# 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?

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 $$\begin{equation*} \liminf_{n \to \infty} J(u_n, v_n) \ge J(u,v) . \end{equation*}$$