0
$\begingroup$

If $A$ is a balanced subset of a vector space $V$ then conv$(A)$ is balanced.

Proof:

Let $C=\{ax+by:a,b\ge0,a+b=1,x,y\in A\}$

Is enough to show that $C= $ conv$(A)$

How can we prove that $C$ is convex $?$

$A$ is balanced if $\lambda A\subset A$ for all $ \lambda\in \mathbb{C }\;,\; |\lambda|\le 1$

Any hints would be appreciated.

$\endgroup$
1
  • 2
    $\begingroup$ Your $C$ is not defined correctly. Define $C$ as $C = \{\sum_{i = 1}^{|A|}a_ix_i : a_i \geq 0 \forall i, \sum_{i = 1}^{|A|} a_i = 1, x_i \in A \forall i\}$. Proving that this is convex is very easy. You should also be able to show that this is equal to conv(A) very easily. $\endgroup$
    – TenaliRaman
    Commented Jul 21, 2013 at 7:17

2 Answers 2

Reset to default
2
$\begingroup$

Merely taking combinations of pairs of elements will not be enough in general, for dimension reasons. Thus, take 4 generic lines in $\mathbb{C}^4$. They will span span something 4-dimensional, but the combinations you allowed will at most give you something 3-dimensional.

$\endgroup$
1
$\begingroup$

Let $C=\{ax+by:a,b\ge0,a+b=1,x,y\in A\}$ Is enough to show that $C= $ conv$(A)$ How can we prove that C is convex ?

I'm afraid this strategy is a dead end, because $C$ is not necessarily convex! To see this, you'll need a vector space of dimension at least 3. So if you're trying to visualize the problem using real vector spaces, you'll have to go beyond the plane.

Also, make sure you re-read your textbook's definition of the convex hull of $A$. It should be the set of all convex combinations of points from $A$, and this means all combinations of $n$ points, not just combinations of 2 points. Now, given this definition, you can prove that conv($A$) is balanced in a straightforward manner. Let $x$ be a point in conv($A$), represent it in terms of $A$, and use this representation to prove that $\lambda x$ is also in conv($A$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .