# Existence of nonnegative bounded derivatives with a dense set of zeros

Assume that $f: \mathbb [0,1] \rightarrow \mathbb R$ is a bounded derivative,
that is $f$ is bounded and $f=F'$ for some differentiable $F: \mathbb [0,1] \rightarrow \mathbb R$. In Wikipedia see here, there is a construction nontrivial (not identically equal $0$) bounded derivative $f$ for which the set $\{x\in [0,1]: f(x)=0 \}$ is dense in $[0,1]$.

Edit. Let $f$ be nontrivial bounded derivative with dense set of zeros. How to prove modifying $f$ that there is a nonzero nonnegative bounded derivative $g$ with the dense set $\{x\in [0,1]: g(x)=0 \}$ in $[0,1]$ ?

• I am confused. Isn't the $f$ in the wikipedia article the example you are looking for? – Harald Hanche-Olsen Jan 8 '14 at 16:27
• Thanks. I thout about another example, the article in Wikipedia which I cited but I not read solve its question. However (also in Wikipedia) is staded that "typical" bounded derivative with dense set of zeros has both positive and negative values. I wanted to modify such a function in such a way to obtain a positive bounded derivative with dense set of zeros. – A.B Jan 8 '14 at 16:59
• Any derivative that is defined throughout an interval satisfies the intermediate value property on that interval. This is a fairly well known result in real analysis, but not something I'm going to try proving in this comment. However, using this property, it follows that if a derivative defined on an interval has has a dense (in that interval) set of positive values and a dense (in that interval) set of negative values, then the derivative must also have a dense (in that interval) set of zero values. – Dave L. Renfro Jan 8 '14 at 17:12
• FYI, here's a proof I posted back in 2006 that derivatives have the intermediate value property. There are probably nicer looking (LaTeX'ed) proofs in math StackExchange, but some of my tangential technical comments might still be of interest. – Dave L. Renfro Jan 8 '14 at 17:25
• FYI, there exists $f:{\mathbb R} \rightarrow {\mathbb R}$ such that $f$ is a derivative, and $f(x) \geq 0$ for each $x \in {\mathbb R},$ and $\{x \in {\mathbb R}: f(x) = 0\}$ is dense in $\mathbb {R},$ and $\{x \in {\mathbb R}: f(x) > 0 \}$ is dense in $\mathbb {R}.$ This post and this other post might be of use, especially my comments in the references section of the second post. – Dave L. Renfro Jan 8 '14 at 20:10