# Schauder fixed point extended

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?

• 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? – Aweygan Feb 26 at 19:54
• $K$ 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. – R. N. Marley Feb 26 at 20:12
• Well does either the map $T$ or $h^{-1}Th$ satisfy the compactness criterion? – Aweygan Feb 26 at 21:07
• Yes, $T$ is compact and I am wondering if $h^{-1}Th$ is also compact Or not. – R. N. Marley Feb 27 at 10:25

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.
• 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. – R. N. Marley Feb 27 at 17:06
• What do you mean by $T$ is compact? I'm just reading the Schauder fixed point theorem from Wikipedia. – Aweygan Feb 27 at 17:10
• $T:\overline{\Omega}\rightarrow X$ (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. – R. N. Marley Feb 27 at 17:22