Is the functional $I(u) = \int_{\Bbb{R}^N}h(x) |u|^q \ dx $ weakly lower semicontinuous? Is the functional
$$
I(u) = \int_{\Bbb{R}^N}h(x) |u|^q \ dx 
$$
weakly lower semicontinuous for $N \geq 3$ and $1 < q < 2$?
This is part of an exercise that asks to solve the problem
$$
\begin{cases}
- \Delta u + u = h(x)|u|^{q - 2}u \quad \text{ in } \Bbb{R}^N, \\
u \in H^1 (\Bbb{R}^N)
\end{cases}
$$
where $1 < q < 2$ and $h \in L^{\frac{2^*}{2^* - q}} \cap L^\infty$ is nonnegative, using the method of minimization.
I showed that weak solutions are critical points of the functional 
$$
J(u) = ||u||^2 - \int_{\Bbb{R}^N}h(x)|u|^q \ dx.
$$
Now I am trying to show that $J$ is weakly lower semicontinuous and coercive. Coerciveness is easy, as is the weak lower semicontinuity of the norm. Hence, it remains to show that $I$ is weakly lower semicontinuous.
Thanks in advance and kind regards.
 A: I believe the following works. Because strongly lower semicontinuous convex functionals are weakly lower semicontinuous, assume that $u_n\rightarrow u\in H^1(\mathbb{R}^N)$. Using the fact that $|a|^q\leq |b|^{q}+q|a|^{q-2}a(a-b) $, which follows from the convexity of $x\mapsto |x|^q$ we see that
$$h(x)|u(x)|^q\leq h(x)|u_n(x)|^q+qh(x)|u(x)|^{q-1}|u_n(x)-u(x)|$$
which implies that
$$I(u)\leq I(u_n)+q\int_{\mathbb{R}^n}h(x)|u(x)|^{q-1}|u(x)-u_n(x)|\,dx$$
If we can show that $\lim_{n\rightarrow\infty}\int_{\mathbb{R}^n}h(x)|u|^{q-1}|u-u_n|\,dx=0$ then we are done by taking the liminf of both sides of the above. To see this, note that $|u-u_n|$ goes to $0$ in $L^{2^*}$ by Sobolev embedding, so we use Hölder's inequality
$$\int_{\mathbb{R}^n}h(x)|u(x)|^{q-1}|u(x)-u_n(x)|\,dx\leq ||h|u|^{q-1}||_{L^{\frac{2^*}{2^*-1}}(\mathbb{R}^N)}||u-u_n||_{L^{2^*}(\mathbb{R}^N)}$$
which goes to $0$ as long as $h|u|^{q-1}\in L^{\frac{2^*}{2^*-1}}(\mathbb{R}^N)$. But this is true because $|u|^{q-1}\in L^{\frac{2^*}{q-1}}(\mathbb{R}^N)$ and
$$\frac{2^* -q }{2^*}+\frac{q-1}{2^*}=\frac{2^*-1}{2^*}$$
