Better proof for: if $u_n \to u$ in $L^2$ then $F(u_n) \to F(u)$

Question

Let $F(v) = \int_{A} v^2(x)J(x)dx$ where $J$ is bounded.

If $u_n \to u$ in $L^2$ then I want to show that $F(u_n) \to F(u)$.

The proof is $$F(u_n) - F(u) = \int_A (u_n^2 - u^2)J \leq C\int_A (u_n^2 -u^2)$$ Now add and subtract $u_nu$: $$F(u_n) - F(u) \leq C\int_A (u_n^2 -u_nu + u_nu- u^2)=C\int_A (u_n(u_n -u) + (u_n- u)u)\\\leq C|u_n||u_n-u| + C|u||u_n-u|$$ with the norms in $L^2$.This tends to 0.

I wonder is there an easier proof of this statement??

Answer 1

The only thing which could simplify the proof probably is the following: if $u_n\to u$ in $L^2$ then $u_n^2 \to u^2$ in $L^1$ and then one can pass to the limit directly in $F$...

Answer 2

You should not forget to apply absolute values from the beginning, since you would only deduce that $\limsup_n F(u_n)-F(u) \leq 0$. However, as you essentially noticed, $$ \begin{split} |F(u_n)-F(u)| &\leq \|J\|_\infty \int_A |u_n^2-u^2| = \|J\|_\infty \int_A |u_n+u| |u_n-u| \\ &\leq \|J\|_\infty \left( \int_A |u_n| |u_n-u| + \int_A |u||u_n-u| \right) \\ &\leq \|J\|_\infty \left( \|u_n\|_2 \|u_n-u\|_2 + \|u\|_2 \|u_n-u\|_2 \right) \end{split} $$ and you conclude because any convergent sequence is bounded. I do not know if there is an easier proof: it seems to me that this proof is really elementary, and makes use of the minimal assumptions and tools for the conclusion to be correct.