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Show that a convex set in a real vector space is symmetric if and only if the set is balanced.

For the backward direction, i.e. if the convex set is balanced in real vector space, then that it is symmetric is easy to show.

For the forward direction, let the set be $S$. Intuitively, if $S$ is convex and symmetric, then $0 \in S$. Then I don't know how to proceed. Can anyone give some hint?

Remark: A set $S$ is balanced if and only if $\alpha S \subseteq S$ for any scalar $\alpha \in [-1,1].$

Idonknow
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2 Answers2

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Let the convex set be $S$. If $S$ is balanced, by taking $\alpha = -1,$ we have $(-1)S \subset S,$ which is symmetric. Suppose $S$ is symmetric. If $x \in S,$ by symmetry, $-x \in S.$ Note that $0 = \frac{1}{2}(-x) + \frac{1}{2}(x)$ and $[-x,x] \subset S.$ It follows that $0 \in S.$ For any $\lambda \in [0,1],$ note that $\lambda x = (1 - \lambda)0 + \lambda x \in [0,x] \subset S.$ For any $\lambda \in [-1,0),$ by the previous result, we have $- \lambda x \in S.$ By symmetry, $(-1)(-\lambda x) = \lambda x \in S.$

Idonknow
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To get $0\in S$ formally: take any $x\in S$; by symmetry, $-x\in S$; by convexity, $0 = \frac12 x + \frac12 (-x)\in S$.

Now, to show balance, you want to take any $x\in S$ and any scalar $\lambda$ with $|\lambda|\le 1$ and get $\lambda x\in S$. Can you write $\lambda x$ as a convex combination of things you know to be in $S$?

  • If $0 \leq \lambda \leq 1$, then $\lambda x \in [0,x]$. If $-1 \leq \lambda <0$, then $\lambda x=\frac{-\lambda}{2}x+(1+\frac{\lambda}{2})x$. – Idonknow Dec 18 '13 at 07:59
  • For $0\le\lambda\le1$, I agree. For $-1\le\lambda<0$, the equation you've written says $\lambda x = x$; and indeed, it purports to write $\lambda x$ as a convex combination of $x$ and $x$, which would only be possible if $\lambda x\in [x,x]$. –  Dec 18 '13 at 14:41
  • Got any hint for $-1 \leq \lambda<0$? – Idonknow Dec 18 '13 at 14:56
  • Draw a picture. Put the points you know on it: $x$, $0$, $\lambda x$. To do that, you'll have to figure out how these points are arranged with respect to each other. I'm sure you know they're collinear. What order are they in? What does $\lambda<0$ tell you about the position of $\lambda x$ with respect to $0$ and $x$? What does $-1\le\lambda$ tell you? Once you've got your picture, look at it with the idea of finding points in $S$ on either side of $\lambda x$. –  Dec 18 '13 at 15:03
  • For $-1 \leq \lambda <0$, we have $\lambda x \in [-x,x]$ (due to symmetric and convex property of $S$). So, $\lambda x=\frac{-\lambda +1}{2}(-x)+\frac{\lambda +1}{2}(x)$, correct? – Idonknow Dec 18 '13 at 18:52
  • That's the idea. You've got a sign error there, but otherwise yes. (And I wouldn't write that $\lambda x\in[-x,x]$ is due to $S$ being convex and symmetric: $\lambda x\in[-x,x]$ is true regardless; what's due to $S$ being convex and symmetric is that $[-x,x]\subseteq S$.) –  Dec 18 '13 at 19:00