Is $f(u) := \int_\Omega u^2(x)h(x)$ weakly lower semicontinuous in $L^2(\Omega)$?

Question

Define $f(u) := \int_\Omega u^2(x)h(x)$ weakly lower semicontinuous in $L^2(\Omega)$, where $h \in L^\infty(\Omega)$, but nothing is known about the sign of $h$?

I do not believe it is weakly lower semi-continuous. What do you think?

Answer 1

Without further assumptions on $h$ this is not correct:

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Consider for example $\Omega = \mathbb{R}$ endowed with the Lebesgue measure and

$$u_n(x) := 1_{[n,n+1)}(x).$$

Then $u_n$ converges weakly to $u:=0$ in $L^2(\Omega)$. If we choose $h:=-1$, then this means that

$$f(u) = 0 > -1 = \liminf_{n \to \infty} f(u_n).$$

This shows that $f$ is not weakly lower semicontinuous.