-1
$\begingroup$

Let

  • $d\in\mathbb N$
  • $\lambda$ be the Lebesgue measure on $\mathbb R^d$
  • $\Omega\subseteq\mathbb R^d$ be open

Why can we use partial integration to obtain $$\int_\Omega\phi\Delta\psi\;{\rm d}\lambda=-\int_\Omega\nabla\phi\cdot\nabla\psi\;{\rm d}\lambda\tag 1$$ for all $\phi,\psi\in C_c^\infty(\Omega)$?


The multi-dimensional partial integration is a special case of the divergence theorem. But the divergence theorem in the version that I know only can be applied on compact subsets $K$ of $\mathbb R^d$ with smooth boundary $\partial K$. Smooth boundary means that for all $p\in\partial K$, there is an open set $U\subseteq\mathbb R^d$ with $p\in U$ such that there is a continuously differentiable $F:U\to\mathbb R$ with

  1. $K\cap U=\left\{x\in U:F(x)\le 0\right\}$ and
  2. $F'(x)\ne 0$ for all $x\in U$

Let's so what we've got in the situation of $\phi,\psi\in C_c^\infty(\Omega)$. Suppose we can find a compact $K$ with smooth boundary such that $K\subseteq\Omega$ and $\operatorname{supp}\Delta\psi\subset K$. Then it's easy to conclude that $$\int_\Omega\nabla\phi\cdot\nabla\psi+\phi\Delta\psi\;{\rm d}\lambda=\int_K\nabla\phi\cdot\nabla\psi+\phi\Delta\psi\;{\rm d}\lambda=\int_{\partial K}\phi\frac{\partial\psi}{\partial\nu}\;{\rm d}o=0$$ by the divergence theorem, since $\phi$ vanishes at $\partial K$.

So, can we always find such a $K$? If not, how can we prove $(1)$ instead?

$\endgroup$
1
$\begingroup$

You do not need to find such a set $K \subset \Omega$!

At first, let us assume that $\Omega \subset \mathbb{R}^d$ is bounded. Then there is an open ball $B_R(0) \subset \mathbb{R}^d$ such that $\Omega \subset B_{R/2}(0)$. For every function $\varphi \in C_c^{\infty}(\Omega)$, we define an extension $\tilde{\varphi} \colon B_R(0) \to \mathbb{R}$ by \begin{align} \tilde{\varphi}(x) &:= \varphi(x) &\text{ for } x \in \Omega, \\ \tilde{\varphi}(x) &:= 0 &\text{ otherwise.} \end{align} Then $\tilde{\varphi} \in C_{c}^{\infty}(B_R(0))$.

Let $\varphi, \psi \in C_c^{\infty}(\Omega)$. As above, we construct extensions $\tilde{\varphi}, \tilde{\psi} \in C_c^{\infty}(B_R(0))$. As the ball $B_R(0)$ has a smooth boundary, we are allowed to integrate by parts. So we obtain that \begin{equation} \int_{\Omega} \varphi \Delta \psi = \int_{B_R(0)} \tilde{\varphi} \Delta \tilde{\psi} = - \int_{B_R(0)} \nabla \tilde{\varphi} \cdot \nabla \tilde{\psi} = - \int_{\Omega} \nabla \varphi \cdot \nabla \psi. \end{equation} This proves the claim for bounded domains $\Omega$.

If $\Omega \subset \mathbb{R}^d$ is not bounded, then there is no ball $B_R(0) \subset \mathbb{R}^d$ such that $\Omega \subset B_{R/2}(0)$. However, our construction from above works! This is because for $\varphi \in C_c^{\infty}(\Omega)$ the support of $\varphi$ is a compact set; in particular, $\text{supp} \, \varphi \subset \mathbb{R}^d$ is bounded. So we find a ball $B_R(0) \subset \mathbb{R}^d$ such that $\text{supp} \, \varphi \subset B_{R/2}(0)$. Now, you can define the same extension $\tilde{\varphi} \colon B_R(0) \to \mathbb{R}$ as above. This proves the claim for unbounded domains $\Omega$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.