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](http://www.mat.uniroma2.it/~porretta/papers/Blanchard-Porretta-JDE.pdf), 



> 
> 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.




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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.



Want to show $\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle = \int_\Omega |u(T)| - \int_\Omega |u(0)|$
This is a classical trick: the function $\frac{T\_{\epsilon}(.)}{\epsilon}$ is an approximation to the sign function so at least formally the result is clear.


In order to work ou the details, define
$$
S\_{\epsilon}(z)=\int\_0^z\frac{T\_{\epsilon}(s)}{\epsilon}ds.
$$
You can compute explicitly $S\_{\epsilon}(z)$ so I won't do it here, but once you do that you realize $S\_{\epsilon}(.)$ is an approximation to $|.|$ (with uniform convergence when $\epsilon\to 0$). Now for fixed $\epsilon>0$ you have that $T\_{\epsilon}$ is bounded and Lipschitz, and $S\_{\epsilon}$ is $C^1$ and uniformly Lipschitz (actually linear with slope $\pm 1$ outside $(-\epsilon,\epsilon)$). Since you're assuming that $u \in L^2(0,T;H^1(\Omega))$ and $u\_t \in L^2(0,T;H^{-1}(\Omega))$ you can legitimately write for fixed $\epsilon >0$
\begin{align\*}
\frac{1}{\epsilon} \int\_0^T \langle u\_t(t), T\_\epsilon(u(t)) \rangle\_{H^{-1}(\Omega), H^1(\Omega)} & = \int\_0^T\frac{d}{dt}\left(\int\_{\Omega}S\_{\epsilon}\circ u\,dx\right)\,dt\\
& =\int\_\Omega S\_{\epsilon}(u(T))\,dx - \int\_\Omega S\_{\epsilon}(u(0))\,dx
\end{align\*}
(you have all you need to justify the chain rule $\partial\_t \left(S\_{\epsilon}(u)\right)=S\_{\epsilon}'(u)\,\partial\_t u=\frac{T\_{\epsilon}(u)}{\epsilon}\partial\_t u$ with the usual duality pairings etc...).
Now since $S\_{\epsilon}(.)\to |.|$ uniformly when $\epsilon\to 0$ the convergence of the r.h.s. should be a simple exercise (you may want to use here $L^2(\Omega)\subset L^1(\Omega)$ for bounded $\Omega$).


