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Sorry for the awkward title.

I'm looking for an example of a normed space $X$ and a point $x$ on $X$'s unit ball such that $x$ is an extreme point of the unit ball when considering $X$ as a vector space over the reals, but not when considering $X$ as a vector space over the complex field - or vice versa.

Or, if such a setting could not exist, some intuitive justification for that.

Thanks in advance.

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As for me this question is not well formulated. If initially $X$ is a vector space over $\mathbb{R}$ you need to make complexification to consider it over $\mathbb{C}$. If initially your vector space is over $\mathbb{C}$ you just need to restrict scalar multiplication, to reals. So you need to specify initial field of scalars. –  no identity Feb 5 '12 at 14:31
    
Untrue. You can, for example, examine $\mathbb{C}^n$ as a vector field above $\mathbb{R}$. It changes it's structure of course (for example, $e_1$ and $i\cdot e_1$ are no longer co-linear) - but it's legitimate none the less... –  Shai Deshe Feb 5 '12 at 15:45
    
You just showed an example when no complexification needed. In general, you must make this proceure since you don't know anything about internal structure of $X$. –  no identity Feb 5 '12 at 16:49

1 Answer 1

up vote 3 down vote accepted

I suspect you haven't thought carefully enough about this. By definition, a point $x$ in some convex set $C \subseteq X$ is extreme if it cannot be written as $x = ty + (1-t)z$ for $y,z \in C$ and $t \in (0,1)$. In other words, the property depends only on the structure of $X$ as a real vector space, no complex numbers will ever enter the picture. So, at least for restriction of scalars it makes no difference. If, instead, you start with a real vector space $X$ and take its complexification $X^{\mathbb{C}} = X \otimes_{\mathbb{R}} \mathbb{C}$, we have $X^{\mathbb{C}} \cong X \oplus iX$ (as real vector spaces) and an embedding $i: X \hookrightarrow X^{\mathbb{C}}$ by $x \mapsto (x,0)$. In particular, scalar multiplication by real numbers on $X$ is unchanged (for that reason, it is called an extension of scalars), so again a point $x$ in $C$ is extreme if and only if $i(x)$ is extreme in $i(C)$. If you had something else in mind, please edit your question appropriately.

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You are right, I haven't thought carefully enough about this –  Shai Deshe Feb 5 '12 at 15:46

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