Is this element of $H^1(\Omega)^*$ actually in $L^2(\Omega)$?

Question

Let $\Omega$ be a smooth bounded domain.

Let $v \in H^2(\Omega)$ satisfy $-\Delta v = 0$ on $\Omega$ with $\partial_\nu v = g$ where $g \in H^{1/2}(\partial\Omega)$ is normal derivative data.

Define $F:H^1(\Omega) \to \mathbb{R}$ by $$\langle F, \varphi\rangle = \int_\Omega \nabla v \nabla \varphi.$$ We know $F \in H^1(\Omega)^*$. But is actually $F \in L^2(\Omega)$?

I ask because $v \in H^2$, so I expect $F$ to be nicer.. how to prove if true?

Answer 1

By definition, you have $$\int_\Omega \nabla v \cdot \nabla \varphi \, \mathrm{d}x = \int_{\partial\Omega} g \, \varphi \, \mathrm{d}s$$ for all $\varphi \in H^1(\Omega)$. Hence, you have the representation $$\langle F, \varphi \rangle = \int_{\partial\Omega} g \, \varphi \, \mathrm{d}s.$$ This shows clearly that $F$ does not belong to $L^2(\Omega)$, unless $g = 0$.