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?
On
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 $d>1$.
First note that $f(u)\|\nabla v\|^2 = \|\sqrt{f(u)}v\|^2$. Therefore your integral is $J(u,v)=\|\sqrt{f(u)}v\|_{L^2}^2$. Now it is known that $x_n \xrightarrow[weak]{X} x \implies \liminf \|x_n\|_X \geq \|x\|_X$ holds in every Hilbert space $X$.
Therefore we aim to prove $\Phi: H^1\times H^1 \to L^2, (u,v) \mapsto g(u)\nabla v$ is (sequentially) continuous from the weak to the weak topology whenever $g:\mathbb{R}\to\mathbb{R}$ is continuous and bounded. In fact we break it down into three steps:
$\Psi: H^1 \to Hom(L^2,L^2), \Psi_u(v):=g(u)v$ is continuous from the weak topology to the strong operator topology (aka topology of pointwise convergence)
$im(\Psi)$ is an equicontinuous sets of self-adjoint operators.
The evaluation map $Hom(L^2,L^2) \times L^2 \to L^2$ -- at least when restricted to subsets of the form $E\times L^2$ with $E$ equicontinuous and self-adjoint -- is continuous w.r.t. the strong operator topology on $Hom(L^2,L^2)$ and the weak topology on (both) $L^2$.
The map $\Phi$ is then the composition $$H^1\times H^1 \xrightarrow{\Psi\times id} Hom(L^2,L^2)\times H^1 \xrightarrow{id\times\nabla} Hom(L^2,L^2)\times L^2 \xrightarrow{eval} L^2$$ and therefore continuous in the way we want it to be.
Alright, let's get started.
Let $u_n \xrightarrow[weak]{H^1} u$ be arbitrary. We use that $H^1$ is compactly embedded in $L^2$ so that $u_n \xrightarrow{L^2} u$. Now we can extract an a.e. convergent subsequence $u_{n_k}$ of $u_n$. Because $g$ is continuous, $g(u_{n_k})\to g(u)$ a.e. as well. If $v\in L^2$ is fixed, then $\|\Psi_{u_{n_k}}(v)-\Psi_u(v)\|_{L^2}^2 = \int_\Omega (g(u_{n_k})-g(u))^2 v^2 \to 0$ by dominated convergence. We have in fact shown: Every subsequence of $\Psi_{u_n}$ contains a subsubsequence which converges to $\Psi_u$. Therefore $\Psi_{u_n}$ converges to $\Psi_u$.
Follows directly from $\|\Phi_u\|_{op} = \|g(u)\|_\infty \leq \|g\|_\infty$ and $im(g)\subseteq\mathbb{R}$.
Can be seen as follows: If $\Phi_n \xrightarrow{s.o.t} \Phi$ are self-adjoint, $v_n \xrightarrow{weak L^2} v$ and $w\in L^2$ are arbitrary, then $$\Phi_n(v_n)-\Phi(v) = (\Phi_n(v_n)-\Phi_n(v)) + (\Phi_n(v)-\Phi(v))$$ The second summand goes to zero because $\Phi_n\to\Phi$ in the strong operator topology. The first summand weakly goes to zero because $$\langle \Phi_n(v_n-v),w\rangle = \langle v_n-v,\Phi_n^\ast(w)\rangle = \langle v_n-v,\Phi^\ast(w)\rangle + \langle v_n-v,(\Phi_n-\Phi)^\ast(w)\rangle = \langle v_n-v,\Phi(w)\rangle + \langle v_n-v,(\Phi_n-\Phi)(w)\rangle$$ The first summand here goes to zero because $v_n\to v$ in the weak topology. The second summand goes to zero because $v_n-v$ is bounded and $\Phi_n(w) \to \Phi(w)$ in norm.