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Let $E$ be a dual Banach space and $C$ a nonempty convex weak* compact subset of $E$. Let $G$ be a group of weak* continuous linear isometries on $E$. Suppose that $g(C)\subset C$ for all $g\in G$.

A fixed point for $G$ is an element $x$ of $C$ such that $g(x)=x$ for any $g\in G$.

What conditions on $G$ assure the existence of a fixed point for $G$?

The only condition which I know is noncontracting (=distal), see Fixed point theory, Granas/Dugundji, page 173. I need other conditions.

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4 Answers 4

Kakutani's fixed point theorem says that it is enough for $G$ to be equicontinuous on $C$. Now equicontinuity in the weak$^*$ topology might be too restrictive, but the pre adjoints of of the elements of $G$ are equicontinuous in the normed topology and this can sometimes (always?) be used to find a fixed point of $G$ in $C$. This is what Rudin does in his book Functional Analysis to prove the existence of Haar measure on a compact group.

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I think if $G$ is amenable, you always get a fixed point. Check the definition of amenability in Wikipedia: http://en.wikipedia.org/wiki/Amenable_group [Even though the definition is given for discrete groups, it generalizes to second countable, locally compact groups: see Bob Zimmer's book semisimple groups and ergodic theory"

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Finally, Bourbaki "topological vector spaces" seems to answer completely the question if $C$ has a denumerable type. None condition is needed. A such group has a fixed point!

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Is this an answer or a comment to your original question? –  Yemon Choi Oct 26 '10 at 19:13
    
This is an answer to the initial question in the case where $C$ is second countable for the topology of the norm. –  BigBill Oct 26 '10 at 19:56

Let X be the predual of E. If the dual of every separable subspace of X is separable, then C contains a point that is fixed by EVERY weak* continuous affine isometry of C into C. This is Theorem 2 in the following paper: http://math.gmu.edu/~tlim/pams81.pdf -TCL

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Thank you very much. It is a surprising result. Actually, I am interested in the case where $E$ is not separable (unfortunately for me). What is known about my question? –  BigBill Nov 8 '10 at 18:08
    
If $G$ is commutative and $E$ is uniformly convex (or uniformly smooth), then $G$ has a common fixed point. Other weaker conditions like weak* normal structure on $C$ will also suffice. –  TCL Nov 10 '10 at 21:00
    
Thank you. A last question: What do you think about the case where $E$ is the space $B(\ell_\infty)$ of bounded linear operators on $\ell_\infty$ (its predual is the space $\ell_\infty\hat{\otimes}\ell_1$ where $\hat{\otimes}$ denotes the projective tensor product) and a noncommutative discrete group $G$? Same question with $E=B_{w*}(\ell_\infty)$ the space of weak* continuous bounded operators on $\ell_\infty$. –  BigBill Nov 11 '10 at 22:55
    
I don't know the answer to your last question. I will look into it. BTW commutativity is not needed in my previous comment, the group of isometries is left-reversible semigroup, so the theorem in math.gmu.edu/~tlim/pams74.pdf applies. –  TCL Nov 13 '10 at 15:31

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