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Consider the Heaviside function that is undefined in zero, i.e. $$H(t)=\begin{cases} 1&t>0 \\ 0&t< 0\end{cases}$$ Now consider a sequence of $H^1(\Omega)$-functions $u_n\to u$ in the norm of the sobolev space and $u_n(x) \in [-1,1]$ for a.e. $x\in \Omega$ (bounded domain). Let $f$ be a smooth testfunction. Can we say that $$\int_\Omega f\cdot H(u_n) \nabla u_n \to \int_\Omega f \cdot H(u) \nabla u$$

Of course one would try to apply Lebesgue's dominated convergence theorem (for a subsequence and a subsequence of that). The issue is that I'm not sure to find an integrable bound. The set were $u$ is zero and consequently $H(u)$ is undefined could have positive measure. However on that set I'm not sure how to find a good bound. On the other hand, if the set where $u$ is zero we find that the gradient is $0$ too and we can bound it on that set by $1$.

Having that said, I would think of estimating $|H(u_n)\nabla u_n|^2\leq |\nabla u_n|^2$. The latter sequence converges in $L^1(\Omega)$ (by assumption). Hence a generalized version of Lebesgues theorem yields that $H(u_n)\nabla u_n \to H(u)\nabla u$ in $L^2$ and consequently we can pass to the limit in the integral.

Is my reasoning correct? Does anyone have an idea on how to attack the problem without the gradient?

Quickbeam2k1
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Your reasoning is correct. But the problem simplifies if $H(u)\nabla u$ is written as $\nabla u^+$, where $u^+=\max(u,0)$. (The equality of these two things is standard: e.g., Proposition 3.22 of Chapter 3 of Lecture Notes on PDEs by Hunter.) The sequence $(u_n^+)$ is bounded in $H^1$, hence has a weakly convergent subsequence $u_{n_k}^+\to v$. On the other hand, for its subsequence we have convergence a.e. to $u^+$. Hence $v=u^+$.

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  • The Lemma you need is also known as Stampacchia's Lemma and in the mean time I'm ashamed of having asked the question :) – Quickbeam2k1 Jun 15 '14 at 19:30