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I was thinking last days about the following problem - the sum of Darboux is a Darboux function?

Do You know a proof or counter example?

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No this is not true, as darboux functions are really a general class of functions, you can proof that you can write any (really any) real valued function as the sum of two Darboux functions, look here

The example of such a function from the link is, with $$\omega(x)=\limsup_{n\to \infty} \frac{1}{n} \sum_{i=1}^n a_i$$ where $x\in(0,1)$ has the dyadic expansion $x=0.a_1 a_2 \dots$ $\omega$ takes every value between $0$ and $1$ in any subinterval of $(0,1)$ and has the intermediaty value property. Now we denote $$g(x)=\begin{cases} 0 & \omega (x)=x\\ \omega(x) & \text{else}\\ \end{cases}$$ This function still hase the intermediate value property but $h(x)=g(x)-x$ doesn't have the intermediate value property.

The proof is here on page 6 theorem 4.1

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  • $\begingroup$ Hm... but I want to hear english proof if possible $\endgroup$ – Jonny Mar 14 '13 at 21:51
  • $\begingroup$ here page 6 theorem 4.1 $\endgroup$ – Dominic Michaelis Mar 14 '13 at 21:54
  • $\begingroup$ @Jonny added the example to my post $\endgroup$ – Dominic Michaelis Mar 14 '13 at 22:06
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There's a simpler example. Let $g,h: \mathbb{R} \to \mathbb{R}$ be given by

$$g(x)=\begin{cases} {\sin x^{-1}} & x \neq 0 \\ 1 & x=0\\ \end{cases}$$

$$h(x)=\begin{cases} {-\sin x^{-1}} & x \neq 0 \\ 0 & x=0\\ \end{cases}$$

These are Darboux. However,

$$g(x) + h(x) =\begin{cases} 0 & x \neq 0 \\ 1 & x=0\\ \end{cases}$$

is not.

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