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Fixed points of a weakly continuous map f:H->H satisfying f(f(x))=x
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Jack Stone
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Joined: 08 Jul 2006
Posts: 3

PostPosted: Mon Jul 10, 2006 3:12 pm    Post subject: Fixed points of a weakly continuous map f:H->H satisfying f(f(x))=x Reply with quote
Let H be an infinite-dimensional Hilbert space.

Does there exist a mapping f:H--->H so that:

i) f is weakly continuous,
ii) f(f(x))=x for all x in H,
iii) f has no fixed points (for no x f(x)=x)

or

every mapping f:H--->H satisfying i) and ii) must have a fixed point?
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Lee Rudolph
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Joined: 28 Apr 2005
Posts: 566

PostPosted: Mon Jul 10, 2006 9:19 pm    Post subject: Re: Fixed points of a weakly continuous map f:H->H satisfying f(f(x))=x Reply with quote
Jack Stone <stone1292@mail.com> writes:

Quote:
Let H be an infinite-dimensional Hilbert space.

Does there exist a mapping f:H--->H so that:

i) f is weakly continuous,
ii) f(f(x))=x for all x in H,
iii) f has no fixed points (for no x f(x)=x)

or

every mapping f:H--->H satisfying i) and ii) must have a fixed point?

Let h:H-->S be a diffeomorphism from H to the unit sphere S in H;
it's known that such an h exists. Let f = h^{-1} o A o h, where
A is the antipodal map on S, and there you are.

Lee Rudolph
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G. A. Edgar
science forum Guru


Joined: 29 Apr 2005
Posts: 470

PostPosted: Tue Jul 11, 2006 12:47 pm    Post subject: Re: Fixed points of a weakly continuous map f:H->H satisfying f(f(x))=x Reply with quote
Quote:

Let h:H-->S be a diffeomorphism from H to the unit sphere S in H;
it's known that such an h exists.

But not weakly continuous...

The closed unit ball of H is weakly compact and convex, so every weakly
continuous map has a fixed point. This doesn't solve the
original question, but does imply that f weakly continuous with no
fixed points cannot map any ball into itself.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
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