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Let $f \in W^{1,p}\bigl([0,1], \mathbb{R} \bigr)$ with $p \in (1,\infty)$. Then, we can assume that $f$ is absolutely continuous with the classical derivative a.e. equal to the weak derivative, which we denote as $f' \in L^p[0,1]$.

Now, assume that $f(t)f'(t) \leq 0$ for a.e. $t \in [0,1]$. Then, does the product rule apply to $[f(t)]^2$ in the sense that $[f(t)]^2$ is a.e. differentiable in the classical sense and satisfies \begin{equation} \bigl( f^2 \bigr)' = 2 ff' \text{ almost everywhere on } [0,1]? \end{equation}

Or do we at least have any inequality bewteen $\bigl( f^2 \bigr)'$ and $2 ff'$?

Could anyone please clarify my confusion?

Keith
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    I am not so familiar with Sobolev spaces, but $\bigl( f^2 \bigr)'(x) = 2 f(x)f'(x)$ holds at every point $x$ where the classical derivative $f'(x)$ exists. – Martin R Feb 02 '24 at 17:55
  • @MartinR Well...I am dumb... thank you. – Keith Feb 03 '24 at 00:29
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    Product rule/Leibniz rule is a classical result, and it can be proven for Sobolev functions as well, see, e.g. the book by Evans. – daw Feb 04 '24 at 08:41

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