Annihilator of the image of $\Delta$ is a subset of the space of harmonic functions.

Question

Let $\Omega\subset\mathbb{R}^{2}$ be a polygon. Let $N_{q}$ be the subspace of all functions $v\in L^{q}(\Omega)$ such that $$ \int_{\Omega}fv\ dxdy=0 $$ for all $f\in L^{q}(\Omega)$ such that there exists $u\in W_{p}^{2}(\Omega)$ satisfying $$ \begin{cases} \Delta u=f & \text{in }\Omega\\ u=0 & \text{on }\partial\Omega \end{cases}. $$ Then $N_{q}$ is a subspace of harmonic functions.

This is from the book of Elliptic Problems in Nonsmooth Domains. The author said this is obvious. I think this is not obvious. How I can do this? If you need more information about this, let me know.

Answer 1

Boundary condition

The statement "$u=0$ on $\partial \Omega$" is ambiguous. There are two natural but inequivalent ways to interpret it:

• $u\in W^{2,p}_0(\Omega)$ (function is "clamped" on the boundary, with its gradient); or
• $u\in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)$ (function is $0$ on the boundary, but not its gradient).

I interpret it in the first sense, as $u\in W^{2,p}_0(\Omega)$.

Proof of the claim

To show that $v\in N_q$ is harmonic, it suffices to have $\int_\Omega (\Delta u)v = 0$ for all $u\in C_c^\infty(\Omega)$, thanks to Weyl's lemma. And we have that by the given assumption, so this part is indeed sort-of obvious.

For the converse, suppose $v\in L^q$ is harmonic. We know that $\int_\Omega (\Delta u)v = 0$ for all $u\in C_c^\infty(\Omega)$. Since the functional $u\mapsto \int_\Omega (\Delta u)v$ is continuous in the $W^{2,p}$-norm, the equality $\int_\Omega (\Delta u)v = 0$ extends to the closure of $C_c^\infty(\Omega)$ in $W^{2,p}(\Omega)$, which is precisely $W^{2,p}_0(\Omega)$.

Remarks

Intuitively, the claim is: since $\Delta$ is self-adjoint, its kernel is the orthogonal complement of its range. Or rather the annihilator of the range since we are not talking about an operator from a Hilbert space to itself.

The self-adjointness is supposed to come from integration by parts, $$\int_\Omega (\Delta u) v = - \int_\Omega \nabla u\nabla v = \int_\Omega u \Delta v\tag1$$ The property (1) holds for smooth compactly supported functions. The left hand side of (1) also makes sense for $u\in W^{2,p}(\Omega)$ and $v\in L^q(\Omega)$, but the other terms do not (or, they do in some distributional sense which means we declare (1) to be true by definition, at least when functions vanish on the boundary).