – Did Jan 5 '14 at 14:50
  • $\begingroup$ 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. $\endgroup$ – derivative Jan 5 '14 at 14:51
  • 3
    $\begingroup$ @derivative The claim is not that the union is convex. $\endgroup$ – Did Jan 5 '14 at 15:01
3
$\begingroup$

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.

||cite|improve this answer
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Silent Jan 5 '14 at 14:58
  • $\begingroup$ Something is wrong with your answer which not shown properly. please edit it. $\endgroup$ – Majid Oct 14 '16 at 13:39
  • $\begingroup$ @MajidRahimi No. $\endgroup$ – Did Oct 14 '16 at 13:45
  • $\begingroup$ see this image: i.stack.imgur.com/QjquF.png $\endgroup$ – Majid Oct 14 '16 at 13:49
  • $\begingroup$ @MajidRahimi Yes, this is source for my answer. And? $\endgroup$ – Did Oct 14 '16 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.