Weak lower semicontinuity of functional with two arguments Let $\Omega$ be a bounded domain (smooth if necessary) 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, which can make as nice 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 general case?
 A: This functional is sequentially weakly lower semicontinuous under fairly mild assumptions on $f$.
We need that $f$ is non-negative, continuous and bounded from above.
Let $u_n \rightharpoonup u$ in $H^1(\Omega)$ and $v_n \rightharpoonup v$ in $H_0^1(\Omega)$.
Rellich-Kondrachov implies that $u_n \to u$ in $L^2(\Omega)$ (we might need a mild assumption on $\Omega$ here).
Hence, $f(u_n) \to f(u)$ (along a subsequence) a.e.
For $\varepsilon > 0$ we can use Egorov's theorem to get a subset $E \subset \Omega$
of measure smaller than $\varepsilon$ such that
$f(u_n) \to f(u)$ uniformly on $\Omega \setminus E$.
On $\Omega \setminus E$ we can use
\begin{equation*}
 \int_{\Omega \setminus E} f(u_n) \, |\nabla v_n|^2 \, \mathrm{d}x
 =
 \int_{\Omega \setminus E} f(u) \, |\nabla v_n|^2 \, \mathrm{d}x.
 +
 \int_{\Omega \setminus E} (f(u_n) - f(u)) \, |\nabla v_n|^2 \, \mathrm{d}x
\end{equation*}
For the first addend, we can use weak lower semicontinuity,
whereas the second addend goes to zero.
On $E$ we use the simple estimate
\begin{equation*}
 \int_{E} f(u_n) \, |\nabla v_n|^2 \, \mathrm{d}x \ge 0.
\end{equation*}
Alltogether, we get
\begin{equation*}
 \liminf_{n \to \infty} J(u_n, v_n)
 \ge
 \int_{\Omega \setminus E} f(u) \, |\nabla v|^2 \, \mathrm{d}x
 =
 J(u,v)
 -
 \int_{E} f(u) \, |\nabla v|^2 \, \mathrm{d}x
 .
\end{equation*}
Since $f(u) \,  |\nabla v|^2 \in L^1(\Omega)$, we have
\begin{equation*}
 \int_{E} f(u) \, |\nabla v|^2 \, \mathrm{d}x
 \to
 0
\end{equation*}
for $\varepsilon \searrow 0$.
This shows
\begin{equation*}
 \liminf_{n \to \infty} J(u_n, v_n)
 \ge
 J(u,v)
 .
\end{equation*}
