If $u,v \in H^1_0(\Omega)$ and $\Delta u \in L^2(\Omega)$, can we write $\int_\Omega \nabla u \cdot \nabla v = \int_\Omega (-\Delta u)v $?

Question

Let $u, v \in H^1_0(\Omega)$ with $\Delta u \in L^2(\Omega)$ on a bounded Lipschitz domain $\Omega$ (it is only Lipschitz!)

Does this hold:

$$\langle -\Delta u, v \rangle = \int_\Omega \nabla u \cdot \nabla v = \int_\Omega (-\Delta u)v $$ where we have an actual integral on the right with the Laplacian.

The first equality is the definition of the weak Laplacian. Since $\Delta u \in L^2$ and $v \in H^1_0$, I thought we can just write it as an integral using Gelfand triple. However, this is Green's identity, for which we typically need $H^2$ for $u$.

What do you think?

Answer 1

I think your statement is true for any bounded open domain. A detail proof can be found in the Lemma 4.4 of the book [1] by Alexander Grigor’yan.

[1] Grigor’yan, Alexander, Heat kernel and analysis on manifolds, AMS/IP Studies in Advanced Mathematics 47. Providence, RI: American Mathematical Society (AMS); Somerville, MA: International Press (ISBN 978-0-8218-4935-4/hbk). xvii, 482 p. (2009). ZBL1206.58008.