3
$\begingroup$

Let $ 1 < p < 2 $ and $ u \in W_0^{1,p}( \Omega) $, where $ \Omega $ is some smooth open set of $ \mathbb{R}^N,\ N \geq 2. $ Suppose that $$ \int_{\Omega} \nabla u \nabla \phi dx = 0,\ \forall\ \phi \in C_0^{\infty}( \Omega). $$ Is it true that $ u = 0? $

The use of approximation of $ u $ by smooth functions is not useful. Any idea is welcome.

$\endgroup$
3
  • $\begingroup$ I believe it is. By "integral by parts" we have $\Delta u =0$ as distribution. Therefore, $u$ is harmonic and, as consequence, smooth. A smooth function in $W_0^{1,p}(\Omega)$ must be zero. $\endgroup$
    – Hugo
    Jun 29, 2021 at 11:42
  • $\begingroup$ I'm not totally sure about my last statement. If $u$ is smooth and belongs to $W_0^{1,p}(\Omega)$, must $u$ be zero? $\endgroup$
    – Hugo
    Jun 29, 2021 at 11:46
  • $\begingroup$ @HugoCBotós it is true that $H^1(\Omega) = H^1_0(\Omega) \oplus \{u\in H^1 : \Delta u = 0\}$, this might be what you meant, however I am not sure about p<2. $\endgroup$
    – Lassadar
    Jun 29, 2021 at 12:27

1 Answer 1

6
$\begingroup$

This is true for bounded domains $\Omega$ with smooth boundary ($C^1$ is sufficient also), as a consequence of the invertibility of the Laplacian as a map $$ \Delta : W^{1,p}_0(\Omega) \to W^{-1,p}(\Omega) $$ for all $1<p<\infty.$ These follow from the Calderón and Zygmund $L^p$ estimates, and I've written some further details and included relevant references in this answer.

The above does require some fairly heavy machinery however, and unfortunately I don't think this can be avoided. The necessity of requiring some regularity of the boundary is shown by Hajłasz in Theorem 1 of his paper A counterexample to the $L^p$ Hodge decomposition; there he constructs a bounded domain $\Omega \subset \Bbb R^2$ satisfying the cone condition, along with a non-trivial harmonic function $u$ which lies in $W^{1,p}_0(\Omega)$ for all $1 \leq p < \frac43.$ This suggests you do need to use the boundary regularity in a non-trivial way, which is why a direct approximation argument doesn't work.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .