# Weakly harmonic function is identically zero

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.

• 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.
– Hugo
Jun 29, 2021 at 11:42
• 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?
– Hugo
Jun 29, 2021 at 11:46
• @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. Jun 29, 2021 at 12:27

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 $$1 These follow from the Calderón and Zygmund $$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.