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

$\endgroup$
4
  • 2
    $\begingroup$ 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, 2014 at 14:49
  • $\begingroup$ The definition is odd. Is $(\alpha_i)$ fixed once and for all (LHS) or not (RHS)? $\endgroup$
    – Did
    Jan 5, 2014 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, 2014 at 14:51
  • 3
    $\begingroup$ @derivative The claim is not that the union is convex. $\endgroup$
    – Did
    Jan 5, 2014 at 15:01

1 Answer 1

Reset to default
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.

$\endgroup$
1
  • $\begingroup$ 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, 2014 at 14:58

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.