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From Wikipedia: Darboux functions are a quite general class of functions. It turns out that any real-valued function f on the real line can be written as the sum of two Darboux functions. This implies in particular that the class of Darboux functions is not closed under addition.

(Darboux functions are simply those that satisfy the intermediate value property).

Proof? I'm looking for this out of interest, and couldn't find one - a hint, nudge, reference or link are also sufficient. If you're feeling brave: can we extend this, say to $f:\mathbb{C} \to \mathbb{R}$? Proof or counter-example, of course.

EDIT: Further question: If a function $f:[a,b] \to \mathbb{R}$ is differentiable on $[a,b]$ then its derivative $f'$ is Darboux on $[a,b]$. Can any real function be written as $f'+g'$ for some $f,g$? (no)

ShakesBeer
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    See this. For the edit, the answer is no. Otherwise, every function would be a derivative, which isn't true. – David Mitra Sep 27 '14 at 09:11
  • Not to be an arse (well, ok, I guess precisely to be an arse), but let me google that for you. The very first PDF hit contains a reference to the paper by Sierpinski where the result is proven. – Jonas Dahlbæk Sep 27 '14 at 09:11
  • @user161825 I said a link/reference was fine and clearly I'd looked, just simply not well enough. No need to be an arse... – ShakesBeer Sep 27 '14 at 09:13
  • Well, since I supplied you with what you wanted, I would only classify my comment as that of a half-arse. – Jonas Dahlbæk Sep 27 '14 at 09:16
  • @user161825: Have you read that paper?I came to this on the very first page: "Definition 2. If $f : I \to \mathbb R$ is a function, and $I \subset \mathbb R$ is an interval, $f$ has the Intermediate Value property if $f(I)$ is an interval." That looked like nonsense to me, so I gave up. – TonyK Sep 27 '14 at 09:17
  • @TonyK That is precisely correct. – ShakesBeer Sep 27 '14 at 09:18
  • Do you mean the definition is precisely correct? It's nonsense is what it is. – TonyK Sep 27 '14 at 09:19
  • @TonyK I could ask you the same thing. What they use in the proof is that if $f(I)=\mathbb R$ for any interval I, then $f$ is Darboux, and this implication is trivial. Also, this took me 1 minute to figure out, and I have never before met the term Darboux function. – Jonas Dahlbæk Sep 27 '14 at 09:20
  • @user161825: But that's not what the definition says. $I=[0,2]$ is an interval, and $f: x \mapsto {x}$ (where ${x}$ denotes fractional part of $x$) has the property that $f(I)$ is an interval. So according to the definition, $f$ has the Intermedate Value property. – TonyK Sep 27 '14 at 09:24
  • @TonyK your $f$ does. The whole thing with the intervals is not the usual definition, but it quickly implies the usual definition. – ShakesBeer Sep 27 '14 at 09:26
  • My $f$ doesn't have the Intermediate Value property! For instance, $f(0.9) = 0.9$ and $f(1.1) = 0.1$, but there is no $x \in [0.9,1.1]$ such that $f(x) = 0.5$. – TonyK Sep 27 '14 at 09:28
  • @TonyK the IVP states that there must be some $x \in [0,2]$ such that ${x}=0.5$... Wikipedia actually has both definitions http://en.wikipedia.org/wiki/Intermediate_value_theorem – ShakesBeer Sep 27 '14 at 09:30
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    No, you have it wrong. The Intermediate Value property says that for any interval $I$ in the domain of the function, $f(I)$ is an interval. Which is not true of my function $f$. – TonyK Sep 27 '14 at 09:32
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    @TonyK They explicitly state afterwards that the Darboux property is stronger than what they define as the intermediate value property. So I don't understand what the problem is. What you call the intermediate value property is what they call the Darboux property. – Jonas Dahlbæk Sep 27 '14 at 09:33
  • @TonyK your definition is much stronger than what people usually call the IVP. Note that in the case $f(x)={x}$, $f([0.9,1.1])$ is not an interval anyway. – ShakesBeer Sep 27 '14 at 09:36
  • @Shakespeare: According to your own Wikipedia link, the Darboux property and the Intermediate Value property are the same thing. – TonyK Sep 27 '14 at 09:37
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    @TonyK In the article it is never claimed that their two definitions are equivalent, on the contrary it is stated (in the first line after the definitions are given) that the properties are different from each other. – Jonas Dahlbæk Sep 27 '14 at 09:41
  • Even on its own terms, the definition doesn't make sense. It has to say "...$f$ has the Intermediate Value property on $I$ if $f(I)$ is an interval." I wouldn't agree with that definition, but at least it wouldn't be nonsense. – TonyK Sep 27 '14 at 09:48
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    @TonyK There is no reason to write 'on $I$', because $f:I\rightarrow \mathbb R$ is a function on $I$ already, so it would be like writing $f$ has the IVP on the domain of $f$, which is not necessary. As the definition stands, it is quite meaningful. – Jonas Dahlbæk Sep 27 '14 at 10:09

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Further question: $f' + g' = (f+g)'$ is a derivative and is therefore a Darboux function. So no, not every real function can be written as $f'+g'$.

TonyK
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