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Piero D'Ancona
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The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^-1(\Sigma)$$u^{-1}(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one defines the product $F'(u)\nabla u$ to be 0, irrespective of the fact that $F'(u)$ is defined or not a such points. See e.g. Leoni, Morini: JEMS 9 pp 219-252

The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^-1(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one defines the product $F'(u)\nabla u$ to be 0, irrespective of the fact that $F'(u)$ is defined or not a such points. See e.g. Leoni, Morini: JEMS 9 pp 219-252

The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^{-1}(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one defines the product $F'(u)\nabla u$ to be 0, irrespective of the fact that $F'(u)$ is defined or not a such points. See e.g. Leoni, Morini: JEMS 9 pp 219-252

Source Link
Piero D'Ancona
  • 9k
  • 1
  • 33
  • 57

The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^-1(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one defines the product $F'(u)\nabla u$ to be 0, irrespective of the fact that $F'(u)$ is defined or not a such points. See e.g. Leoni, Morini: JEMS 9 pp 219-252