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Is there any result in fixed point theory which will give the existence of a fixed point for a continuous function defined on a non-compact, closed and bounded convex set?

Praveen
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  • What about the unit ball in $\ell^2$? – Alex Ravsky Dec 20 '14 at 07:25
  • I expect that many good topological spaces have a continuous map into itself without fixed points. Maybe there exists a topological space $X$ admitting only identical and constant continuous maps into itself, but this space should be rather exotic. – Alex Ravsky Dec 20 '14 at 07:30
  • @ Alex Ravsky : I think you mean closed unit ball $\bar D$ in $\ell^2$. Since we have $(e_i)_{i=1}^{\infty} \subset \bar D$ and this sequence doesn't have any convergent subsequence, $\therefore \bar D$ is not sequentially compact, and hence not compact. So we have a closed, bounded and non-compact space $\bar D$ in $\ell^2$. Also using triangle inequality of norm the space $\bar D$ is convex. But what about fixed-points of any continuous function on $\bar D$? – Praveen Dec 20 '14 at 08:03
  • Yes, this is a first non-trivial relevant question, which came to my mind. The answer should be known, but I don't know it. – Alex Ravsky Dec 20 '14 at 08:11
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    Duplicate of http://math.stackexchange.com/questions/275890/how-to-define-continuous-f-b-rightarrow-b-without-fixed-points ? – Moishe Kohan Dec 20 '14 at 14:50
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    Write Schauder fixed point theorem in Google. – Tomás Dec 20 '14 at 15:37

1 Answers1

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The contraction mapping theorem seems to do the job, without needing closedness nor convexity.

RikOsuave
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  • Its only for contraction mapping, but see the question, I am asking a general case "for any continuous function". – Praveen Dec 20 '14 at 07:12
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    @supremum In general, continuous functions don't have fixed points, so some additional hypotheses are required. –  Dec 20 '14 at 07:19
  • @ Behaviour : I am asking for the hypotheses when the function is defined on a particular kind of space(non-compact,closed,bounded and convex). Hope you understand what I am asking. – Praveen Dec 20 '14 at 07:28
  • @Tomás: That case is ruled out by the conditions of the question, because in a finite dimensional vector space, every closed bounded set is compact. – Lee Mosher Dec 20 '14 at 14:32
  • You are right @supremum . I deleted my comment and posted another in the question. Take a look on it. – Tomás Dec 20 '14 at 15:40