TITLE: Weak lower semicontinuity of functional with two arguments
QUESTION [6 upvotes]: Let $\Omega$ be a bounded and smooth domain and let $J:H^1(\Omega) \times H^1_0(\Omega) \to \mathbb{R}$ be defined by
$$J(u,v) = \int_\Omega f(u)|\nabla v|^2$$
where $f\colon \mathbb{R} \to \mathbb{R}$ is a smooth function, bounded above and below away from zero (other assumptions can be added as necessary).
Under what conditions on $f$ do I get that $J$ is weakly lower semicontinuous? 
Obviously if $f \equiv 1$ then it is true, but what about the more genera case?

REPLY [2 votes]: Here is the proof of @JohannesHahn in a more elementary way.
The idea is to show that
$$
(u,v) \mapsto \sqrt{ f(u)}\nabla v
$$
is weakly sequentially continuous from $H^1(\Omega)\times H^1(\Omega)$ to $L^2(\Omega)$.
Let me denote $g(u):=\sqrt{ f(u)}$. Let $(u_n,v_n) \rightharpoonup (u,v)$ in $H^1(\Omega)\times H^1(\Omega)$.
Then (after possibly extracting subsequences) $u_n \to u$ in $L^2(\Omega)$ and $u_n(x)\to u(x)$ pointwise almost everywhere. 
Let $w\in L^2(\Omega)$. By dominated convergence,
$$
\int_\Omega |g(u_n)-g(u)|^2 w^2 \to 0
$$
Then we have
$$
\int_\Omega (g(u_n)\nabla v_n - g(u)\nabla v)w  \ dx 
=\int_\Omega (g(u_n) - g(u))\nabla v_n w\ dx + \int_\Omega g(u)(\nabla v_n - \nabla v)w \ dx.
$$
The first integral goes to zero, since $(g(u_n)-g(u))w\to0$ in $L^2(\Omega)$. The second integral vanishes due to the weak convergence $\nabla v_n - \nabla v\rightharpoonup 0$ in $L^2(\Omega)$.
It follows
$$
g(u_n)\nabla v_n \rightharpoonup g(u)\nabla v
$$
in $L^2(\Omega)$
as required. (Here the usual subsequence argument has to be used: Every subsequence of $(g(u_n)\nabla v_n)$ contains a subsequence converging weakly to $g(u)\nabla v$.)

Note that a similar argument does not work when applied directly to the functional:
$$
\int_\Omega f(u_n)|\nabla v_n|^2 - f(u)|\nabla v|^2\ dx
= 
\int_\Omega (f(u_n)-f(u))|\nabla v_n|^2\ dx + \int_\Omega f(u)(|\nabla v_n|^2- |\nabla v|^2)\ dx.
$$
While the $\liminf$ of the second integral is non-negative, nothing can be said about the first integral: $f(u_n)-f(u)$ does not converge to zero in $L^\infty(\Omega)$ if $\Omega\subset \mathbb R^d$ for $d>1$.