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I have this lemma+proof and i dont understand why it follows from $J'(u_n)\rightarrow 0$ that $-\Delta_p u_n- f(x,u_n)\rightarrow 0$ such that

$J(u)=\frac1p\int_{\Omega} |\bigtriangledown u|^p dx-\int_{\Omega} F(x,u) dx, u\in W^{1,p}_0(\Omega) $ where $F(x,t)=\int_0^t f(x,s)ds$ enter image description here

This lemma is from the paper

enter image description here

Help me please

Thank you

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up vote 1 down vote accepted

This is really about what the Frechet derivative $J'(u)$ is. For given $u\in W^{1,p}_0$ it is by definition the unique element in the dual space $(W^{1,p}_0)^*=W^{-1,p'}$ such that $$ J(u+h)=J(u)+\left<J'(u),h\right>_{(W^{1,p}_0)^*,W^{1,p}_0}+o(|h|_{W^{1,p}_0}). $$ Here you can easily compute for fixed $u\in W^{1,p}_0$ and arbitrary "test-function" (or first order perturbation) $$ \frac{1}{p}\int_{\Omega}|\nabla(u+h)|^p=\frac{1}{p}\int_{\Omega}|\nabla u|^p+\int_{\Omega}|\nabla u|^{p-2}\nabla u\cdot\nabla h+o(|\nabla h|_{L^p}) $$ and $$ \int_{\Omega}F(x,u+h)\,dx=\int_{\Omega}F(x,u)\,dx+\int_{\Omega}\partial_u F(x,u)h\,dx+o(| h|_{L^p}) $$ hence by definition of $f=\partial_u F$ $$ \left<J'(u),h\right>_{(W^{1,p}_0)^*,W^{1,p}_0}=\int_\Omega|\nabla u|^{p-2}\nabla u\cdot\nabla h \,dx-\int_{\Omega}f(x,u) h\,dx. $$ The first term in the right-hand side is nothing but the weak formulation of the (negative) $p$-Laplacian, i-e assuming that everything is smooth enough and integrating by parts $$ \int_\Omega|\nabla u|^{p-2}\nabla u\cdot\nabla h=-\int_\Omega\operatorname{div}(|\nabla u|^{p-2}\nabla u) h=-\int_{\Omega}(\Delta_p u)h. $$ This means that $J'(u)=-\Delta_p u-f(x,u)$ as elements of $(W^{1,p}_0)^*$, which is exactly your statement.

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