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> Suppose $f:I\rightarrow\mathbb R$ has antiderivatives on an interval $I\subset\mathbb R$. Then $f$ can be decomposed as $f=g+h$, where both $g,h:I\rightarrow\mathbb R$ have antiderivatives. In addition, $g$ is bounded below and $h$ is bounded above.
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I have been long thinking about this statement, but it proved elusive; I was neither able to show it is true in the general case, nor come up with a counterexample. So I am posting it here in the hope that somebody with much deeper real analysis knowledge than me might come up with an answer or an idea.



Decompose a function having antiderivatives into bounded components
In general, $f$ does not admit such a decomposition. If it does, $f=g+h=(g+m)+(h-m)$, so we can assume $g\ge0$ and $h\le0$, that is, any antiderivative $F$ of $f$ is bounded variation. But in general this is not the case for an everywhere derivable function , e.g $F(x)=x^2\sin( 1/x^2)$ for $x\neq0$, $F(0)=0$.


