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January 26, 2014 1 comment

We say that ${f: \Bbb{R} \rightarrow \Bbb{R}}$ is an additive function if

$\displaystyle f(x+y)=f(x)+f(y),\ \forall x,y \in \Bbb{R}.$

1. Prove that there exist additive functions which are discontinuous with or without the Darboux Property.

2. Prove that for every additive function ${f}$ there exist two functions ${f_1,f_2:\Bbb{R} \rightarrow \Bbb{R}}$ which are additive, have the Darboux Property, and ${f=f_1+f_2}$.

The second part is similar to Sierpinski’s Theorem which states that every real function can be written as the sum of two real functions with Darboux property.

(A function ${g:I \rightarrow \Bbb{R}}$ has the Darboux property if for every ${[a,b]\subset I}$, ${g([a,b])}$ is an interval.)

Darboux Functions with no Iterate Fixed Points

It is well known that there exist functions ${f:[0,1] \rightarrow [0,1]}$ which have Darboux property and they have no fixed points. An example can be found in an earlier post of mine. Here is a generalization of that result.

There exist functions ${f:[0,1] \rightarrow [0,1]}$ which have Darboux property and for which none of its iterates has a fixed point, i.e. ${\underbrace{f\circ ..\circ f}_{n \text{ times}}(x)\neq x}$ for every ${x \in [0,1]}$ and for every ${n \geq 1}$.

Contest problem

There exist functions $f:[0,1]\to [0,1]$ with Darboux property, such that there exist sets $A,B$ with $A\cap B=\emptyset,\ A\cup B=[0,1]$ and $f(A)\subseteq B,\ f(B)\subseteq A$.
For any function $f:\mathbb{R} \to \mathbb{R}$ there exist two functions $f_1, \ f_2$ such that $f=f_1+f_2$ and $f_1,f_2$ have the Darboux property.
A function has the Darboux property if for any interval $I\subseteq \mathbb{R}$ we have $f(I)$ is also an interval. This is slightly different from continuity and intermediate value property. Cotinuity implies Darboux and Darboux implies Intermediate value property.