# Mixed strong-weak convergence problem

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?

• Could you specify what you know on $J$ ? Mar 15, 2021 at 18:07
• @Velobos $J$ is a $\mathcal{C}^1$ functional such that $|J(u_n)|\leq C$ for all $n$ (C is a positive constant) and $J^{\prime}(u_n)\to 0$ in $W^{-1, 2}(\Omega)$. Mar 16, 2021 at 7:23
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 $$f_n = J'(u_n).$$