Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and let $u \in L^2(0,T;H^1(\Omega))$ with $u_t \in L^2(0,T;H^{-1}(\Omega))$. Define the truncation function$$T_\epsilon(x) = \begin{cases} -\epsilon &: x \in (-\infty, -\epsilon]\\ x & x \in (-\epsilon, \epsilon)\\ \epsilon &: x \in [\epsilon, \infty) \end{cases}.$$ According to the second to last page in this paper,

clearly $$\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle_{H^{-1}(\Omega), H^1(\Omega)} = \int_\Omega |u(T)| - \int_\Omega |u(0)|$$

but I don't see this. We can write the LHS as $$\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle_{H^{-1}(\Omega), H^1(\Omega)} = \lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \left(\frac{d}{dt}\int uT_\epsilon(u) - \langle (T_\epsilon u)_t, u \rangle_{H^{-1}(\Omega), H^1(\Omega)}\right)$$ and the first term on the RHS is good (after using the FTOC) and we want to show the second term on the RHS vanishes. But I am not sure that the second term even makes sense since $u_t$ is a distribution not a function.

I tried to do this via a density argument (so let $u_n \to u$ where $u_n$ are smooth) but I was unable to pass to the limit on the left hand side of the desired equation.

Crossposted from MSE (https://math.stackexchange.com/questions/843760/want-to-show-lim-epsilon-to-0-frac1-epsilon-int-0t-langle-u-tt-t). There is an answer there but it is incorrect since the answerere assumes $u_t(t) \in L^2(\Omega)$ which is not true.