Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(Note: I have a very little knowledge in the area related to this question. (My research is more related to Combinatorics.) So, I apologize in advance if I say something wrong or trivial.)

One version of the Brouwer fixed point theorem states that any continuous function from the closed unit ball in n-dimensional Euclidean space to itself must have a fixed point. Moreover, it is possible that many fixed points exist.

One specific question I would like to ask is whether there exists a unique fixed point if we restrict the function to be convex. Any pointers to something related or counter examples would be appreciated.

More generally, what are the restrictions on the functions that are sufficient to guarantee the uniqueness. The only result I'm aware of is Banach fixed point theorem which says that if the function is a contraction mapping (thus continuous) then there is a unique fixed point. Are there other restrictions like this?

share|improve this question
1  
Defiinition of convex in dimension $\ge 2$ ? –  Gerald Edgar Aug 7 '11 at 12:21
    
Concerning your last paragraph question, you can of course replace contraction by the property that $d(f(x),f(y))<d(x,y)$ for all $x,y$. –  Benoît Kloeckner Aug 7 '11 at 12:28
    
@Gerald Edgar: This is probably too late to clarify your question above as you and others already answered my question. Anyway, what I meant is actually that the mapping is coordinate-wise convex in the sense that, for all $1\leq i\leq n$, it is convex when we fix the values on all coordinates except the $i^{th}$ one. –  Danu Aug 7 '11 at 13:36
1  
I just wanted to note that "unique fixed point" is similar in spirit to "invertible" and what is invertible is the (generalized) derivative of a real-valued convex function. –  Deane Yang Aug 7 '11 at 14:59

2 Answers 2

up vote 4 down vote accepted

As to the specific question in the next to last paragraph, the identity function is convex. Except for the zero dimensional case there are many fixed points.

share|improve this answer

Surely you can come up with a convex function $[0,1] \to [0,1]$, not the identity, with two fixed points.

share|improve this answer
3  
$y=x^2$ seems ok –  Francesco Polizzi Aug 7 '11 at 12:54
    
Thank you for the answer. Since there are two answers for my silly question but I can accept only one answer, I'll accept Jay Kangel's since it came first. Thanks again. –  Danu Aug 7 '11 at 13:41

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.