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Theorem 1. Let $B=\{x\in \mathbb R^n :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^n$ . Any continuous function $f:B\rightarrow B$ has a fixed point.

Theorem 2. Let $X$ be a finite dimensional normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.

Theorem 3. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.

Theorem 4. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, closed, and bounded set. Then given any compact mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.

For some authors Theorem 1 is Brouwer's fixed-point theorem. For others Brouwer's fixed-point theorem is Theorem 2. Actually there is no difference because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball.

My problem is with theorems 3 and 4. For some authors Theorem 3 is Schauder's fixed-point theorem, for others Schauder's fixed-point theorem is Theorem 4.

Are Theorem 3 and Theorem 4 are equivalent? If not, are Theorems 1 and 2 special cases of Theorem 4?

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  • $\begingroup$ "because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball" is not quite true, it could be lower-dimensional. Theorems 1 and 2 are special cases, because in that case, a continuous $f \colon K \to K$ is compact. $\endgroup$ Jul 13, 2013 at 15:39
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    $\begingroup$ What is the nonl tag for? $\endgroup$
    – draks ...
    Jul 13, 2013 at 20:08
  • $\begingroup$ @draks... My guess is a foreshortened nonlinear-something that never really got off the ground. $\endgroup$ Jul 13, 2013 at 22:56

1 Answer 1

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Your statement of Theorem 4 is missing an assumption on $K$, such as being convex, or at least homeomorphic to such a set (convex, closed, bounded). Without such an assumption, rotation of a circle gives a counterexample. Also, I think that in Theorem 4 you want the normed space to be complete, i.e., a Banach space.

Theorem 3 is contained in Theorem 4, because on a compact set every continuous map is compact. Theorem 4 cannot be easily obtained from Theorem 3 (I think) because if we tried to simply replace $K$ with $\overline{f(K)}$ (which is compact), we can't apply Theorem 3 because $\overline{f(K)}$ is not known to be convex.

Both 3 and 4 were stated and proved by Schauder in his 1930 paper Der Fixpunktsatz in Funktionalraümen, which is in open access. Here is Theorem 3:

Satz I. Die stetige Funktionaloperation $F(x)$ bilde die konvexe, abgeschlossene und kompakte Menge $H$ auf sich selbst ab. Dann ist ein Fixpunkt $x_0$, vorhanden, d.h. es gilt $F(x_0)=x_0$.

And this is Theorem 4 (in slightly less general version: the image of $F$ is assumed compact instead of relatively compact; possibly because the latter concept wasn't in use).

Satz II. In einem "B"-Raume sei eine konvexe und abgeschlossene Menge $H$ gegeben. Die stetige Funktionaloperation $F(x)$ bilde $H$ auf sich selbst ab. Ferner sei die Menge $F(H)\subset H$ kompakt. Dann ist ein Fixpunkt vorhanden.

("B"-Raume is what is now called a Banach space.) So, it is correct to call both Theorem 3 and Theorem 4 "Schauder's fixed-point theorem".

And yes, Theorems 1 and 2 follow by specialization of Theorem 3 or 4 to finite dimensions.

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  • $\begingroup$ In fact, we don't need completeness. Let $X$ be a locally convex topological vector space, $K\subset X$ nonempty convex subset, $C\subset K$ compact, $f:K\to C$ continuous, then $f$ admits a fixed point. $\endgroup$
    – Doug
    Apr 30, 2022 at 8:32

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