Let $X$ be a Banach space. Is there always a normed vector space $Y$ such that $X$ and $Y^*$ are isometric or isomorphic as topological vector spaces (that is, there exists a linear homeomorphism between $X$ and $Y^*$)?
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Martin Sleziak
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Seirios
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2No. There is no $Y$ such that $Y^* = L_1[0,1]$. See this question http://math.stackexchange.com/questions/137677/what-is-the-predual-of-l1 – Yury Nov 21 '12 at 15:33
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1Also http://math.stackexchange.com/q/107706/ and http://math.stackexchange.com/questions/210043/ – commenter Nov 21 '12 at 15:35
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All these examples need a lot of theory. Do you know if there are elementary arguments? – Seirios Nov 24 '12 at 11:12
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1The question is subtle. I am not aware of any elementary arguments. You need to use some theory to show that a Banach space is not isometric to a dual space. The argument via Krein-Milman is probably the easiest. Non-isometric Banach spaces can be isomorphic (for example $c_0$ and $c$ are isomorphic but they are not isometric because the unit ball of $c$ has lots of extreme points while the one of $c_0$ doesn't have any) and all $L^p$-spaces with $1 \leq p \lt \infty$ are homeomorphic as topological spaces (even locally Lipschitz), but they are not isomorphic as topological vector spaces. – commenter Nov 24 '12 at 20:29
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I just sum up the answers given above to refer to my question as answered:
My favorite argument is the one given by commenter here using Krein-Milmann theorem to prove that $C_0(K)$ has no predual space.
A good reference for such questions seems to be Topics in in Banach Space Theory by Kalton and Albiac. It is used here and here to prove that $C_0(K)$ and $L_1[0,1]$ have no predual space.
