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

A function $f: I \to J$ ($I,J$ intervals) has the Darboux property or the Intermediate value property if for every $a < b \in I$ and for every $\lambda$ between $f(a)$ and $f(b)$ there exists $c \in [a,b]$ such that $f(c)=\lambda$. Equivalently, the image of any interval under $f$ is an interval.

I know that there are functions $f: [0,1] \to [0,1]$ which have the Darboux property, but have no fixed points.

What are some articles which can be taken as references for this non-existence theorem? The only one I found was this, but I guess that there are older articles which deal with this subject. I searched Google and Mathscinet, but didn't find any except the one above (maybe I don't know how to search...).

share|improve this question
add comment

2 Answers

The standard reference for Darboux functions is

Andrew Michael Bruckner and Jack Gary Ceder, Darboux continuity, Jahresbericht der Deutschen Mathematiker-Vereinigung 67 (1965), 93-117. MR 32 #4217l; Zbl 144.30003

http://eudml.org/doc/146526;jsessionid=98522A06CD68A44763F32C1354F068AB

Given its publication date, I don't think you'll find much about fixed points of Darboux functions in this paper, but you should look over the paper anyway, plus it's an excellent survey that should be quite useful in general. However, and this is a useful search tip, you can use the paper to refine your search. Search google using the phrases "Darboux continuity" AND "Bruckner" AND "fixed point". The idea is that the hits you get will be those papers and other items that cite Bruckner/Ceder's paper, and thus you'll exclude the more superficially researched items.

http://www.google.com/search?q=%22Darboux+continuity%22+Bruckner+%22fixed+point%22

share|improve this answer
1  
The reference on Darboux functions is awesome. Thank you. –  Beni Bogosel Apr 10 '12 at 20:31
add comment

In Exercise 9.C of van Rooij - Schikhoff: A Second Course on Real Functions the following example is given.

Take any function $f \colon \mathbb R\to\mathbb R$ such that $f[I]=\mathbb R$ for every non-degenerate interval.

(There are many examples of such functions, see here or here.) Several such examples are also given in the book I've mentioned.

Define $$g(x):= \begin{cases} f(x) & \text{if }f(x)\ne x, \\ x+1 & \text{if }f(x)=x. \end{cases} $$ Prove that $g$ is Darboux continuous.

This example is given there as an example of a function, which is Darboux continuous but does not have connected graph. But this function also has the property that $f(x)\ne x$ for every $x\in\mathbb R$.

But the authors of this book did not explicitly stated where this example is taken from.


I'll add my solution of this part of the exercise.

$g(x)$ is strongly Darboux. Let $I=(a,b)$ and $I_1=(a,(2a+b)/3)$, $I_2=((2a+b)/3,(a+2b)/3)$, $I_3=((a+2b)/3,b)$ be a division of $I$ into three parts. We know that $f[I_1]=\mathbb R$, which implies $g[I_1]\supseteq \mathbb R\setminus I_1$. (If $x\in I_1$ and $f(x)\notin I_1$ then necessarily $f(x)=g(x)$.)

By the same argument $f[I_3]\supseteq \mathbb R\setminus I_3$. Hence $f[I] \supseteq f[I_1]\cup f[I_3] \supseteq (\mathbb R\setminus I_1)\cup(\mathbb R\setminus I_3)=\mathbb R$.


A rather similar example is given in Andrew Michael Bruckner and Jack Gary Ceder, Darboux continuity, Jahresbericht der Deutschen Mathematiker-Vereinigung 67 (1965), 93-117, eudml; as Example 4.1. (In this case, the function is defined on $(0,1)$.)

The references given there are Knaster, B. and Kuratowski, K.: Sur quelues propriétés topologiques des fonctions dérivées, Rend. Circ. Mat. Palermo. 49 (1925), 382-386; and Kuratowski's Topologie II (Warszawa 1952), p.82.

share|improve this answer
    
BTW the name strongly Darboux function is used in my post for function which maps every non-trivial interval to $\mathbb R$. –  Martin Sleziak Jan 10 at 16:04
add comment

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.