2

Let $f$ be a locally integrable function on an open set $G$ of $\mathbb{R}^{n}$ and $ n\geq2 $. Suppose $\theta$ is in $ C^{\infty}(G) $ such that its laplacian $ \Delta \theta=1 $ everywhere in $\mathbb{R}^{n}$, and let $\phi\in C_{c}^{\infty}(G)$. Clearly $\phi \theta\in C_{c}^{\infty}(G).$ We know that the distributional laplacian of $f$ is defined and is given by $$ \langle \Delta f,(\phi \theta)\rangle=\int_{G}f(x)\Delta[\phi(x) \theta(x)]dx. $$

Now, if I write that this integral is equal to $$ \int_{G}f(x)\phi(x)dx+\int_{G}f(x)[\Delta\phi(x)] \theta(x)dx +\sum\limits^{n}_{j=1}\int_{G}f(x)\frac{\partial \phi(x)}{\partial x_{j}}\frac{\partial\theta(x) }{\partial x_{j}} dx,$$ I obtain absurd results by using some specific functions for $\phi$. There must be somthing wrong with the first integral (by the way I used here the fact that $\Delta\theta\equiv1$). What's wrong with that?

M. Rahmat
  • 1,369
  • Is it $\Delta f = 1$ or $\Delta f = \delta$ (Dirac distribution) ? – Jean Marie Dec 15 '18 at 04:15
  • $f$ is a locally integrable function, but not smooth. – M. Rahmat Dec 15 '18 at 04:35
  • Be kind enough to answer my question : is the RHS constant function $1$ or a Dirac peak $\delta$ ? This is not at all the same. – Jean Marie Dec 15 '18 at 05:16
  • Sorry! I don't understand your question! What does it mean that the integral on the right equals 1? Or equals the Dirac distribution? – M. Rahmat Dec 15 '18 at 06:51
  • I mean the RHS of your first equation $\Delta \theta = 1$. – Jean Marie Dec 15 '18 at 09:24
  • Je suis désolé: I steel don't get it. For example take $\theta (x)=\frac{1}{2}x^{2}{1}$, where $x=(x{1},...,x_{n})\in\mathbb{R}^{n}$. Can you figure out the answer to your question through this example? – M. Rahmat Dec 15 '18 at 17:58
  • I understand that you mean 1 and not $\delta$. Because rather often, one meets such a formulation, for example in https://math.stackexchange.com/q/368155 or https://arxiv.org/pdf/1303.2567.pdf – Jean Marie Dec 15 '18 at 18:17
  • Thanks for the references. – M. Rahmat Dec 15 '18 at 18:39

0 Answers0