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The Schauder fixed point theorem states that if $X$ is a Banach space, $K\subset X$ is a convex, bounded and closed subset and $T:K\rightarrow K$ is compact, then $T$ has, at least, one fixed point in $K$. I want to know if this statement still holds true when the subset $K$ is replaced by another subset $D$ that is homeomorphism to $K$.

Roughly speaking, if $h:K\rightarrow D$ is an homeomorphism, then $h^{-1}Th:K\rightarrow K$ and I am able to apply the Schauder fixed point theorem if $h^{-1}Th$ is compact.

Is there any result that guarantee that the compactness of operators is preserved by homeomorphism or something similar?

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  • $\begingroup$ The wording of your question is tripping me up. In the second paragraph, do $D$ and $T:D\to D$ staisfy the hypotheses of the Schauder fixed point theorem? $\endgroup$
    – Aweygan
    Feb 26, 2019 at 19:54
  • $\begingroup$ $K$ satisfies the hypotheses of the Schauder fixed point theorem and $D$ is just an homemorphic set to $K$. The goal is to prove that $T:D\rightarrow D$ also has a fixed point. I am not sure if it is true or not. $\endgroup$
    – Senna
    Feb 26, 2019 at 20:12
  • $\begingroup$ Well does either the map $T$ or $h^{-1}Th$ satisfy the compactness criterion? $\endgroup$
    – Aweygan
    Feb 26, 2019 at 21:07
  • $\begingroup$ Yes, $T$ is compact and I am wondering if $h^{-1}Th$ is also compact Or not. $\endgroup$
    – Senna
    Feb 27, 2019 at 10:25

1 Answer 1

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If $T(D)$ is a compact subset of $D$, then $Th(K)=T(D)$ is a compact subset of $D$, whence $h^{-1}Th(D)$ is a compact subset of $K$. Thus there is some $x_0\in K$ such that $h^{-1}Th(x_0)=x_0$. Put $y_0=h(x_0)$. Then $Ty_0=hh^{-1}Th(x_0)=h(x_0)=y_0$, so $T$ has a fixed point.

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  • $\begingroup$ Your argument is good, but you are assuming that $T(D)$ is compact ?? OR this follows from the fact that $T:K\rightarrow K$ is compact. $\endgroup$
    – Senna
    Feb 27, 2019 at 17:06
  • $\begingroup$ What do you mean by $T$ is compact? I'm just reading the Schauder fixed point theorem from Wikipedia. $\endgroup$
    – Aweygan
    Feb 27, 2019 at 17:10
  • $\begingroup$ $T:\overline{\Omega}\rightarrow X$ (where $\Omega$ is an open and bounded subset of a Banach space $X$) is compact when $T$ is continuous and $\overline{T(\overline{\Omega})}$ is compact. That is the definition that gave to me at university. $\endgroup$
    – Senna
    Feb 27, 2019 at 17:22

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