If $u_m \rightharpoonup u$, how to show using monotonicity that $f(u_m) \rightharpoonup f(u)$? Let 
$$u_m \rightharpoonup u \quad \text{(weakly) in $L^\infty(0,T;L^2(\Omega)) \cap L^2(0,T;H^1(\Omega))$}.$$ We are given $f:\mathbb R \to \mathbb R$, a Lipschitz continuous invertible map which is monotone:
$$(f(x)-f(y))(x-y) \geq 0\quad\text{for all $x, y$}$$
and satisfies $f'> 0$. Suppose we have
$$f(u_m) \rightharpoonup f(v) \quad \text{in $L^2(0,T;H^1(\Omega))$}$$
for some $v \in L^2(0,T;H^1(\Omega))$ (assume that for $u \in L^2(0,T;H^1)$, $f(u)$ and $f^{-1}(u)$ are in $L^2(0,T;H^{1}(\Omega))$.)
Is it possible to show that indeed $v=u$, i.e. $f(u_m) \rightharpoonup f(u)$? I can't seem to do it by using the monotonicity method.
(Despite posting this on MO I still did not solve this problem) 

In case the convergence $u_m\rightharpoonup\,u$ is understood as a weak convergence in $L^2\bigl(0,T;H^1(\Omega)\bigr)$ being as well weak-star in $L^{\infty}\bigl(0,T;L^2(\Omega)\bigr)$, the answer is definitely No. It can't be done, because generally a nonlinear superposition operator cannot be weakly continuous. Consider the following counterexample. Take any function $f\in C^1(\mathbb{R})$ with uniformly bounded strictly positive derivative such that $f(\xi)=2\xi^2$ on $[-1,1]$, and let $u_m(t,x)=\sin{(mt)}\varphi(x)$ with some fixed nonzero $\varphi\in W^{1,4}(\Omega)$ when $\Omega$ is bounded, or $\varphi\in W^{1,4}(\Omega)\cap H^1(\Omega)$ otherwise. It is clear that $u_m\rightharpoonup\,0=u$ where convergence is understood as weak in $L^2\bigl(0,T;H^1(\Omega)\bigr)$ and weak-star in $L^{\infty}\bigl(0,T;L^2(\Omega)\bigr)$, while 
$$
f(u_m)=\bigl(1-\cos{(2mt)}\bigr)\varphi^2\rightharpoonup\,\varphi^2\not\equiv\,0=f(u)
$$
with convergence understood as weak in $L^2\bigl(0,T;H^1(\Omega)\bigr)$.
