I am working in a exercise, to my solution works I need the following affirmation is true:
Let $\varphi : R \rightarrow R$ a convex and smooth function. Let $u \in H^{1}(U)$ a bounded function and $v \in H^{1}_{0}(U)$ a non negative function. Then $\varphi^{'}(u)v \in H^{1}_{0}(U)$.
I am trying to use the definiton , but it's not working. Someone can give me a hint? Thanks in advance
You need some more assumptions. Let us consider the simple case $\varphi(x) = x^2$, then $\varphi'(x) = x$. You want to have $u \, v \in H^1(\Omega)$ for $u, v \in H^1(\Omega)$. Using the product rule one can show that the weak derivative of the product is $\nabla (u \, v) = u \, \nabla v + v \, \nabla u$. However, this function may not belong to $L^2(\Omega)$. You may want to assume $u, v \in H^1(\Omega) \cap L^\infty(\Omega)$.