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Let $\Omega$ be an open bounded domain in $\mathbb{R}^N$ and let $J:W_0^{1, 2}(\Omega)\setminus\{0\}\to\mathbb{R}$ be a $\mathcal{C}^1$ functional. Let $(u_n)_n$ be a sequence such that $$ u_n\to u \mbox{ in } L^2(\Omega), \quad u_n\rightharpoonup u \mbox{ in } \ W_0^{1, 2}(\Omega)\quad\mbox{ and }\quad J^{\prime}(u_n)\to 0 \mbox{ in } W^{-1, 2}(\Omega).$$

My question is: under these hypotheses, it is true that $$J^{\prime}(u_n)(u_n -u)\to 0\quad\mbox{ and }\quad J^{\prime}(u)(u_n -u)\to 0?$$

If yes, could anyone please explain me why?

I guess it follows from the assumptions that $J^{\prime}(u_n)\to 0 \mbox{ in } W^{-1, 2}(\Omega)$ and $u_n\rightharpoonup u \mbox{ in } \ W_0^{1, 2}(\Omega)$, but I don't know how to use these informations.

Could anyone please help? Thank you in advance!

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If $x_n\to x$ in $X$ and $f_n\rightharpoonup f$ in $X^*$ then $f_n(x_n)\to f(x)$. To prove this, use $$ f_n(x_n)- f(x) = f_n(x_n-x) + (f_n-f)(x) $$

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  • $\begingroup$ Sorry, could you give me some more details? $\endgroup$
    – C. Bishop
    Feb 24, 2021 at 14:25
  • $\begingroup$ you need to prove that the two summands converge both to zero. Use boundedness and the convergence assumptions. $\endgroup$
    – daw
    Feb 24, 2021 at 15:56
  • $\begingroup$ I am interested in that notation ⇀. What does it stand for? $\endgroup$ Mar 31, 2021 at 10:57
  • $\begingroup$ weak convergence $\endgroup$
    – daw
    Apr 1, 2021 at 11:21

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