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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?

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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*}

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