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In a topological vector space $X$, a subset $S$ is convex if \begin{equation}tS+(1-t)S\subset S\end{equation} for all $t\in (0,1)$.

$S$ is balanced if \begin{equation}\alpha S\subset S\end{equation} for all $|\alpha|\le 1$.

So if $S$ is balanced then $0\in S$, $S$ is uniform in all directions and $S$ contains the line segment connecting 0 to another point in $S$.

Due to the last condition it seems to me that balanced sets are convex. However I cannot prove this, and there are also evidence suggesting the opposite.

I wonder whether there is an example of a set that is balanced but not convex.

Thanks!

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What "$S$ is balanced" tells you is that the intersection of $S$ with every one-dimensional subspace is either the subspace itself, a disk centered at the origin or $0$. It doesn't tell you anything about affine segments that don't lie in a one-dimensional linear subspace, which is what convexity does. It is somewhat similar to starshaped versus convex. –  t.b. Sep 16 '12 at 14:12
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3 Answers

up vote 4 down vote accepted

Consider the vector space $\mathbb{R}^2$ and the set $S=\{(x,y)|x=0 \text{ or } y=0\}$, i.e. the union of the two axes. Then $S$ is balanced but not convex.

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Let $x=(1,0)$, $\alpha = 2^{-1/2}(1+i)$. Then $x\in S$, $|\alpha|\leq 1$ but $\alpha x\notin S$ –  Norbert Sep 16 '12 at 13:53
    
@Norbert: this is a real vector space. Multiplying by a complex scalar is nonsense. –  Chris Eagle Sep 16 '12 at 13:56
    
If you want complex scalars just replace $\mathbb{R}^2$ by $\mathbb{C}^2$ and take the same example $S = \{(x,y)\mid x = 0\text{ or } y =0\}$ (or any finite union of disks centered at $0$ and lying in distinct one-dimensional subspaces)... –  t.b. Sep 16 '12 at 13:58
    
@ChrisEagle I thought you consider $\mathbb{R}^2$ as $\mathbb{C}$ –  Norbert Sep 16 '12 at 14:06
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The interior of a regular pentagram centered at the origin is balanced but not convex.

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That's not balanced: If $x$ is near one of the corners then $-x$ is not in the pentagram. Similar stars work if you take an even number of corners. –  t.b. Sep 16 '12 at 22:20
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Astroid: $|x|^{2/3}+|y|^{2/3} \le 1$.

http://en.wikipedia.org/wiki/Astroid

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