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Let $E$ be a Vector Space over $\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}= \mathbb{C}$. We say that $U\subset E$ is balanced if $\alpha U\subset U$ whenever $|\alpha|\leq 1$, where $\alpha\in\mathbb{K}$. Also, $U$ is said to be symmetric if $-U\subset U$, where $-U=\{-u:\ u\in U\}$.

When $E$ is a vector space over $\mathbb{R}$, then it is possible to show that a convex set $U$ is symmetric if and only if it is balanced. On the other hand, if we consider the complex vector space $\mathbb{C}$ and we take the set $S=\{a+bi:\ |a|\leq 1,\ |b|\leq 1\}$, we have that $S$ is convex and symmetric but $E$ is not balanced $\Big($multiply $1+i$ by $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ $\Big)$.

My question: Does anyone knows some characterization of these kind of sets in complex vector spaces: Convex and Symmetric implies Balanced? I mean: necessary and sufficient conditions?

Note: Balanced always implies Symmetric.

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The characterization of sets for which "convex and symmetric" implies "balanced"? Logically, these are the sets which satisfy at least one of the properties: (i) not convex; (ii) not symmetric; (iii) balanced. This is a rather strange class of sets, and I don't think there can be a better characterization of it than (i)-(ii)-(iii). –  user53153 Feb 21 '13 at 22:41
    
@5pm, can you give me an example of a Symmetric Balanced Convex set that is not the ball (in $\mathbb{C}$). –  Tomás Feb 21 '13 at 22:55
    
Anyway, I could not undertand your logic @5pm, or maybe it is my english. Because if the set satisfy i), how can it be in the required class? –  Tomás Feb 21 '13 at 22:58
    
There aren't other balanced sets in $\mathbb C$. Note that $\mathbb C$ is a one-dimensional vector space over $\mathbb C$. You should not expect to find interesting geometry in one dimension. // I was reacting to your "kind of sets in complex vector spaces: Convex and Symmetric implies Balanced". The statement "$A\implies B$" is equivalent to saying "either $A$ is false or $B$ is true". For example, the class of sets for which "convex implies bounded" consists of (i) all non-convex sets; (ii) all bounded sets. –  user53153 Feb 21 '13 at 23:03
    
It is true. Well, in $\mathbb{C}^2$ it is impossible to visualize such a thing, but maybe there is some geometrical intuitive set. What do you think? –  Tomás Feb 21 '13 at 23:19

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