# Is this distributional laplacean a measure?

Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain. Suppose that $0<k<t<1$ and $\Phi$ is a mooth function satisfying: $\Phi(x)=0$ for $x\le k$, $\Phi(x)=1$ for $x\ge t$.

Take $u\in C_0^\infty(\overline{\Omega})$ with $u\ge 0$ and $u$ superharmonic ($-\Delta u\ge 0$). Note that $$\Delta (\Phi\circ u)=(\Phi''\circ u)|\nabla u|^2+(\Phi'\circ u)\Delta u,\tag{1}$$

so $$|\Delta (\Phi\circ u)|\le C(|\nabla u|^2+\chi_{u>k}|\Delta u|).\tag{2}$$

Once $u\ge 0$ and $-\Delta u\ge 0$, we have that from $(2)$ that $$\chi_{u>k}|\Delta u|\le \frac{1}{k}u|\Delta u|=-\frac{1}{k}u\Delta u\tag{3}.$$

We combine Green's indetity with $(3)$ to conclude that $$\int_{u>k}|\Delta u|\le \frac{-1}{k}\int _\Omega u\Delta u=\frac{1}{k}\int_\Omega |\nabla u|^2.\tag{4}$$

Therefore, $(1)$ and $(4)$ gives $$\int_\Omega |\Delta (\Phi\circ u)|\le C\left(1+\frac{1}{k}\right)\int_\Omega |\nabla u|^2,$$

or equivalently $$\|\Delta(\Phi\circ u)\|_{\mathcal{M}(\Omega)}\le C\|\nabla u\|_2^2,\ \forall \ u\in C_0^\infty (\overline{\Omega}),\ u\ge 0,\ -\Delta u\ge 0.\tag{5}$$

So my question is: Can we extend $(5)$ for all functions $u\in W_0^{1,2}(\Omega)$ such that $u\ge 0$ and $u$ is superharmonic (the distributional laplacean of $u$ is nonegative)?

Remark 1: $C_0^\infty(\overline{\Omega})$ is the space of all $C^\infty(\overline{\Omega})$ functions, which vanishes on the boundary.

Remark 2: $C>0$ is a constant, which can change in every line and it depends only on $k$, $\|\Phi'\|_\infty$ and $\|\Phi''\|_\infty$.

Remark 3: In this question, I tried to solve this problem, by showing that $(1)$ were true in the sense of measure, also for $u\in W_0^{1,2}(\Omega)$, however, it does not seems to be true.

$\textbf{Update (A Supposed Proof)}$: Assume that $u\in W_0^{1,2}(\Omega)$ satisfies $u\ge 0$, $-\Delta u\ge 0$. Let $\varphi_n$ be the standard mollifier sequence. Extend $u$ by zero outside $\Omega$ (we still using the same notation $u$) and consider the sequence $$u_n(x)=\int_{\mathbb{R}^N} \varphi_n(x-y)u(y)dy,\ x\in \mathbb{R}^N$$

Once $u\in W_0^{1,2}(\mathbb{R}^N)$, we have that $u_n\in C_0^\infty(\overline{\Omega})$, $u_n\ge 0$ and $-\Delta u_n\ge 0$. Moreover, $u_n\to u$ in $W^{1,2}(\Omega)$. From $(5)$, we have that $$\|\Delta (\Phi\circ u_n)\|_{\mathcal{M}(\Omega)}\le C\|\nabla u_n\|_2^2,$$

therefore, we can assume without loss of generality that $\Phi\circ u_n$ converges in the weak star topology, to some measure $\mu\in \mathcal{M}(\Omega)$. Let $v\in W_0^{1,1}(\Omega)$ be the solution of the problem

$$\left\{ \begin{array}{ccc} \Delta v =\mu&\mbox{ in \Omega} \\ v=0 &\mbox{ on \partial \Omega} \end{array} \right.$$

By the weak convergence, we have that $$\int_\Omega \phi(x)\Delta (\Phi\circ u_n)\to \int_\Omega \phi(x) d\mu=\int_\Omega \phi(x)\Delta v, \forall \phi\in C_0(\overline{\Omega}),\tag{6}$$

hence, from $(6)$ and the definition of $v$, we conclude that $$\int_\Omega \Delta \phi(x) (\Phi\circ u_n)\to\int_\Omega \Delta \phi(x) v,\ \forall \phi\in C_0^\infty(\overline{\Omega}) .$$

As $\Phi\circ u_n \to \Phi\circ u$ in $W^{1,2}(\Omega)$, we must concude that $v=\Phi\circ u$ and thus, $\Delta(\Phi\circ u)\in \mathcal{M}(\Omega)$ and $$\|\Delta (\Phi\circ u)\|_{\mathcal{M}(\Omega)}\le C\|\nabla u\|_2^2$$

Is this proof right, can anyone check it for me please?

• Summary: you know that for smooth functions, the mass of $\Delta(\Phi\circ u)$ is controlled by $\|\nabla u\|_{L^2}$ and the mass of $\Delta u$. You want to use a sequence of smooth functions $u_n$ converging to $u$, so that the mass of $\Delta(\Phi\circ u_n)$ remains under control. Indeed, having this would be enough. Derivatives commute with distributional limits, and the distributional limit of measures with uniformly bounded TV is a measure. I think the only gap in your argument is: the extended $u$ picks up some generalized Laplacian along $\partial \Omega$. What to do with it? – user147263 Jul 21 '14 at 0:11
• @Thisismuchhealthier. Thank you for your reply. All calculations are local, i.e., in $\Omega$, so I can't see where the generalized Laplacian along the boundary, would give some trouble. Could you point me out, where is the possible flaw? – Tomás Jul 21 '14 at 1:52
• When you mollify extended $u$, its behavior on the boundary influences what happens inside of $\Omega$. – user147263 Jul 21 '14 at 1:53
• @Thisismuchhealthier. I see, and now I have the problem that $u_n$ is not superharmonic. So, in some way, we will have to consider the normal derivative of $u$ along the boundary. – Tomás Jul 21 '14 at 2:23
• @Thisismuchhealthier. Let $\delta>0$ be small, $\Omega_\delta=\{x\in \Omega: d(x,\partial\Omega)<\delta\}$. Let $u_\delta(x)=u(x)$ for $x\in \Omega\setminus\Omega_\delta$ and $u_\delta(x)=u(\tau_\delta(x))d(x,\partial\Omega)/\delta$ for $x\in \Omega_\delta$, where $\tau_\delta(x)$ satisfies $d(x,\partial(\Omega\setminus\Omega_\delta)=d(x,\tau_\delta(x))$. Is true that $u_\delta \to u$ in $W^{1,2}(\Omega)$? If this is true, I think I can use $u_\delta$ to approximate $u$ by smooth, superharmonic functions. – Tomás Jul 21 '14 at 12:58