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Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain. Assume that $u\in W_0^{1,2}(\Omega)$ is such that $\Delta u$ is a measure.

Let $\Phi$ be a smooth function in $\mathbb{R}$, such that $\Phi'(u)\in W_0^{1,2}(\Omega)$. Is it true that $\Delta (\Phi\circ u)$ is a measure and $$\Delta (\Phi\circ u)=(\Phi''\circ u)|\nabla u|^2+(\Phi'\circ u)\Delta u, \tag{1}$$

If the equality $$\int_\Omega \phi \Delta u=-\int_\Omega \nabla \phi \nabla u,\ \forall \phi\in W_0^{1,2}(\Omega),\tag{2}$$

were true, then I can show that $(1)$ is true, however, I fail to see if $(2)$ is true. Any idea is appreciated.

Remark: I don't know if it matter, but in my particular case, I have that $u$ is superharmonic also.

Remark 2: Equality $(2)$ is true for all $\phi\in C_0^\infty(\Omega)$.

Remark 3: If $(2)$ is true, I think the proof goes as follows: if $\varphi\in C_0^\infty(\Omega)$ then (remember that $\Phi'\circ u\in W_0^{1,2}(\Omega)$, so that $\varphi(\Phi'\circ u)\in W_0^{1,2}(\Omega)$)

\begin{eqnarray} \int_\Omega (\Phi''\circ u)|\nabla u|^2\varphi+\int_\Omega \varphi(\Phi'\circ u)\Delta u &=& \int_\Omega (\Phi''\circ u)|\nabla u|^2\varphi-\int_\Omega \nabla (\varphi(\Phi'\circ u))\nabla u \nonumber \\ &=& \int_\Omega \varphi\nabla (\Phi'\circ u)\nabla u-\int_\Omega\varphi\nabla (\Phi'\circ u)\nabla u-\\ && \int _\Omega (\Phi'\circ u)\nabla u\nabla\varphi \nonumber \\ &=& \int_\Omega \nabla (\Phi\circ u)\nabla \varphi = \int_\Omega (\Phi\circ u)\Delta \varphi \end{eqnarray}

therefore $(1)$ is true.

Update: Motivated by @thisismuch answer, we will include the additional hypothesis: $u\ge 0$, $u$ is superharmonic and there are $0<k<t<1$ such that $\Phi(x)=0$ for $x\le k$ and $\Phi(x)=1$ for $x\ge t$.

Note that $|\Delta u|=-\Delta u$, therefore

\begin{eqnarray} \left|\int \nabla u\nabla ((\Phi'\circ u)\phi)\right| &=& \left|\int_{spt(\phi)\cap \{u>k\}}\nabla u\nabla ((\Phi'\circ u)\phi) \right| \nonumber \\ &=& \left|\int_{spt(\phi)\cap \{u>k\}}\phi (\Phi'\circ u)\Delta u\right| \nonumber \\ &\le& c\|\phi\|_\infty \int_{spt(\phi)\cap \{u>k\}}(-\Delta u) \\ &\le & c\|\phi\|_\infty\int_{spt(\phi)} \frac{u(-\Delta u)}{k} \end{eqnarray}

I would like to apply Green identity in the last equality. Is the following equality true? $$\int_\Omega u\Delta u=-\int_\Omega |\nabla u|^2$$

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  • $\begingroup$ Ops @Thisismuchhealthier. I was doing some wrong calculations. Well, $\Delta u\in W^{-1,2}(\Omega)$ so there is $F\in L^2(\Omega)^N$ such that $$\langle \Delta u, \phi\rangle =\int_\Omega F\cdot \nabla \phi,\ \forall \phi\in W_0^{1,2}$$ Does is help for anything? $\endgroup$ – Tomás Jul 17 '14 at 16:59
  • $\begingroup$ I can't formalize it @Thisismuchhealthier. How can I prove that $\langle \Delta u,\phi\rangle =\int_\Omega \phi\Delta u$? $\endgroup$ – Tomás Jul 17 '14 at 17:21
  • $\begingroup$ @Thisismuchhealthier. Why $\int_\Omega \phi\Delta u$ is continuous in $W^{1,2}_0$? $\endgroup$ – Tomás Jul 17 '14 at 19:04
  • $\begingroup$ I think I am confused @Thisismuchhealthier. We can see $\Delta u$ as a measure, or we can see it, as a bounded linear functinal in $W_0^{1,2}$. My question is: if we see it as a measure and if we integrate $\phi\in W_0^{1,2}$ agaisnt this measure, why is this operation continuous in $W_0^{1,2}$. Sorry if I am writing nonsense here. $\endgroup$ – Tomás Jul 17 '14 at 19:11
  • $\begingroup$ Underneath it all, $\Delta u$ is always the same thing: a distribution $T$ defined by its action on test functions $T(\phi)=\int u\Delta \phi$. This is true for every distribution $u$. But since $u$ is a $W^{1,2}$ function, $\Delta u$ is a particularly nice distribution: its values on test functions are controlled by $W^{1,2}$ of test functions, because $$|T(\phi)| \le \int |\nabla u \nabla \phi|\le \|\nabla u\|_{L^2} \|\nabla \phi\|_{L^2}$$ (for smooth $\phi$). This allows us to extend $T$ to a continuous linear functional on $W^{1,2}_0$. $\endgroup$ – user147263 Jul 17 '14 at 19:22
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I can show (1), but I don't know if $\Delta(\Phi\circ u)$ is a measure.

