Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
"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. –  Daniel Fischer Jul 13 '13 at 15:39
1  
What is the nonl tag for? –  draks ... Jul 13 '13 at 20:08
    
@draks... My guess is a foreshortened nonlinear-something that never really got off the ground. –  Sharkos Jul 13 '13 at 22:56

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.