If given $F(u_k)\to F(u)$ and $F'(u_k)\nabla u_k \to F'(u)\nabla u$ in $L^p$, why we have $F(u) \in W^{1,p}$ and $\nabla F(u)=F'(u)\nabla u?$

Question

If $F$ continuous, $u_k \in C^{\infty}(\overline U)$, $u \in W^{1,p} (U)$.

If given $F(u_k)\to F(u)$ and $F'(u_k)\nabla u_k \to F'(u)\nabla u$ in $L^p$, why we have $$F(u) \in W^{1,p}$$ and $$\nabla F(u)=F'(u)\nabla u?$$

Thanks so much!

Answer 1

You will need more than mere continuity of $F$. What is meant by $F'$? Assuming $F$ is smooth enough (Lipschitz continuous will do) you have the following: if $\phi \in C_0^\infty(U)$, then integration by parts and the usual chain rule imply $$ \int_U F(u_k) \nabla \phi \, dx = - \int_U \nabla (F(u_k)) \phi \, dx = - \int_U (F'(u_k) \nabla u_k) \phi \, dx$$ Under the stated hypotheses, and using the fact that $\phi,\nabla \phi \in L^{p'}$, you may take the limit on both sides to arrive at $$ \int_U F(u) \nabla \phi \, dx = - \int_U (F'(u) \nabla u) \phi \, dx.$$ Since this holds for any $\phi \in C_0^\infty(U)$, by definition the weak gradient of $F(u)$ is $F'(u) \nabla u$.