# Under what conditions can a function $y: \mathbb{R} \to \mathbb{R}$ be expressed as $z z'$?

This is a follow-up to Under what conditions can a function $y: \mathbb{R} \to \mathbb{R}$ be expressed as $\dfrac{z'}{z}$?. It turns out that in that case, \begin{align} \text{$y = \dfrac{z'}{z}$ for some differentiable $z$} & \iff \text{$y$ has an anti-derivative} \\ & \iff \text{$y$ is discontinuous on a meager $F_{\sigma}$-subset of $\mathbb{R}$}. \end{align} I was thus wondering whether the answer is still the same in this case.

• And your take on this would be?
– Did
Jun 16, 2014 at 7:36

Claim: Let $y: \mathbb{R} \to \mathbb{R}$. Then there exists a differentiable $z: \mathbb{R} \to \mathbb{R}$ such that $y = z z'$ if and only if $y$ has a non-negative anti-derivative.
Proof: Suppose that there exists a differentiable $z: \mathbb{R} \to \mathbb{R}$ such that $y = z z'$. Then $$y = z z' = \left( \frac{1}{2} z^{2} \right)'.$$ Hence, $\dfrac{1}{2} z^{2}$ is a non-negative anti-derivative of $y$.
Conversely, suppose that $y$ has a non-negative anti-derivative. By adding the constant function $1$ to it, we obtain a positive anti-derivative $F$ of $y$. Let $z \stackrel{\text{def}}{=} \sqrt{2 F}$. Then $z$ is well-defined (since $F \geq 0$), is positive and is differentiable everywhere (since $F > 0$). By the Chain Rule, $$z' = (\sqrt{2 F})' = \frac{1}{2 \sqrt{2 F}} \cdot 2 F' = \frac{F'}{\sqrt{2 F}} = \frac{y}{z}.$$ Hence, $y = z z'$. $\quad \blacksquare$