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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.

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

1 Answer 1

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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 $

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