Composition of a weakly convergent sequence with a nonlinear function Let $\Omega\subset\mathbb{R}^n $ be bounded smooth domain.
Given a sequence $u_m$ in  Sobolev space $H=\left \{v\in H^2(\Omega ):\frac{\partial v}{\partial n}=0 \text{ on } \partial \Omega  \right \}$ such that $u_m$ is uniformly bounded i.e. $\|u_m\|_{H^2}\leq M$ and given the function $f(u)=u^3-u$.
If I know that $u_m\rightharpoonup u(u\in H)$ in $L^2$ sense  i.e. $\int_{\Omega}u_m v\to \int_{\Omega}u v$ for every $v\in H$. Is it true that $f(u_m)\rightharpoonup f(u)$ (in $L^2$ sense)?
 A: I post this as answer,but I am studying myself Sobolev spaces right now, so I am not sure whether it is correct (in fact it may well be completely wrong). The argument will use embedding theorems for Sobolev spaces.


*

*First we have a sequence that converges weakly in $L^2,$ but is bounded in $H^2 = W^{2,2}.$ Since $W^{2,2}$ is reflexive, the unit ball is weakly compact. So you can extract converging subsequences. But every converging subsequence converges tu $u,$ since we know that $v_m\rightharpoonup v$ in $W^{k,p}$ implies $v_m\rightharpoonup v$ in $L^p.$ A sequence in a compact (if you want also metric) space in which every converging subsequence converges to the same limit point, is itself convergent.

*Now we know that $W^{2,2}$ embeds compactly in $L^{q},$ for $q < \frac{2n}{n - 4},$ and $n \ge 4$. For smaller $n$ we have compact embedding in $C^{k, \gamma},$ for $\gamma \in (0,1)$ and $k< 2 - n/2 - \gamma.$ Compact embeddings tell us that a weakly converging sequence in the departure space is strongly converging in the arrival space. So all in all, for $n <6,$ we know that the sequence is actually strongly converging in $L^6$.

*Part 2 thus tells us that $u_n^3$ is in $L^2$ and this answers some of the comments. Furthermore it is strongy convergent in $L^2.$ So for $n < 6$ we have that actually $f(u_n) \rightarrow f(u)$ strongly in $L^2.$

*Last but not least we consider the border case $n = 6.$ Here we have just continuous embedding of $W^{2,2}$ in $L^6$. So we know $u_n^3$ is in $L^2$,and we know that it has some weakly converging subsequence, since it`s norm is bounded. But is  $u_n^3\rightharpoonup u^3$? I suppose it is,but I am actually having some trouble proving it.
A: For simplicity, I consider only the case $n=3$.
You have $u_m\rightharpoonup u$ in $L^2$, so it is enough to show that $u_m^3\rightharpoonup u^3$ in $L^2$. To show this we will use Sobolev compact embedding theorems, the Bounded Convergence Theorem, and the fact that if each subsequence of $y_m$ contains a strongly/weakly convergent subsequence to some $y$ then the whole sequence converges to $y$.
So, you have that $u_m\rightharpoonup u$ in $L^2$ and $\|u_m\|_{H^2}\leq M$. Take an arbitrary subsequence $u_{m_k}$. Because $H^2\hookrightarrow L^\infty\Rightarrow \|u_{m_k}\|_{L^\infty}\leq cM$, where $c>0$ is the embedding constant. Further, because $u_{m_k}$ is also bounded in $H^1$, which is reflexive, it follows that there exists a further weakly convergent subsequence $u_{m_{k_l}}\rightharpoonup g\in H^1$ in $H^1\hookrightarrow L^2$ (compactly). Because the embedding is compact $u_{m_{k_l}}\to g$ in $L^2$. Now, note that $g\equiv u$, because weak limits are unique, and therefore you conclude from here that actually $u\in L^6$ since $H^1\hookrightarrow L^6$ (continuously). From the $L^2$ convergence you find a further pointwise a.e convergent subsequence, this time denote it $u_s(x)\to u(x)$ for which you still have the $L^\infty$ bound $\|u_s\|\leq cM$. Finally, you note that $u_s^3(x)\to u^3(x)$ a.e and that $\|u_s^3\|_{L^\infty}\leq c^3M^3<\infty$, and apply the Bounded Convergence Theorem to $\{u_s^3\}$ to conclude that $u_s^3\to u^3$ in $L^p,\,\forall 1\leq p<\infty$. Obviously $u_s^3\rightharpoonup u^3$ in $L^2$ also.
As a conclusion, from the very strong assumption, that $\|u_m\|_{H^2}\leq M$ and that $u_m\rightharpoonup u$ in $L^2$ it follows that $u_m\to u$ in $L^p,\,\forall 1\leq p<\infty$ and also $u_m^3\to u^3$ in $L^p,\,\forall 1\leq p<\infty$.

Bounded Convergence Theorem: Suppose $\Omega\subset \mathbb R^n$ is bounded,  $f,f_1,f_2,...,f_n$ is a sequence of measurable functions over $\Omega\subset \mathbb R^n$ and $M>0$ is such that
$$\|f_n\|_{L^\infty(\Omega)}\leq M,\,\forall n\in\mathbb N$$
and $\lim\limits_{n\to\infty}{f_n(x)}=f(x)$ a.e $x\in\Omega$. Then
$$\lim\limits_{n\to\infty}{\|f_n-f\|_{L^p(\Omega)}}=0,\,\forall 1\leq p<\infty$$

