If $u_n \rightharpoonup u\,$ in $\,L^2(\Omega)\,$ and $\,u_n^2 \rightharpoonup \upsilon$ in $L^1(\Omega),\,$ then is $\upsilon=u^2$?

Question

If $u_n \rightharpoonup u$ in $L^2(\Omega)$ and $u_n^2 \rightharpoonup \upsilon$ in $L^1(\Omega)$, then is $\upsilon=u^2$?

We assume that the domain $\Omega$ is bounded.

If not is there any way to ensure this?

Answer 1

The answer is: Not in general.

For example, $\Omega=(0,2\pi)$, $u_n(x)=\sin nx$.

Then $u_n\rightharpoonup 0$, since $$ \int_0^{2\pi} f(x)\,\sin nx\,dx\to 0=u, $$ for all $f\in L^2[0,2\pi]$.

Meanwhile $$ v_n(x)=u_n^2(x)=\sin^2 nx=\frac{1}{2}-\frac{\cos (2nx)}{2}\rightharpoonup \frac{1}{2}=v, $$ and $u^2\ne v$.

Note. If $\{u_n\}$ is bounded and converges against $L^1$ functions (i.e. weak$^*$ $L^\infty$) then $u_n^2$ also converges in the same fashion and the limit is provided by the Young measure of this convergence. Similarly, is $f$ is bounded, then $f(u_n)$ converges weak$^*$ to a bounded function $\bar f$ which can also be described by Young's measure.