Talk:Ryll-Nardzewski fixed-point theorem

if E is a normed vector space and K is nonempty convex subset of E, which is closed for the weak topology, then every group of affine isometries of K has at least one fixed point.

This statement is clearly false: take E=K=R², and take the group generated by the translation by 1 in x-direction, which doesn't have fixed points. So at least we need K to be weakly compact. Even with that change, I still don't see how this theorem generalizes the Brouwer fixed point theorem (as stated in Fixed point theorems in infinite-dimensional spaces), which talks about general continuous maps and not just about affine maps. I would also be much more comfortable with assuming a Banach space E rather than just a normed space in this theorem. AxelBoldt 04:04, 11 June 2006 (UTC)[reply]

Also, our article on infinite-dimensional fixed point theorems states that the Ryll-Nardzewsk theorem is from 1967, which accords with the reference I just added to the article, but does not match the 1964 Bourbaki reference given in the article. AxelBoldt 04:17, 11 June 2006 (UTC)[reply]

Oh, the "closed" should of course be replaced by compact. I remember seeing another so-called Ryll-Nardzewski theorem in Fixed point theory by Andrzej Granas, James Dugundji, which was an actual generalisation. I'll try to have a look when I have time. By the way, I was wondering : how can you have an articlemarked as a stubChinedine 11:56, 11 June 2006 (UTC)[reply]

Stub'd. (Added {{topology-stub}}, but {{mathanalysis-stub}} may be more appropriate.) — Arthur Rubin | (talk) 16:17, 11 June 2006 (UTC)[reply]

I'm still worried about the statement of the theorem. The Mathematical Encyclopaedia [1] requires a Banach space, but does not mention that the maps need to be affine isometries (but only that they form a "non-contracting semigroup of mappings", whatever that is) nor that the set has to be convex. Are we talking about two different fixed point theorems here? AxelBoldt 04:25, 12 June 2006 (UTC)[reply]

See Math project talk page where I saw and replied to your query.--CSTAR 12:42, 12 June 2006 (UTC)[reply]

Ok, I copied your reply here, to keep it all in one place:

As stated it's wrong. The semigroup is required to satisfy another property, that it be "distal". Also I don't think it can be used to prove existence of Haar measure on general locally compact groups, although I think for compact groups yes. I think this is in Rudin's functional analysis book for instance. Also see Frederic Greanleaf's little book (now horribly outdated) on "Amenable Groups".--CSTAR 12:41, 12 June 2006 (UTC)[reply]

I take this back. It's correct. I noticed that it says "affine isometries" which I had missed. A semigroup of isometries on a metric space is automatically distal. The assumption requiring the normed space E to be Banach is not necessary as the following argument shows: Complete the underlying space E; the dual space of the completion is the same as that of E, and the inclusion mapping is continuous with both spaces E and its completion equipped with the weak topology. Therefore the original compact convex set is still compact (and convex of course) in the completion.

However the claim regarding Haar measure is true only for compact groups to the best of my knowledge.--CSTAR 03:04, 8 July 2006 (UTC)[reply]

In fact, the existence of the Haar measure for compact groups follows from an older (and easier) theorem -- because for that application it is enough to deal with norm-compact convex sets. I think this easier fixed point theorem is due to Kakutani (although it is not what is usually called the Kakutani fixed point theorem!). This is nicely explained in Rudin's book. — Preceding unsigned comment added by 196.210.218.159 (talk) 18:31, 4 December 2011 (UTC)[reply]