The equality (1) has to be understood in the sense of distributions, so we must figure out how $(\Phi'\circ u)\Delta u$ acts on test functions. We know how $\Delta u$ acts: $T(\phi)=\int u\Delta \phi$. Since $u$ is a $W^{1,2}$ function, $T$ is a particularly nice distribution: its value on $\phi$ is controlled by the $W^{1,2}$ norm of $\phi$: $$\left|T(\phi)\right| \le \int |\nabla u \nabla \phi|\le \|\nabla u\|_{L^2} \|\nabla \phi\|_{L^2}\tag{A}$$ This allows us to extend $T$ to a continuous linear functional on $W^{1,2}_0$. In fact, this extension is simply $$T(\phi) = -\int \nabla u \nabla \phi,\quad \phi\in W^{1,2}_0 \tag{B}$$ because both sides are continuous on $W^{1,2}_0$ and agree on a dense subset.

Now we can make sense of $(\Phi'\circ u)\Delta u$: it is the distribution $$\phi \mapsto T((\Phi'\circ u) \phi)\tag{C} $$ which is defined because $(\Phi'\circ u) \phi\in W_0^{1,2}$.

Assuming $\Phi''$ is bounded, the term $(\Phi''\circ u)|\nabla u|^2$ in (A) is an $L^1$ function. Now that the right hand side of (1) makes sense, we can apply it to test function $\phi$ and find, using (B)-(C) (as in your Remark 3) that it results in $\int (\Phi\circ u)\Delta \phi$. This qualifies the right hand side of (1) as the distributional Laplacian of $\Phi\circ u$.

But is it a measure? For a distribution to be a measure, it must be controlled by the supremum norm of test functions. The first term on the right in (1) is an $L^1$ function, no problem there. The problem is the second term. We need $$ \left|\int \nabla u\nabla \left((\Phi'\circ u) \phi\right)\right\| \lesssim \sup |\phi| \tag{D}$$ and I don't see why this would be true.

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  • $\begingroup$ Ok, now things are getting clear to me. Suppose that $0<k<t<1$ and let $\Phi$ be a smooth function with $\Phi(x)=0$ for $x\le k$ and $\Phi(x)=1$ for $x\ge t$. Assume also that $u$ is superharmonic. Now, with these conditions, we can prove your estimate $(D)$, therefore, it is a measure. Am I right? $\endgroup$ – Tomás Jul 17 '14 at 22:56
  • $\begingroup$ @Tomás I don't see how to prove (D) even with these conditions. $\endgroup$ – user147263 Jul 17 '14 at 23:03
  • $\begingroup$ Please, take a look in my update. $\endgroup$ – Tomás Jul 17 '14 at 23:54
  • $\begingroup$ @Tomás I have an issue with the transition to $\int \phi (\Phi'\circ u)\Delta u$. What does this integral mean? The Laplacian $\Delta u$ is a Radon measure, something we can integrate against continuous functions. But $u$ is not necessarily continuous. We can't integrate, say, an arbitrary element of $L^\infty$ against a singular measure. I think the best way forward is to mollify $u$ and try to pass to the limit. That ought to work... $\endgroup$ – user147263 Jul 18 '14 at 0:07
  • $\begingroup$ With these new conditions, I can prove that for all $u\in C_0^\infty(\overline{\Omega})$ with $u\geq 0$ and $u$ superharmonic, $$\|\Phi\circ u\|_{\mathcal{M}(\Omega)}\le C\|\nabla u\|_2,$$ From here I can't find a good argument of density to conclude, for the case where $u\in W_0^{1,2}(\Omega)$. Do you have any idea? $\endgroup$ – Tomás Jul 18 '14 at 0:46

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