Let $\Omega$ be an open bounded domain in $\mathbb{R}^N$. Let $J \;$ be a $\mathcal{C}^1$ class functional and $(u_n)_n\subset W_0^{1, 2}(\Omega)$. Suppose that $$ u_n\rightharpoonup u \quad\mbox{ in } W_0^{1, 2}(\Omega)\quad\mbox{ and }\quad J^{\prime}(u_n)\to 0 \quad \mbox{ in } \; W^{-1, 2}(\Omega).$$
How to show that $$J^{\prime}(u_n)(u_n - u)\to 0?$$
On my notes, it seems to be an immediate thing, but I don't know how to justify it. Could anyone please explain me why the above convergence holds?
Thank you in advance!
This follows from the general fact that in a Banach space $X,$ if $x_n \rightharpoonup x$ weakly in $X$ and $f_n \to f$ strongly in $X',$ then $f_n(x_n) \to f(x)$ as $n \to \infty.$ Indeed we can write $$ |f_n(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n-x)|, $$ and by the principle of weak boundedness, $\lVert x_n\rVert_X \leq C$ for all $n$ so we have $|f_n(x_n)-f(x_n)| \leq C \lVert f_n - f\rVert_{X'} \to 0$ as $n \to \infty.$ Also by weak convergence $|f(x_n-x)| \to 0$ as $n \to \infty,$ so it follows that $f_n(x_n) \to f(x)$ as $n \to \infty.$
In your setting take $X = W^{1,2}_0(\Omega)$ $x_n = u_n$ and $f_n = J'(u_n).$