A couple of inequalities (explanation needed), how to show $\liminf_{k \to 0}k^{-1}\int(\nabla u - \nabla v)\nabla (T_k(b(u)-b(v))) \geq 0$ Let $T_k(x) = \max\{-k, \min(x,k)\}$, a truncation function at levels $k$ and $-k$. 
Let $b$ be a Lipschitz increasing function with $b(0)=0$ and $b'$ and $b^{-1}$ also Lipschitz.
I have seen it claimed that
$$\limsup_{k \to 0}\frac{1}{k}\int_0^t \int_\Omega (f_1-f_2)T_k(b(u)-b(v)) \leq \int_0^t \int_{\Omega}|f_1-f_2|$$
and
$$\liminf_{k \to 0}\frac{1}{k}\int_0^t\int_{\Omega}(\nabla u - \nabla v)\nabla (T_k(b(u)-b(v))) \geq 0.$$
Here $f_i$ are $L^\infty(0,T;L^\infty)$ and $u$ and $v$ are in $L^2(0,T;H^1)$.

My attempt:

For the first inequality: Isn't this trivial or am I missing something? We have $\frac{1}{k}T_k \leq 1$ so isn't it immediate? Why the $\limsup$???
For the second inequality:
I was thinking of using the dominated convergence theorem. It is true that for all points $x,t$ such that $b(u)-b(v) \neq 0$, we have
$$\lim_{k \to 0}\frac{1}{k}(\nabla u - \nabla v)\nabla (T_k(b(u)-b(v))) = \lim_{k \to 0}\frac{1}{k}(\nabla u - \nabla v)T_k'(b(u)-b(v))\nabla (b(u)-b(v)) = 0$$
because $\lim \frac{1}{k}T'_k(x) = 0$ when $x \neq 0$. 
So the limit of the integrand converges (but I don't know where the set $\{b(u)-b(v) \neq 0\}$ is null!). But I can't use DCT since I cannot bound the integrand uniformly in $k$.  Moreover the author uses a $\liminf$  so I guess this is not the right way.
 A: For your first question, we have (as you alread observed) that $$\frac{1}{k}\int_0^t\int_\Omega (f_1-f_2)T_k(b(u)-b(v))\le\int_0^t\int_\Omega |f_1-f_2|.$$
From here, we need to take $\limsup$, because we don't know if the left hand side converge.
With respect to the second question, I will give a partial answer. Assume that $b^{-1}$ and $b'$ is Lipschitz (note that in the paper you are reading, they assume that $b^{-1}$ is Lipschitz, however, there is nothing about $b'$). Let $A_k=\{x\in\Omega:\ |b(u)-b(v)|<k\}$. Note that for all $k$, $$\tag{1}\int_0^t\int_\Omega (\nabla u-\nabla v)\nabla (T_k(b(u)-b(v)))=\int_0^t \int_{A_k}(\nabla u-\nabla v)( b'(u)\nabla u- b'(v)\nabla v).$$
On the other hand, as $(\nabla u-\nabla v)(\nabla u-\nabla v)\ge 0$, we conclude that (remember that $b'\ge 0$) $b'(u)\nabla u(\nabla u-\nabla v)\ge b'(u)\nabla v(\nabla u-\nabla v)$. We combine this inequality with $(1)$ to obtain
$$\tag{2}
 \frac{1}{k}\int_0^t\int_\Omega (\nabla u-\nabla v)\nabla (T_k(b(u)-b(v))) \ge \frac{1}{k}\int_0^t\int_{A_k} (\nabla u-\nabla v)\nabla v(b'(u)-b'(v)).$$
If the right hand side of $(2)$ converge to zero, we are done. Note that 
$$\left|\frac{1}{k}\int_0^t\int_\Omega (\nabla u-\nabla v)\nabla v(b'(u)-b'(v))\right|\le \frac{1}{k}\int_0^t\int_{A_k} c|\nabla u-\nabla v||\nabla v||u-v|,\tag{3}$$
where $c$ is the Lipschitz constant of $b'$. As $b^{-1}$ is Lipschitz, there is a constant $C$ such that $C|u-v|\le |b(u)-b(v)|$, therefore, in the set $A_k$, we must have $|u-v|\le \frac{k}{C}$. Thus, from $(3)$
$$\left|\frac{1}{k}\int_0^t\int_\Omega (\nabla u-\nabla v)\nabla v(b'(u)-b'(v))\right|\le \int_0^t\int_{A_k} cC|\nabla u-\nabla v||\nabla v|.$$
To finish, we can just apply Lebesgue theorem.
I hope this calculation may be of some help.
